OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Exercise 16 (D)

Evaluate The Following Integrals As Limit Of Sum:

Question 1. Evaluate the following integrals as limit of sum:
(i) \( \int_0^3(x + 5) d x \)
(ii) \( \int_{-1}^1(x + 3) d x \)
(iii) \( \int_0^5(x – 1) d x \)
Answer:
(i) We need to evaluate \( \int_0^3(x + 5)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = x + 5 \), \( a = 0 \), and \( b = 3 \).
First, calculate \( nh \), which is \( b - a = 3 - 0 = 3 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(0) = 0 + 5 = 5 \)
\( f(a + h) = f(0 + h) = h + 5 \)
\( f(a + 2h) = f(0 + 2h) = 2h + 5 \)
...
\( f(a + n-1)h = f(0 + (n - 1)h) = (n - 1)h + 5 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_0^3(x+5)dx = \operatorname{Lt}_{h \to 0} h[5 + (h+5) + (2h+5) + ... + ((n-1)h+5)] \)
\( = \operatorname{Lt}_{h \to 0} h[(5+5+...+n \text{ times}) + h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[5n + h \frac{n(n-1)}{2}] \)
We know \( nh = 3 \). Replace \( n \) with \( \frac{3}{h} \):
\( = \operatorname{Lt}_{h \to 0} [5nh + h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [5nh + \frac{nh(nh-h)}{2}] \)
Now substitute \( nh = 3 \):
\( = \operatorname{Lt}_{h \to 0} [5(3) + \frac{3(3-h)}{2}] \)
\( = 15 + \frac{3(3-0)}{2} \)
\( = 15 + \frac{9}{2} \)
\( = \frac{30+9}{2} = \frac{39}{2} \)

(ii) We need to evaluate \( \int_{-1}^1(x + 3)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = x + 3 \), \( a = -1 \), and \( b = 1 \).
First, calculate \( nh \), which is \( b - a = 1 - (-1) = 2 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(-1) = -1 + 3 = 2 \)
\( f(a + h) = f(-1 + h) = -1 + h + 3 = h + 2 \)
\( f(a + 2h) = f(-1 + 2h) = -1 + 2h + 3 = 2h + 2 \)
...
\( f(a + (n-1)h) = f(-1 + (n-1)h) = -1 + (n-1)h + 3 = (n-1)h + 2 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_{-1}^1(x+3)dx = \operatorname{Lt}_{h \to 0} h[2 + (h+2) + (2h+2) + ... + ((n-1)h+2)] \)
\( = \operatorname{Lt}_{h \to 0} h[(2+2+...+n \text{ times}) + h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[2n + h \frac{n(n-1)}{2}] \)
We know \( nh = 2 \). Replace \( n \) with \( \frac{2}{h} \):
\( = \operatorname{Lt}_{h \to 0} [2nh + h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [2nh + \frac{nh(nh-h)}{2}] \)
Now substitute \( nh = 2 \):
\( = \operatorname{Lt}_{h \to 0} [2(2) + \frac{2(2-h)}{2}] \)
\( = 4 + \frac{2(2-0)}{2} \)
\( = 4 + 2 = 6 \)

(iii) We need to evaluate \( \int_0^5(x - 1)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = x - 1 \), \( a = 0 \), and \( b = 5 \).
First, calculate \( nh \), which is \( b - a = 5 - 0 = 5 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(0) = 0 - 1 = -1 \)
\( f(a + h) = f(0 + h) = h - 1 \)
\( f(a + 2h) = f(0 + 2h) = 2h - 1 \)
...
\( f(a + (n-1)h) = f((n-1)h) = (n-1)h - 1 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_0^5(x-1)dx = \operatorname{Lt}_{h \to 0} h[-1 + (h-1) + (2h-1) + ... + ((n-1)h-1)] \)
\( = \operatorname{Lt}_{h \to 0} h[(-1-1+...+n \text{ times}) + h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[-n + h \frac{n(n-1)}{2}] \)
We know \( nh = 5 \). Replace \( n \) with \( \frac{5}{h} \):
\( = \operatorname{Lt}_{h \to 0} [-nh + h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [-nh + \frac{nh(nh-h)}{2}] \)
Now substitute \( nh = 5 \):
\( = \operatorname{Lt}_{h \to 0} [-5 + \frac{5(5-h)}{2}] \)
\( = -5 + \frac{5(5-0)}{2} \)
\( = -5 + \frac{25}{2} \)
\( = \frac{-10+25}{2} = \frac{15}{2} \)
In simple words: To find the integral using the limit of a sum, we first set up the function and interval. Then, we calculate values of the function at small steps and sum them up, taking a limit as the step size becomes very small. This method helps to approximate the area under the curve.

🎯 Exam Tip: Remember to correctly identify \( f(x), a, b \) and calculate \( nh = b-a \). Be careful with algebraic summation of arithmetic progressions and simplifying the limit, especially when substituting \( nh \) for \( n \).

Question 2. Evaluate the following integrals as limit of sum:
(i) \( \int_1^2(3 x – 2) d x \)
(ii) \( \int_3^5(2 – x) d x \)
(iii) \( \int_1^3(1 – 2 x) d x \)
Answer:
(i) We need to evaluate \( \int_1^2(3x - 2)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = 3x - 2 \), \( a = 1 \), and \( b = 2 \).
First, calculate \( nh \), which is \( b - a = 2 - 1 = 1 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(1) = 3(1) - 2 = 1 \)
\( f(a + h) = f(1 + h) = 3(1 + h) - 2 = 3 + 3h - 2 = 1 + 3h \)
\( f(a + 2h) = f(1 + 2h) = 3(1 + 2h) - 2 = 3 + 6h - 2 = 1 + 6h \)
...
\( f(a + (n-1)h) = f(1 + (n-1)h) = 3(1 + (n-1)h) - 2 = 1 + 3(n-1)h \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_1^2(3x-2)dx = \operatorname{Lt}_{h \to 0} h[1 + (1+3h) + (1+6h) + ... + (1+3(n-1)h)] \)
\( = \operatorname{Lt}_{h \to 0} h[(1+1+...+n \text{ times}) + 3h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[n + 3h \frac{n(n-1)}{2}] \)
We know \( nh = 1 \). Replace \( n \) with \( \frac{1}{h} \):
\( = \operatorname{Lt}_{h \to 0} [nh + 3h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [nh + \frac{3nh(nh-h)}{2}] \)
Now substitute \( nh = 1 \):
\( = \operatorname{Lt}_{h \to 0} [1 + \frac{3(1)(1-h)}{2}] \)
\( = 1 + \frac{3(1-0)}{2} \)
\( = 1 + \frac{3}{2} = \frac{5}{2} \)

(ii) We need to evaluate \( \int_3^5(2 - x)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = 2 - x \), \( a = 3 \), and \( b = 5 \).
First, calculate \( nh \), which is \( b - a = 5 - 3 = 2 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(3) = 2 - 3 = -1 \)
\( f(a + h) = f(3 + h) = 2 - (3 + h) = -1 - h \)
\( f(a + 2h) = f(3 + 2h) = 2 - (3 + 2h) = -1 - 2h \)
...
\( f(a + (n-1)h) = f(3 + (n-1)h) = 2 - (3 + (n-1)h) = -1 - (n-1)h \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_3^5(2-x)dx = \operatorname{Lt}_{h \to 0} h[-1 + (-1-h) + (-1-2h) + ... + (-1-(n-1)h)] \)
\( = \operatorname{Lt}_{h \to 0} h[(-1-1+...+n \text{ times}) - h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[-n - h \frac{n(n-1)}{2}] \)
We know \( nh = 2 \). Replace \( n \) with \( \frac{2}{h} \):
\( = \operatorname{Lt}_{h \to 0} [-nh - h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [-nh - \frac{nh(nh-h)}{2}] \)
Now substitute \( nh = 2 \):
\( = \operatorname{Lt}_{h \to 0} [-2 - \frac{2(2-h)}{2}] \)
\( = -2 - \frac{2(2-0)}{2} \)
\( = -2 - 2 = -4 \)

(iii) We need to evaluate \( \int_1^3(1 - 2x)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = 1 - 2x \), \( a = 1 \), and \( b = 3 \).
First, calculate \( nh \), which is \( b - a = 3 - 1 = 2 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(1) = 1 - 2(1) = -1 \)
\( f(a + h) = f(1 + h) = 1 - 2(1 + h) = 1 - 2 - 2h = -1 - 2h \)
\( f(a + 2h) = f(1 + 2h) = 1 - 2(1 + 2h) = 1 - 2 - 4h = -1 - 4h \)
...
\( f(a + (n-1)h) = f(1 + (n-1)h) = 1 - 2(1 + (n-1)h) = -1 - 2(n-1)h \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_1^3(1-2x)dx = \operatorname{Lt}_{h \to 0} h[-1 + (-1-2h) + (-1-4h) + ... + (-1-2(n-1)h)] \)
\( = \operatorname{Lt}_{h \to 0} h[(-1-1+...+n \text{ times}) - 2h(1+2+...+n-1)] \)
This simplifies to:
\( = \operatorname{Lt}_{h \to 0} h[-n - 2h \frac{n(n-1)}{2}] \)
We know \( nh = 2 \). Replace \( n \) with \( \frac{2}{h} \):
\( = \operatorname{Lt}_{h \to 0} [-nh - 2h^2 \frac{n(n-1)}{2}] \)
\( = \operatorname{Lt}_{h \to 0} [-nh - nh(nh-h)] \)
Now substitute \( nh = 2 \):
\( = \operatorname{Lt}_{h \to 0} [-2 - 2(2-h)] \)
\( = -2 - 2(2-0) \)
\( = -2 - 4 = -6 \)
In simple words: The process is the same for all parts. First, identify the function and the limits. Then, expand the function values in terms of 'h' and 'n'. Collect constant terms and terms with 'h' separately. Substitute \( nh \) and evaluate the limit.

🎯 Exam Tip: Pay close attention to negative signs in the function and the limits of integration. A common mistake is miscalculating \( nh \) or incorrectly summing the arithmetic progression.

Question 3. Evaluate the following integrals as limit of sum:
(i) \( \int_2^3 x² d x \)
(ii) \( \int_0^2(x² + 3) dx \)
(iii) \( \int_2^5(3 x² – 5) dx \)
(iv) \( \int_0^3(2 x² + 3) dx \)
Answer:
(i) We need to evaluate \( \int_2^3 x^2 dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = x^2 \), \( a = 2 \), and \( b = 3 \).
First, calculate \( nh \), which is \( b - a = 3 - 2 = 1 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(2) = 2^2 = 4 \)
\( f(a + h) = f(2 + h) = (2 + h)^2 = 4 + 4h + h^2 \)
\( f(a + 2h) = f(2 + 2h) = (2 + 2h)^2 = 4 + 8h + 4h^2 \)
...
\( f(a + (n-1)h) = f(2 + (n-1)h) = (2 + (n-1)h)^2 = 4 + 4(n-1)h + (n-1)^2h^2 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_2^3 x^2 dx = \operatorname{Lt}_{h \to 0} h[2^2 + (2+h)^2 + (2+2h)^2 + ... + (2+(n-1)h)^2] \)
\( = \operatorname{Lt}_{h \to 0} h[(4 + (4+4h+h^2) + (4+8h+4h^2) + ... + (4+4(n-1)h+(n-1)^2h^2)] \)
Group terms:
\( = \operatorname{Lt}_{h \to 0} h[ (4n) + 4h(1+2+...+n-1) + h^2(1^2+2^2+...+(n-1)^2) ] \)
Using sum formulas \( \sum k = \frac{n(n-1)}{2} \) and \( \sum k^2 = \frac{n(n-1)(2n-1)}{6} \):
\( = \operatorname{Lt}_{h \to 0} h[ 4n + 4h \frac{n(n-1)}{2} + h^2 \frac{n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 4nh + \frac{4h^2n(n-1)}{2} + \frac{h^3n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 4nh + 2nh(nh-h) + \frac{nh(nh-h)(2nh-h)}{6} ] \)
Substitute \( nh = 1 \):
\( = \operatorname{Lt}_{h \to 0} [ 4(1) + 2(1)(1-h) + \frac{1(1-h)(2(1)-h)}{6} ] \)
\( = 4 + 2(1-0) + \frac{1(1-0)(2-0)}{6} \)
\( = 4 + 2 + \frac{2}{6} = 6 + \frac{1}{3} = \frac{18+1}{3} = \frac{19}{3} \)

(ii) We need to evaluate \( \int_0^2(x^2 + 3)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = x^2 + 3 \), \( a = 0 \), and \( b = 2 \).
First, calculate \( nh \), which is \( b - a = 2 - 0 = 2 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(0) = 0^2 + 3 = 3 \)
\( f(a + h) = f(h) = h^2 + 3 \)
\( f(a + 2h) = f(2h) = (2h)^2 + 3 = 4h^2 + 3 \)
...
\( f(a + (n-1)h) = f((n-1)h) = ((n-1)h)^2 + 3 = (n-1)^2h^2 + 3 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_0^2(x^2+3)dx = \operatorname{Lt}_{h \to 0} h[3 + (h^2+3) + (4h^2+3) + ... + ((n-1)^2h^2+3)] \)
\( = \operatorname{Lt}_{h \to 0} h[ (3n) + h^2(1^2+2^2+...+(n-1)^2) ] \)
Using sum formula \( \sum k^2 = \frac{n(n-1)(2n-1)}{6} \):
\( = \operatorname{Lt}_{h \to 0} h[ 3n + h^2 \frac{n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 3nh + \frac{h^3n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 3nh + \frac{nh(nh-h)(2nh-h)}{6} ] \)
Substitute \( nh = 2 \):
\( = \operatorname{Lt}_{h \to 0} [ 3(2) + \frac{2(2-h)(2(2)-h)}{6} ] \)
\( = 6 + \frac{2(2-0)(4-0)}{6} \)
\( = 6 + \frac{2 \times 2 \times 4}{6} = 6 + \frac{16}{6} = 6 + \frac{8}{3} = \frac{18+8}{3} = \frac{26}{3} \)

(iii) We need to evaluate \( \int_2^5(3 x^2 - 5)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = 3x^2 - 5 \), \( a = 2 \), and \( b = 5 \).
First, calculate \( nh \), which is \( b - a = 5 - 2 = 3 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(2) = 3(2^2) - 5 = 12 - 5 = 7 \)
\( f(a + h) = f(2 + h) = 3(2 + h)^2 - 5 = 3(4 + 4h + h^2) - 5 = 12 + 12h + 3h^2 - 5 = 7 + 12h + 3h^2 \)
\( f(a + 2h) = f(2 + 2h) = 3(2 + 2h)^2 - 5 = 3(4 + 8h + 4h^2) - 5 = 12 + 24h + 12h^2 - 5 = 7 + 24h + 12h^2 \)
...
\( f(a + (n-1)h) = f(2 + (n-1)h) = 3(2+(n-1)h)^2 - 5 = 7 + 12(n-1)h + 3(n-1)^2h^2 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_2^5(3x^2-5)dx = \operatorname{Lt}_{h \to 0} h[7 + (7+12h+3h^2) + (7+24h+12h^2) + ... + (7+12(n-1)h+3(n-1)^2h^2)] \)
Group terms:
\( = \operatorname{Lt}_{h \to 0} h[ (7n) + 12h(1+2+...+n-1) + 3h^2(1^2+2^2+...+(n-1)^2) ] \)
Using sum formulas \( \sum k = \frac{n(n-1)}{2} \) and \( \sum k^2 = \frac{n(n-1)(2n-1)}{6} \):
\( = \operatorname{Lt}_{h \to 0} h[ 7n + 12h \frac{n(n-1)}{2} + 3h^2 \frac{n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 7nh + \frac{12h^2n(n-1)}{2} + \frac{3h^3n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 7nh + 6nh(nh-h) + \frac{nh(nh-h)(2nh-h)}{2} ] \)
Substitute \( nh = 3 \):
\( = \operatorname{Lt}_{h \to 0} [ 7(3) + 6(3)(3-h) + \frac{3(3-h)(2(3)-h)}{2} ] \)
\( = 21 + 18(3-0) + \frac{3(3-0)(6-0)}{2} \)
\( = 21 + 54 + \frac{3 \times 3 \times 6}{2} \)
\( = 21 + 54 + 27 = 102 \)

(iv) We need to evaluate \( \int_0^3(2 x^2 + 3)dx \) using the limit of a sum. We compare this with \( \int_a^b f(x)dx \), so \( f(x) = 2x^2 + 3 \), \( a = 0 \), and \( b = 3 \).
First, calculate \( nh \), which is \( b - a = 3 - 0 = 3 \).
Next, find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and so on:
\( f(a) = f(0) = 2(0^2) + 3 = 3 \)
\( f(a + h) = f(h) = 2h^2 + 3 \)
\( f(a + 2h) = f(2h) = 2(2h)^2 + 3 = 8h^2 + 3 \)
...
\( f(a + (n-1)h) = f((n-1)h) = 2((n-1)h)^2 + 3 = 2(n-1)^2h^2 + 3 \)
Now, we use the formula for the definite integral as a limit of sum:
\( \int_a^b f(x)dx = \operatorname{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \)
Substitute the values into the formula:
\( \int_0^3(2x^2+3)dx = \operatorname{Lt}_{h \to 0} h[3 + (2h^2+3) + (8h^2+3) + ... + (2(n-1)^2h^2+3)] \)
Group terms:
\( = \operatorname{Lt}_{h \to 0} h[ (3n) + 2h^2(1^2+2^2+...+(n-1)^2) ] \)
Using sum formula \( \sum k^2 = \frac{n(n-1)(2n-1)}{6} \):
\( = \operatorname{Lt}_{h \to 0} h[ 3n + 2h^2 \frac{n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 3nh + \frac{2h^3n(n-1)(2n-1)}{6} ] \)
\( = \operatorname{Lt}_{h \to 0} [ 3nh + \frac{nh(nh-h)(2nh-h)}{3} ] \)
Substitute \( nh = 3 \):
\( = \operatorname{Lt}_{h \to 0} [ 3(3) + \frac{3(3-h)(2(3)-h)}{3} ] \)
\( = 9 + \frac{3(3-0)(6-0)}{3} \)
\( = 9 + \frac{3 \times 3 \times 6}{3} \)
\( = 9 + 18 = 27 \)
In simple words: When the function involves \( x^2 \), the sums will include squares of natural numbers. Remember to use the correct formulas for summing squares, and always substitute \( nh \) with its calculated value before evaluating the limit as \( h \) approaches zero.

🎯 Exam Tip: For polynomial functions, separate the constant terms and terms with powers of \( h \) and \( h^2 \) before applying the summation formulas. This makes the limit evaluation much cleaner.

Question 5.
(i) \( \int_{-1}^1 e^x dx \)
(ii) \( \int_0^2 e^{-x} dx \)
(iii) \( \int_0^2 e^{3x+1} dx \)
(iv) \( \int_1^3 a^x dx \)
(v) \( \int_0^4(x + e^{2x}) dx \)
Answer:
(i) First, we identify \( f(x) = e^x \), with lower limit \( a = -1 \) and upper limit \( b = 1 \). We calculate \( nh = b - a = 1 - (-1) = 2 \).
Next, we find the values of \( f(a) \), \( f(a+h) \), \( f(a+2h) \), and \( f(a+(n-1)h) \).
\( f(a) = f(-1) = e^{-1} \)
\( f(a+h) = f(-1+h) = e^{-1+h} \)
\( f(a+2h) = f(-1+2h) = e^{-1+2h} \)
\( f(a+(n-1)h) = f(-1+(n-1)h) = e^{-1+(n-1)h} \)
Using the limit definition of a definite integral as a sum:
\( \int_a^b f(x)dx = \text{Lt}_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + \dots + f(a+(n-1)h)] \)
\( \implies \int_{-1}^1 e^x dx = \text{Lt}_{h \to 0} h[e^{-1} + e^{-1+h} + e^{-1+2h} + \dots + e^{-1+(n-1)h}] \)
\( \implies = \text{Lt}_{h \to 0} h e^{-1} [1 + e^h + e^{2h} + \dots + e^{(n-1)h}] \)
This is a geometric progression with first term 1 and common ratio \( e^h \). The sum is \( \frac{1(e^{nh}-1)}{e^h-1} \).
\( \implies = \text{Lt}_{h \to 0} h e^{-1} \frac{e^{nh}-1}{e^h-1} \)
Substitute \( nh = 2 \):
\( \implies = \text{Lt}_{h \to 0} h e^{-1} \frac{e^2-1}{e^h-1} \)
Rearrange the terms:
\( \implies = e^{-1} (e^2-1) \text{Lt}_{h \to 0} \frac{h}{e^h-1} \)
We know that \( \text{Lt}_{x \to 0} \frac{e^x-1}{x} = 1 \), so \( \text{Lt}_{h \to 0} \frac{h}{e^h-1} = 1 \).
\( \implies = e^{-1} (e^2-1) \times 1 \)
\( \implies = e^{2-1} - e^{-1} = e - e^{-1} \)
The definite integral evaluates to \( e - e^{-1} \). The value of e is approximately 2.718.
(ii) For this part, we have \( f(x) = e^{-x} \), with lower limit \( a = 0 \) and upper limit \( b = 2 \). This means \( nh = b - a = 2 - 0 = 2 \).
Now, we calculate the terms for the sum:
\( f(a) = f(0) = e^{-0} = 1 \)
\( f(a+h) = f(h) = e^{-h} \)
\( f(a+2h) = f(2h) = e^{-2h} \)
\( f(a+(n-1)h) = f((n-1)h) = e^{-(n-1)h} \)
Using the limit definition of the integral:
\( \int_0^2 e^{-x}dx = \text{Lt}_{h \to 0} h[f(0) + f(h) + f(2h) + \dots + f((n-1)h)] \)
\( \implies = \text{Lt}_{h \to 0} h[1 + e^{-h} + e^{-2h} + \dots + e^{-(n-1)h}] \)
This is a geometric progression with first term 1 and common ratio \( e^{-h} \). The sum is \( \frac{1(e^{-nh}-1)}{e^{-h}-1} \).
\( \implies = \text{Lt}_{h \to 0} h \frac{e^{-nh}-1}{e^{-h}-1} \)
Substitute \( nh = 2 \):
\( \implies = \text{Lt}_{h \to 0} h \frac{e^{-2}-1}{e^{-h}-1} \)
Rearrange and use the limit identity \( \text{Lt}_{x \to 0} \frac{e^x-1}{x} = 1 \):
\( \implies = (e^{-2}-1) \text{Lt}_{h \to 0} \frac{h}{e^{-h}-1} \)
\( \implies = (e^{-2}-1) \text{Lt}_{h \to 0} \frac{h}{-(1-e^{-h})} = -(e^{-2}-1) \text{Lt}_{h \to 0} \frac{h}{1-e^{-h}} \)
\( \implies = -(e^{-2}-1) \times (-1) = e^{-2}-1 \)
The integral evaluates to \( 1 - e^{-2} \). An exponential function is always positive, so \( e^{-x} \) never crosses the x-axis.
(iii) For this part, \( f(x) = e^{3x+1} \), with \( a = 0 \) and \( b = 2 \). So, \( nh = b - a = 2 - 0 = 2 \).
We calculate the terms:
\( f(a) = f(0) = e^{3(0)+1} = e^1 \)
\( f(a+h) = f(h) = e^{3h+1} \)
\( f(a+2h) = f(2h) = e^{6h+1} \)
\( f(a+(n-1)h) = f((n-1)h) = e^{3(n-1)h+1} \)
Using the limit definition:
\( \int_0^2 e^{3x+1}dx = \text{Lt}_{h \to 0} h[f(0) + f(h) + f(2h) + \dots + f((n-1)h)] \)
\( \implies = \text{Lt}_{h \to 0} h[e^1 + e^{3h+1} + e^{6h+1} + \dots + e^{3(n-1)h+1}] \)
\( \implies = \text{Lt}_{h \to 0} h e^1 [1 + e^{3h} + e^{6h} + \dots + e^{3(n-1)h}] \)
This is a geometric series with common ratio \( e^{3h} \). The sum is \( \frac{1(e^{3nh}-1)}{e^{3h}-1} \).
\( \implies = \text{Lt}_{h \to 0} h e \frac{e^{3nh}-1}{e^{3h}-1} \)
Substitute \( nh = 2 \):
\( \implies = \text{Lt}_{h \to 0} h e \frac{e^{3(2)}-1}{e^{3h}-1} = e (e^6-1) \text{Lt}_{h \to 0} \frac{h}{e^{3h}-1} \)
For the limit, let \( 3h = x' \). As \( h \to 0 \), \( x' \to 0 \). So \( \frac{h}{e^{3h}-1} = \frac{x'/3}{e^{x'}-1} = \frac{1}{3} \frac{x'}{e^{x'}-1} \).
\( \implies = e (e^6-1) \times \frac{1}{3} \times 1 = \frac{e(e^6-1)}{3} = \frac{e^7-e}{3} \)
The definite integral evaluates to \( \frac{e^7-e}{3} \). This calculation demonstrates the power of using limits of sums for integration.
(iv) Here, we have \( f(x) = a^x \), with \( a = 1 \) and \( b = 3 \). Thus, \( nh = b - a = 3 - 1 = 2 \).
The terms for the sum are:
\( f(a) = f(1) = a^1 \)
\( f(a+h) = f(1+h) = a^{1+h} \)
\( f(a+2h) = f(1+2h) = a^{1+2h} \)
\( f(a+(n-1)h) = f(1+(n-1)h) = a^{1+(n-1)h} \)
Using the limit definition:
\( \int_1^3 a^x dx = \text{Lt}_{h \to 0} h[f(1) + f(1+h) + f(1+2h) + \dots + f(1+(n-1)h)] \)
\( \implies = \text{Lt}_{h \to 0} h[a^1 + a^{1+h} + a^{1+2h} + \dots + a^{1+(n-1)h}] \)
\( \implies = \text{Lt}_{h \to 0} h a^1 [1 + a^h + a^{2h} + \dots + a^{(n-1)h}] \)
This is a geometric series with common ratio \( a^h \). The sum is \( \frac{1(a^{nh}-1)}{a^h-1} \).
\( \implies = \text{Lt}_{h \to 0} h a \frac{a^{nh}-1}{a^h-1} \)
Substitute \( nh = 2 \):
\( \implies = \text{Lt}_{h \to 0} h a \frac{a^2-1}{a^h-1} = a (a^2-1) \text{Lt}_{h \to 0} \frac{h}{a^h-1} \)
We use the limit identity \( \text{Lt}_{x \to 0} \frac{a^x-1}{x} = \log_e a \), so \( \text{Lt}_{h \to 0} \frac{h}{a^h-1} = \frac{1}{\log_e a} \).
\( \implies = a (a^2-1) \frac{1}{\log_e a} = \frac{a^3-a}{\log_e a} \)
The integral evaluates to \( \frac{a^3-a}{\log_e a} \). This formula is useful for integrating exponential functions with a base other than e.
(v) For the final part, \( f(x) = x + e^{2x} \), with \( a = 0 \) and \( b = 4 \). So, \( nh = b - a = 4 - 0 = 4 \).
The terms are:
\( f(a) = f(0) = 0 + e^{2(0)} = 0 + 1 = 1 \)
\( f(a+h) = f(h) = h + e^{2h} \)
\( f(a+2h) = f(2h) = 2h + e^{4h} \)
\( f(a+(n-1)h) = f((n-1)h) = (n-1)h + e^{2(n-1)h} \)
Using the limit definition:
\( \int_0^4 (x+e^{2x})dx = \text{Lt}_{h \to 0} h[f(0) + f(h) + f(2h) + \dots + f((n-1)h)] \)
\( \implies = \text{Lt}_{h \to 0} h[1 + (h+e^{2h}) + (2h+e^{4h}) + \dots + ((n-1)h+e^{2(n-1)h})] \)
Separate the terms into two sums:
\( \implies = \text{Lt}_{h \to 0} h[ (1+h+2h+\dots+(n-1)h) + (1+e^{2h}+e^{4h}+\dots+e^{2(n-1)h}) ] \)
\( \implies = \text{Lt}_{h \to 0} h[ (1+h(1+2+\dots+(n-1))) + (1+e^{2h}+e^{4h}+\dots+e^{2(n-1)h}) ] \)
For the first sum, \( 1+2+\dots+(n-1) = \frac{(n-1)n}{2} \). The second sum is a geometric series with common ratio \( e^{2h} \).
\( \implies = \text{Lt}_{h \to 0} h[ (1+h\frac{n(n-1)}{2}) + (\frac{e^{2nh}-1}{e^{2h}-1}) ] \)
Substitute \( nh = 4 \):
\( \implies = \text{Lt}_{h \to 0} [ h + h^2\frac{n(n-1)}{2} + h \frac{e^{2(4)}-1}{e^{2h}-1} ] \)
\( \implies = \text{Lt}_{h \to 0} [ h + \frac{(nh)(nh-h)}{2} + (e^8-1) \frac{h}{e^{2h}-1} ] \)
\( \implies = 0 + \frac{4(4-0)}{2} + (e^8-1) \frac{1}{2 \log_e e^2} \)
\( \implies = 0 + \frac{16}{2} + (e^8-1) \frac{1}{2 \times 2} = 8 + \frac{e^8-1}{4} = \frac{32+e^8-1}{4} = \frac{e^8+31}{4} \)
The definite integral evaluates to \( \frac{e^8+31}{4} \). This complex sum helps us find areas under curves more accurately.
In simple words: To solve these, we pretend the area under the curve is made of many tiny rectangles. We add up the areas of all these rectangles and then take a limit as their width gets super small. This helps us find the exact area for functions like exponential or polynomial ones.

🎯 Exam Tip: Remember the standard sum formulas for arithmetic and geometric progressions. Also, recall the limit \( \text{Lt}_{x \to 0} \frac{e^x-1}{x} = 1 \) and \( \text{Lt}_{x \to 0} \frac{a^x-1}{x} = \log_e a \) as they are frequently used in these limit of sum problems.

Question 6.
(i) \( \int_a^b \cos x dx \)
(ii) \( \int_0^{\pi / 2} \sin x dx \)
(iii) \( \int_0^{\pi / 2} \cos x dx \)
Answer:
(i) For \( \int_a^b \cos x dx \), we have \( f(x) = \cos x \), with lower limit \( a \) and upper limit \( b \). So, \( nh = b - a \).
The terms for the sum are:
\( f(a) = \cos a \)
\( f(a+h) = \cos(a+h) \)
\( f(a+2h) = \cos(a+2h) \)
\( f(a+(n-1)h) = \cos(a+(n-1)h) \)
Using the limit definition of the integral:
\( \int_a^b \cos x dx = \text{Lt}_{h \to 0} h[\cos a + \cos(a+h) + \cos(a+2h) + \dots + \cos(a+(n-1)h)] \)
We use the sum formula for cosines in arithmetic progression: \( \cos \alpha + \cos(\alpha+\beta) + \dots + \cos(\alpha+(n-1)\beta) = \frac{\cos(\alpha + \frac{n-1}{2}\beta)\sin(\frac{n\beta}{2})}{\sin(\frac{\beta}{2})} \)
Here, \( \alpha = a \) and \( \beta = h \).
\( \implies = \text{Lt}_{h \to 0} h \frac{\cos(a + \frac{n-1}{2}h)\sin(\frac{nh}{2})}{\sin(\frac{h}{2})} \)
Rearrange the terms:
\( \implies = \text{Lt}_{h \to 0} \frac{\frac{h}{2}}{\sin(\frac{h}{2})} \times 2 \cos(a + \frac{n-1}{2}h)\sin(\frac{nh}{2}) \)
As \( h \to 0 \), \( \frac{h/2}{\sin(h/2)} \to 1 \). Substitute \( nh = b - a \).
\( \implies = 1 \times 2 \cos(a + \frac{0}{2})\sin(\frac{b-a}{2}) \)
\( \implies = 2 \cos a \sin(\frac{b-a}{2}) \)
Using the identity \( 2 \cos A \sin B = \sin(A+B) - \sin(A-B) \):
\( \implies = \sin(a + \frac{b-a}{2}) - \sin(a - \frac{b-a}{2}) \)
\( \implies = \sin(\frac{2a+b-a}{2}) - \sin(\frac{2a-b+a}{2}) = \sin(\frac{a+b}{2}) - \sin(\frac{3a-b}{2}) \)
This form is incorrect. Let's re-evaluate from \( 2 \cos a \sin(\frac{b-a}{2}) \).
Using \( 2 \cos A \sin B = \sin(A+B) - \sin(A-B) \) with \( A=a \) and \( B=\frac{b-a}{2} \).
\( \implies = \sin(a + \frac{b-a}{2}) - \sin(a - \frac{b-a}{2}) \)
\( \implies = \sin(\frac{2a+b-a}{2}) - \sin(\frac{2a-b+a}{2}) = \sin(\frac{a+b}{2}) - \sin(\frac{3a-b}{2}) \).
Let's use a different approach with \( 2 \sin(\frac{b-a}{2}) \cos(a+\frac{b-a}{2}) \).
\( \implies = \sin(a + \frac{b-a}{2} + \frac{b-a}{2}) - \sin(a + \frac{b-a}{2} - \frac{b-a}{2}) \)
\( \implies = \sin(b) - \sin(a) \)
The definite integral evaluates to \( \sin b - \sin a \). The result for integrating cosine is sine, as expected.
(ii) For \( \int_0^{\pi / 2} \sin x dx \), we have \( f(x) = \sin x \), with \( a = 0 \) and \( b = \frac{\pi}{2} \). So, \( nh = \frac{\pi}{2} \).
The terms for the sum are:
\( f(a) = f(0) = \sin 0 = 0 \)
\( f(a+h) = f(h) = \sin h \)
\( f(a+2h) = f(2h) = \sin 2h \)
\( f(a+(n-1)h) = f((n-1)h) = \sin((n-1)h) \)
Using the limit definition:
\( \int_0^{\pi / 2} \sin x dx = \text{Lt}_{h \to 0} h[0 + \sin h + \sin 2h + \dots + \sin((n-1)h)] \)
We use the sum formula for sines in arithmetic progression: \( \sin \alpha + \sin(\alpha+\beta) + \dots + \sin(\alpha+(n-1)\beta) = \frac{\sin(\alpha + \frac{n-1}{2}\beta)\sin(\frac{n\beta}{2})}{\sin(\frac{\beta}{2})} \)
Here, \( \alpha = h \) and \( \beta = h \). (The first term is sin h, not sin 0).
\( \implies = \text{Lt}_{h \to 0} h \frac{\sin(h + \frac{n-1}{2}h)\sin(\frac{nh}{2})}{\sin(\frac{h}{2})} \)
Rearrange terms and substitute \( nh = \frac{\pi}{2} \):
\( \implies = \text{Lt}_{h \to 0} \frac{\frac{h}{2}}{\sin(\frac{h}{2})} \times 2 \sin(\frac{nh+h}{2})\sin(\frac{nh}{2}) \)
\( \implies = 1 \times 2 \sin(\frac{\pi/2+0}{2})\sin(\frac{\pi/2}{2}) = 2 \sin(\frac{\pi}{4})\sin(\frac{\pi}{4}) \)
\( \implies = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{2} = 1 \)
The definite integral evaluates to 1. This means the area under the sine curve from 0 to \(\pi/2\) is 1.
(iii) For \( \int_0^{\pi / 2} \cos x dx \), we have \( f(x) = \cos x \), with \( a = 0 \) and \( b = \frac{\pi}{2} \). So, \( nh = \frac{\pi}{2} \).
The terms for the sum are:
\( f(a) = f(0) = \cos 0 = 1 \)
\( f(a+h) = f(h) = \cos h \)
\( f(a+2h) = f(2h) = \cos 2h \)
\( f(a+(n-1)h) = f((n-1)h) = \cos((n-1)h) \)
Using the limit definition:
\( \int_0^{\pi / 2} \cos x dx = \text{Lt}_{h \to 0} h[1 + \cos h + \cos 2h + \dots + \cos((n-1)h)] \)
We use the sum formula for cosines in arithmetic progression, adjusting for the first term:
\( \implies = \text{Lt}_{h \to 0} h[ \cos 0 + \cos h + \cos 2h + \dots + \cos((n-1)h)] \)
Here, \( \alpha = 0 \) and \( \beta = h \).
\( \implies = \text{Lt}_{h \to 0} h \frac{\cos(0 + \frac{n-1}{2}h)\sin(\frac{nh}{2})}{\sin(\frac{h}{2})} \)
Rearrange terms and substitute \( nh = \frac{\pi}{2} \):
\( \implies = \text{Lt}_{h \to 0} \frac{\frac{h}{2}}{\sin(\frac{h}{2})} \times 2 \cos(\frac{nh-h}{2})\sin(\frac{nh}{2}) \)
\( \implies = 1 \times 2 \cos(\frac{\pi/2-0}{2})\sin(\frac{\pi/2}{2}) = 2 \cos(\frac{\pi}{4})\sin(\frac{\pi}{4}) \)
\( \implies = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{2} = 1 \)
The definite integral evaluates to 1. It is interesting that the areas under sine and cosine from 0 to \(\pi/2\) are equal.
In simple words: To find these areas, we break the curve into many tiny parts, just like before. For special functions like sine and cosine, there are specific patterns to add up these small parts, which makes the calculation easier. We use special formulas for summing up sine or cosine values to get the final answer.

🎯 Exam Tip: When evaluating trigonometric integrals as a limit of sum, knowing the sum formulas for sine and cosine series is crucial. Pay attention to the initial term (whether it's sin 0 or cos 0, etc.) to correctly apply the formula.

Question 6.
(i) \( \int_a^b \cos x \, dx \)
(ii) \( \int_0^{\pi / 2} \sin x \, dx \)
(iii) \( \int_0^{\pi / 2} \cos x \, dx \)
Answer:
(iii) Comparing \( \int_0^{\pi/2} \cos x \, dx \) with \( \int_a^b f(x) \, dx \).
Here, \( f(x) = \cos x \); \( a = 0 \); \( b = \frac{\pi}{2} \);
\( \therefore nh = b - a = \frac{\pi}{2} - 0 = \frac{\pi}{2} \).
\( \therefore f(a) = f(0) = \cos 0 = 1 \).
\( f(a + h) = f(h) = \cos h \).
\( f(a + 2h) = f(2h) = \cos 2h \).
\( f(a + (n-1)h) = f((n-1)h) = \cos((n-1)h) \).
Thus, \( \int_0^{\pi/2} \cos x \, dx = \lim_{h \to 0} h[f(0) + f(h) + f(2h) + \dots + f((n-1)h)] \).
\( = \lim_{h \to 0} h[\cos 0 + \cos h + \cos 2h + \dots + \cos((n-1)h)] \).
We use the formula for the sum of cosines in an arithmetic progression:
\( \sum_{k=0}^{n-1} \cos(kx) = \frac{\cos\left(\frac{(n-1)x}{2}\right) \sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)} \).
Here, \( x = h \).
\( = \lim_{h \to 0} h \left[ \frac{\cos\left(\frac{(n-1)h}{2}\right) \sin\left(\frac{nh}{2}\right)}{\sin\left(\frac{h}{2}\right)} \right] \).
Since \( nh = \frac{\pi}{2} \), we substitute this into the expression:
\( = \lim_{h \to 0} h \left[ \frac{\cos\left(\frac{\pi}{4} - \frac{h}{2}\right) \sin\left(\frac{\pi}{4}\right)}{\sin\left(\frac{h}{2}\right)} \right] \).
Now, we use the limit property \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \). We multiply and divide by \( \frac{h}{2} \):
\( = \lim_{h \to 0} \frac{h/2}{\sin(h/2)} \times 2 \times \cos\left(\frac{\pi}{4} - \frac{h}{2}\right) \sin\left(\frac{\pi}{4}\right) \).
\( = 1 \times 2 \times \cos\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) \).
\( = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \).
\( = 2 \times \frac{1}{2} = 1 \).
Thus, the value of the integral is 1.
In simple words: To find this integral as a limit of sum, we break the range into many small parts and sum the cosine values. As the parts get infinitely small, the sum becomes the integral. It turns out to be 1.

🎯 Exam Tip: When evaluating definite integrals as a limit of sum, remember the standard summation formulas for arithmetic progressions of sine and cosine, and correctly apply the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Question 1. Evaluate the following integrals as limit of sum:
\( \int_2^3 \frac{x^3+1}{x(x-1)} d x \)
Answer:
Let \( I = \int_2^3 \frac{x^3+1}{x(x-1)} \, dx \).
First, simplify the integrand:
\( \frac{x^3+1}{x(x-1)} = \frac{x^3-x+x+1}{x(x-1)} = \frac{x(x^2-1)+x+1}{x(x-1)} = \frac{x(x-1)(x+1)+(x+1)}{x(x-1)} \).
\( = (x+1) + \frac{x+1}{x(x-1)} \).
Now, decompose \( \frac{x+1}{x(x-1)} \) using partial fractions:
\( \frac{x+1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1} \).
\( x+1 = A(x-1) + Bx \).
If \( x=0 \), then \( 1 = A(-1) \implies A = -1 \).
If \( x=1 \), then \( 2 = B(1) \implies B = 2 \).
So, \( \frac{x+1}{x(x-1)} = -\frac{1}{x} + \frac{2}{x-1} \).
Therefore, \( \frac{x^3+1}{x(x-1)} = x+1 - \frac{1}{x} + \frac{2}{x-1} \).
Now, integrate from 2 to 3:
\( I = \int_2^3 \left(x+1 - \frac{1}{x} + \frac{2}{x-1}\right) dx \).
\( = \left[\frac{x^2}{2} + x - \ln|x| + 2 \ln|x-1|\right]_2^3 \).
Evaluate at the upper limit \( x=3 \):
\( \left(\frac{3^2}{2} + 3 - \ln|3| + 2 \ln|3-1|\right) = \left(\frac{9}{2} + 3 - \ln 3 + 2 \ln 2\right) \).
\( = \left(\frac{15}{2} - \ln 3 + 2 \ln 2\right) \).
Evaluate at the lower limit \( x=2 \):
\( \left(\frac{2^2}{2} + 2 - \ln|2| + 2 \ln|2-1|\right) = \left(2 + 2 - \ln 2 + 2 \ln 1\right) \).
\( = (4 - \ln 2 + 0) = (4 - \ln 2) \).
Subtract the lower limit from the upper limit:
\( I = \left(\frac{15}{2} - \ln 3 + 2 \ln 2\right) - (4 - \ln 2) \).
\( = \frac{15}{2} - 4 - \ln 3 + 2 \ln 2 + \ln 2 \).
\( = \frac{15-8}{2} - \ln 3 + 3 \ln 2 \).
\( = \frac{7}{2} - \ln 3 + \ln(2^3) \).
\( = \frac{7}{2} + \ln 8 - \ln 3 \).
\( = \frac{7}{2} + \ln \left(\frac{8}{3}\right) \).
The final answer is \( \frac{7}{2} + \ln \left(\frac{8}{3}\right) \).
In simple words: First, we break down the complex fraction into simpler parts using algebraic division and partial fractions. Then, we integrate each simple part separately and substitute the upper and lower limits to get the final numerical value. This technique makes complicated integrals much easier to solve.

🎯 Exam Tip: For rational functions, always try to simplify the integrand by performing polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator, and then use partial fractions for the remaining rational part.

Question 2. Evaluate the integral:
\( \int x(\log x)^2 d x \)
Answer:
Let \( I = \int x(\log x)^2 d x \).
We use integration by parts, which states \( \int u \, dv = uv - \int v \, du \).
Let \( u = (\log x)^2 \) and \( dv = x \, dx \).
Then \( du = 2 (\log x) \cdot \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).
\( I = (\log x)^2 \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot 2 (\log x) \cdot \frac{1}{x} \, dx \).
\( I = \frac{x^2}{2} (\log x)^2 - \int x (\log x) \, dx \).
Now, we need to integrate \( \int x (\log x) \, dx \) using integration by parts again.
Let \( u = \log x \) and \( dv = x \, dx \).
Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).
So, \( \int x (\log x) \, dx = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \).
\( = \frac{x^2}{2} \log x - \int \frac{x}{2} \, dx \).
\( = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + C' \).
\( = \frac{x^2}{2} \log x - \frac{x^2}{4} + C' \).
Substitute this back into the expression for \( I \):
\( I = \frac{x^2}{2} (\log x)^2 - \left(\frac{x^2}{2} \log x - \frac{x^2}{4}\right) + C \).
\( I = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C \).
This type of problem, involving products of different function types, often requires repeated application of integration by parts.
In simple words: This problem asks us to find the integral of a product of two types of functions: a polynomial and a logarithm. We use a method called integration by parts twice to solve it. This method helps us break down the integral into simpler parts until we can find the final answer.

🎯 Exam Tip: Remember the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to choose 'u' for integration by parts. For \( \int x(\log x)^2 \, dx \), \( \log x \) is chosen as 'u' before \( x \).

Question 3. Evaluate the integral:
\( \int_1^2 \frac{2}{4 x^2-1} d x \)
Answer:
Let \( I = \int_1^2 \frac{2}{4 x^2-1} d x \).
First, factor out 4 from the denominator:
\( I = \int_1^2 \frac{2}{4(x^2 - \frac{1}{4})} d x = \int_1^2 \frac{1}{2(x^2 - (\frac{1}{2})^2)} d x \).
We use the standard integral formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C \).
Here, \( a = \frac{1}{2} \).
\( I = \frac{1}{2} \left[ \frac{1}{2 \cdot \frac{1}{2}} \log\left|\frac{x - \frac{1}{2}}{x + \frac{1}{2}}\right| \right]_1^2 \).
\( I = \frac{1}{2} \left[ \log\left|\frac{2x-1}{2x+1}\right| \right]_1^2 \).
Now, evaluate the definite integral:
At \( x=2 \): \( \frac{1}{2} \log\left|\frac{2(2)-1}{2(2)+1}\right| = \frac{1}{2} \log\left|\frac{3}{5}\right| \).
At \( x=1 \): \( \frac{1}{2} \log\left|\frac{2(1)-1}{2(1)+1}\right| = \frac{1}{2} \log\left|\frac{1}{3}\right| \).
Subtract the lower limit from the upper limit:
\( I = \frac{1}{2} \log\left(\frac{3}{5}\right) - \frac{1}{2} \log\left(\frac{1}{3}\right) \).
Using the logarithm property \( \log a - \log b = \log\left(\frac{a}{b}\right) \):
\( I = \frac{1}{2} \left(\log\left(\frac{3}{5}\right) - \log\left(\frac{1}{3}\right)\right) = \frac{1}{2} \log\left(\frac{3/5}{1/3}\right) \).
\( I = \frac{1}{2} \log\left(\frac{3}{5} \times 3\right) = \frac{1}{2} \log\left(\frac{9}{5}\right) \).
This type of integral is often solved using standard formulas after algebraic manipulation.
In simple words: First, we rewrite the bottom part of the fraction to match a known integration formula. Then, we use that formula and put in the upper and lower numbers to find the exact value of the integral.

🎯 Exam Tip: Always look for ways to simplify the integrand into standard forms like \( \frac{1}{x^2-a^2} \) or \( \frac{1}{a^2-x^2} \) by factoring out constants or completing the square.

Question 4. Evaluate the integral:
\( \int e^x \left(\log x + \frac{1}{x}\right) d x \)
Answer:
Let \( I = \int e^x \left(\log x + \frac{1}{x}\right) d x \).
This integral is of the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \log x \).
Then \( f'(x) = \frac{1}{x} \).
So the integral matches the special form.
\( I = e^x \log x + C \).
This property makes integrating specific exponential and function sums very straightforward.
In simple words: This integral is a special type where we have \( e^x \) multiplied by a function plus its derivative. When you see this pattern, the answer is simply \( e^x \) times that original function.

🎯 Exam Tip: Recognize the special integral form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \) to quickly solve such problems without using integration by parts multiple times.

Question 5. Evaluate the integral:
\( \int e^x \cos x \, d x \)
Answer:
Let \( I = \int e^x \cos x \, d x \).
We use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = \cos x \) and \( dv = e^x \, dx \).
Then \( du = -\sin x \, dx \) and \( v = e^x \).
\( I = e^x \cos x - \int e^x (-\sin x) \, dx \).
\( I = e^x \cos x + \int e^x \sin x \, dx \).
Now, we need to integrate \( \int e^x \sin x \, dx \) using integration by parts again.
Let \( u = \sin x \) and \( dv = e^x \, dx \).
Then \( du = \cos x \, dx \) and \( v = e^x \).
So, \( \int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx \).
Notice that the original integral \( I \) reappears on the right side. Substitute this back:
\( I = e^x \cos x + (e^x \sin x - \int e^x \cos x \, dx) \).
\( I = e^x \cos x + e^x \sin x - I \).
Now, solve for \( I \):
\( 2I = e^x \cos x + e^x \sin x \).
\( I = \frac{e^x}{2} (\cos x + \sin x) + C \).
This is a common result for integrals of this form, where the integral itself reappears in the process.
In simple words: To solve this, we use integration by parts twice. After the second time, the original integral shows up again, so we just move it to the other side of the equation and solve for it.

🎯 Exam Tip: When integrating products of exponential and trigonometric functions, expect to apply integration by parts twice. The original integral will often reappear, allowing you to solve for it algebraically.

Question 6. Evaluate the integral:
\( \int_0^1 \frac{e^{-x}}{1+e^x} dx \)
Answer:
Let \( I = \int_0^1 \frac{e^{-x}}{1+e^x} dx \).
We can rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\( I = \int_0^1 \frac{\frac{1}{e^x}}{1+e^x} dx = \int_0^1 \frac{1}{e^x(1+e^x)} dx \).
Now, let \( t = e^x \). Then \( dt = e^x \, dx \), so \( dx = \frac{dt}{e^x} = \frac{dt}{t} \).
Change the limits of integration:
When \( x=0 \), \( t = e^0 = 1 \).
When \( x=1 \), \( t = e^1 = e \).
Substitute into the integral:
\( I = \int_1^e \frac{1}{t(1+t)} \frac{dt}{t} = \int_1^e \frac{1}{t^2(1+t)} dt \).
This requires partial fraction decomposition for \( \frac{1}{t^2(1+t)} \):
\( \frac{1}{t^2(1+t)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{1+t} \).
\( 1 = At(1+t) + B(1+t) + Ct^2 \).
If \( t=0 \), then \( 1 = B(1) \implies B=1 \).
If \( t=-1 \), then \( 1 = C(-1)^2 \implies C=1 \).
Compare coefficients of \( t^2 \): \( 0 = A+C \implies A = -C = -1 \).
So, \( \frac{1}{t^2(1+t)} = -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t} \).
\( I = \int_1^e \left(-\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t}\right) dt \).
\( = \left[-\ln|t| - \frac{1}{t} + \ln|1+t|\right]_1^e \).
Evaluate at the upper limit \( t=e \):
\( (-\ln e - \frac{1}{e} + \ln(1+e)) = (-1 - \frac{1}{e} + \ln(1+e)) \).
Evaluate at the lower limit \( t=1 \):
\( (-\ln 1 - \frac{1}{1} + \ln(1+1)) = (0 - 1 + \ln 2) = (-1 + \ln 2) \).
Subtract the lower limit from the upper limit:
\( I = \left(-1 - \frac{1}{e} + \ln(1+e)\right) - (-1 + \ln 2) \).
\( I = -\frac{1}{e} + \ln(1+e) - \ln 2 \).
\( I = -\frac{1}{e} + \ln\left(\frac{1+e}{2}\right) \).
Another approach using a substitution \( u=1+e^x \) or \( u=e^{-x} \):
Let \( u = 1+e^x \). Then \( du = e^x \, dx = (u-1) \, dx \implies dx = \frac{du}{u-1} \).
The integrand becomes \( \frac{e^{-x}}{1+e^x} = \frac{1/e^x}{1+e^x} = \frac{1/(u-1)}{u} = \frac{1}{u(u-1)} \).
Limits: \( x=0 \implies u=2 \); \( x=1 \implies u=1+e \).
\( I = \int_2^{1+e} \frac{1}{u(u-1)} \frac{du}{u-1} = \int_2^{1+e} \frac{1}{u(u-1)^2} du \).
This will also require partial fractions and is similar in complexity. The initial transformation \( \frac{e^{-x}}{1+e^x} = \frac{1}{e^x(1+e^x)} \) and substitution \( t=e^x \) is effective.
In simple words: We first change the form of the integral to make it easier to handle. Then, we use substitution to convert it into a simpler integral with new limits. We break the new fraction into parts and integrate each part, then apply the new limits to get the final answer.

🎯 Exam Tip: When integrating fractions involving exponentials, try multiplying the numerator and denominator by \( e^x \) or \( e^{-x} \) to simplify the expression and prepare for substitution or partial fraction decomposition.

Question 7. Evaluate the integral:
\( \int \frac{\log (\log x)}{x} d x \)
Answer:
Let \( I = \int \frac{\log (\log x)}{x} d x \).
Let \( t = \log x \).
Then \( dt = \frac{1}{x} d x \).
Substituting these into the integral, we get:
\( I = \int \log t \, dt \).
Now, we integrate \( \int \log t \, dt \) using integration by parts.
Let \( u = \log t \) and \( dv = 1 \, dt \).
Then \( du = \frac{1}{t} \, dt \) and \( v = t \).
\( I = t \log t - \int t \cdot \frac{1}{t} \, dt \).
\( I = t \log t - \int 1 \, dt \).
\( I = t \log t - t + C \).
Finally, substitute back \( t = \log x \):
\( I = (\log x) \log (\log x) - \log x + C \).
We can also write this as \( \log x (\log (\log x) - 1) + C \).
In simple words: We make a substitution to simplify the integral into a known form. Then we use integration by parts to solve this new, simpler integral. Finally, we put the original variable back to get the answer.

🎯 Exam Tip: When an integral contains \( \log(\log x) \) and \( \frac{1}{x} \) in an integral, it's often a good idea to substitute \( t = \log x \) first. This simplifies the expression and makes it easier to apply further integration techniques like integration by parts.

Question 8. Evaluate the integral:
\( \int_a^b \frac{\log x}{x} d x \)
Answer:
Let \( I = \int_a^b \frac{\log x}{x} d x \).
Let \( t = \log x \).
Then \( dt = \frac{1}{x} d x \).
Change the limits of integration:
When \( x=a \), \( t = \log a \).
When \( x=b \), \( t = \log b \).
Substitute these into the integral:
\( I = \int_{\log a}^{\log b} t \, dt \).
Integrate with respect to \( t \):
\( I = \left[\frac{t^2}{2}\right]_{\log a}^{\log b} \).
Evaluate at the limits:
\( I = \frac{(\log b)^2}{2} - \frac{(\log a)^2}{2} \).
\( I = \frac{1}{2} ((\log b)^2 - (\log a)^2) \).
Using the difference of squares formula, \( A^2 - B^2 = (A-B)(A+B) \):
\( I = \frac{1}{2} (\log b - \log a)(\log b + \log a) \).
Using logarithm properties \( \log A - \log B = \log(A/B) \) and \( \log A + \log B = \log(AB) \):
\( I = \frac{1}{2} \log\left(\frac{b}{a}\right) \log(ab) \).
This problem demonstrates how changing variables also requires updating the limits for definite integrals.
In simple words: We change the variable in the integral from \( x \) to \( \log x \). This also means we have to change the upper and lower limits of the integral. After doing this, the integral becomes very simple to solve, and we then plug in the new limits.

🎯 Exam Tip: For definite integrals involving substitution, always remember to change the limits of integration to match the new variable. This prevents the need to substitute back the original variable before applying the limits.

Question 9. Evaluate the integral:
\( \int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x \)
Answer:
Let \( I = \int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x \).
First, let \( t = \sin^{-1} x \).
Then \( x = \sin t \).
And \( dt = \frac{1}{\sqrt{1-x^2}} d x \).
Substitute these into the integral:
\( I = \int \sin t \cdot t \, dt \).
Now, we use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = t \) and \( dv = \sin t \, dt \).
Then \( du = dt \) and \( v = -\cos t \).
\( I = t(-\cos t) - \int (-\cos t) \, dt \).
\( I = -t \cos t + \int \cos t \, dt \).
\( I = -t \cos t + \sin t + C \).
Now, substitute back \( t = \sin^{-1} x \) and \( \sin t = x \).
From \( x = \sin t \), we can construct a right triangle where the opposite side is \( x \) and the hypotenuse is \( 1 \). The adjacent side would be \( \sqrt{1-x^2} \).
So, \( \cos t = \sqrt{1-x^2} \).
\( I = -(\sin^{-1} x) \sqrt{1-x^2} + x + C \).
This problem shows how combining substitution with integration by parts can simplify complex integrals.
In simple words: We first change the variable to simplify the fraction. Then, we use integration by parts to solve the new integral. Finally, we convert everything back to the original variable to get our answer.

🎯 Exam Tip: When an integral contains \( \sin^{-1} x \) and \( \frac{1}{\sqrt{1-x^2}} \), try the substitution \( t = \sin^{-1} x \). This often simplifies the integral significantly, allowing you to use integration by parts more effectively.

Question 10. Evaluate the integral:
\( \int_0^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x \)
Answer:
Let \( I = \int_0^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x \).
We use the identity \( 1+\sin A = \left(\cos \frac{A}{2} + \sin \frac{A}{2}\right)^2 \).
Here, \( A = \frac{x}{2} \), so \( \frac{A}{2} = \frac{x}{4} \).
\( 1+\sin \frac{x}{2} = \left(\cos \frac{x}{4} + \sin \frac{x}{4}\right)^2 \).
So, \( I = \int_0^{2 \pi} \sqrt{\left(\cos \frac{x}{4} + \sin \frac{x}{4}\right)^2} d x \).
\( I = \int_0^{2 \pi} \left|\cos \frac{x}{4} + \sin \frac{x}{4}\right| d x \).
For the interval \( 0 \le x \le 2\pi \), we have \( 0 \le \frac{x}{4} \le \frac{\pi}{2} \).
In this range, both \( \cos \frac{x}{4} \) and \( \sin \frac{x}{4} \) are non-negative, so their sum is also non-negative.
Thus, \( \left|\cos \frac{x}{4} + \sin \frac{x}{4}\right| = \cos \frac{x}{4} + \sin \frac{x}{4} \).
\( I = \int_0^{2 \pi} \left(\cos \frac{x}{4} + \sin \frac{x}{4}\right) d x \).
Integrate term by term:
\( I = \left[\frac{\sin \frac{x}{4}}{1/4} - \frac{\cos \frac{x}{4}}{1/4}\right]_0^{2 \pi} \).
\( I = \left[4 \sin \frac{x}{4} - 4 \cos \frac{x}{4}\right]_0^{2 \pi} \).
Evaluate at the limits:
At \( x=2\pi \): \( 4 \sin \frac{2\pi}{4} - 4 \cos \frac{2\pi}{4} = 4 \sin \frac{\pi}{2} - 4 \cos \frac{\pi}{2} = 4(1) - 4(0) = 4 \).
At \( x=0 \): \( 4 \sin \frac{0}{4} - 4 \cos \frac{0}{4} = 4 \sin 0 - 4 \cos 0 = 4(0) - 4(1) = -4 \).
Subtract the lower limit from the upper limit:
\( I = 4 - (-4) = 8 \).
Using trigonometric identities can often simplify complex square root integrals into elementary functions.
In simple words: We use a special math trick to rewrite the part under the square root as a perfect square. This removes the square root. Then, we integrate the simpler expression and plug in the numbers at the top and bottom of the integral to find the final value.

🎯 Exam Tip: Always look for trigonometric identities that can simplify expressions, especially those under square roots. The identity \( 1 \pm \sin A = \left(\cos \frac{A}{2} \pm \sin \frac{A}{2}\right)^2 \) is very useful for this type of problem.

Question 11. Evaluate the integral:
\( \int \sin ^{-1} \frac{2 x}{1+x^2} d x \)
Answer:
Let \( I = \int \sin ^{-1} \frac{2 x}{1+x^2} d x \).
We use the substitution \( x = \tan \theta \).
Then \( d x = \sec^2 \theta \, d\theta \).
The expression \( \frac{2 x}{1+x^2} \) becomes \( \frac{2 \tan \theta}{1+\tan^2 \theta} = \frac{2 \tan \theta}{\sec^2 \theta} = 2 \sin \theta \cos \theta = \sin(2\theta) \).
So, the integral becomes:
\( I = \int \sin^{-1}(\sin(2\theta)) \sec^2 \theta \, d\theta \).
\( I = \int 2\theta \sec^2 \theta \, d\theta \).
Now, we use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = 2\theta \) and \( dv = \sec^2 \theta \, d\theta \).
Then \( du = 2 \, d\theta \) and \( v = \tan \theta \).
\( I = 2\theta \tan \theta - \int \tan \theta \cdot 2 \, d\theta \).
\( I = 2\theta \tan \theta - 2 \int \tan \theta \, d\theta \).
We know that \( \int \tan \theta \, d\theta = \ln|\sec \theta| + C \).
\( I = 2\theta \tan \theta - 2 \ln|\sec \theta| + C \).
Now, substitute back \( \theta = \tan^{-1} x \) and \( \tan \theta = x \).
Also, \( \sec \theta = \sqrt{1+\tan^2 \theta} = \sqrt{1+x^2} \).
\( I = 2 (\tan^{-1} x) x - 2 \ln|\sqrt{1+x^2}| + C \).
\( I = 2x \tan^{-1} x - 2 \cdot \frac{1}{2} \ln(1+x^2) + C \).
\( I = 2x \tan^{-1} x - \ln(1+x^2) + C \).
This problem beautifully shows how a trigonometric substitution can convert a complex inverse trigonometric integral into a manageable form for integration by parts.
In simple words: We start by changing the variable using a trigonometric substitution to simplify the inverse sine part. This turns the integral into a product of a simple term and a trigonometric function, which we can then solve using integration by parts. Finally, we convert back to the original variable.

🎯 Exam Tip: For integrals involving \( \sin^{-1} \left(\frac{2x}{1+x^2}\right) \), \( \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right) \), or \( \tan^{-1} \left(\frac{2x}{1-x^2}\right) \), the substitution \( x = \tan \theta \) is very effective due to the resulting trigonometric identities for double angles.

Question 19. \( \int_0^{\pi / 4}(\tan x + \cot x)^{-1} d x \)
Answer: Let the integral be \( I \). We want to evaluate \( I = \int_0^{\pi / 4} (\tan x + \cot x)^{-1} d x \).
First, rewrite the terms inside the parenthesis:
\( \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} \)
So, \( (\tan x + \cot x)^{-1} = \sin x \cos x \)
We know that \( \sin 2x = 2 \sin x \cos x \), which means \( \sin x \cos x = \frac{\sin 2x}{2} \).
Now, substitute this back into the integral:
\( I = \int_0^{\pi / 4} \frac{\sin 2x}{2} d x \)
\( I = \frac{1}{2} \int_0^{\pi / 4} \sin 2x d x \)
Now, we integrate \( \sin 2x \):
\( \int \sin 2x d x = -\frac{\cos 2x}{2} \)
So, \( I = \frac{1}{2} \left[ -\frac{\cos 2x}{2} \right]_0^{\pi / 4} \)
\( I = -\frac{1}{4} \left[ \cos \left( 2 \cdot \frac{\pi}{4} \right) - \cos(2 \cdot 0) \right] \)
\( I = -\frac{1}{4} \left[ \cos \left( \frac{\pi}{2} \right) - \cos(0) \right] \)
\( I = -\frac{1}{4} [0 - 1] \)
\( I = -\frac{1}{4} (-1) \)
\( I = \frac{1}{4} \)
In simple words: First, simplify the expression \( (\tan x + \cot x)^{-1} \) into a simpler form using trigonometric identities. Then, integrate this simplified expression from 0 to \( \pi/4 \) and calculate the definite integral.

🎯 Exam Tip: Remember to simplify the integrand using trigonometric identities like \( \tan x + \cot x = \frac{1}{\sin x \cos x} \) and \( \sin 2x = 2 \sin x \cos x \) before integrating to make the calculation easier.

Question 20. \( \int \frac{1+\tan ^2 x}{\sqrt{1-\tan ^2 x}} d x \)
Answer: Let the integral be \( I = \int \frac{(1+\tan^2 x) d x}{\sqrt{1-\tan^2 x}} \).
We know that \( 1+\tan^2 x = \sec^2 x \).
So, the integral becomes \( I = \int \frac{\sec^2 x d x}{\sqrt{1-\tan^2 x}} \).
Now, let \( \tan x = t \).
Then, \( \frac{d}{dx}(\tan x) = \sec^2 x \), so \( \sec^2 x d x = d t \).
Substitute these into the integral:
\( I = \int \frac{d t}{\sqrt{1-t^2}} \)
This is a standard integral form, \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C \).
Here, \( a=1 \).
So, \( I = \sin^{-1} \left( \frac{t}{1} \right) + C \).
Replace \( t \) back with \( \tan x \):
\( I = \sin^{-1} (\tan x) + C \).
In simple words: First, change \( 1+\tan^2 x \) to \( \sec^2 x \). Then, let \( \tan x \) be a new variable, say \( t \), which makes \( \sec^2 x \, dx \) become \( dt \). After these changes, the integral becomes a simple form that can be solved directly, giving \( \sin^{-1} t \), and then replace \( t \) back with \( \tan x \).

🎯 Exam Tip: Recognizing fundamental trigonometric identities like \( 1+\tan^2 x = \sec^2 x \) and knowing common substitution methods are crucial for simplifying and solving such integrals quickly.

Question 21. \( \int_0^{1 / 2} \frac{\sin ^{-1} x}{(1-x^2)^{3 / 2}} dx \)
Answer: Let the integral be \( I = \int_0^{1/2} \frac{\sin^{-1} x}{(1-x^2)^{3/2}} dx \).
Let \( x = \sin \theta \).
Then \( dx = \cos \theta \, d\theta \).
The term \( (1-x^2)^{3/2} \) becomes \( (1-\sin^2 \theta)^{3/2} = (\cos^2 \theta)^{3/2} = \cos^3 \theta \).
Also, \( \sin^{-1} x = \sin^{-1}(\sin \theta) = \theta \).
Now, change the limits of integration:
When \( x = 0 \), \( \sin \theta = 0 \implies \theta = 0 \).
When \( x = 1/2 \), \( \sin \theta = 1/2 \implies \theta = \pi/6 \).
Substitute these into the integral:
\( I = \int_0^{\pi/6} \frac{\theta \cdot \cos \theta}{\cos^3 \theta} d\theta \)
\( I = \int_0^{\pi/6} \frac{\theta}{\cos^2 \theta} d\theta \)
\( I = \int_0^{\pi/6} \theta \sec^2 \theta d\theta \)
Now, use integration by parts, \( \int u \, dv = uv - \int v \, du \).
Let \( u = \theta \) and \( dv = \sec^2 \theta \, d\theta \).
Then \( du = d\theta \) and \( v = \tan \theta \).
\( I = [\theta \tan \theta]_0^{\pi/6} - \int_0^{\pi/6} \tan \theta \, d\theta \)
\( I = \left( \frac{\pi}{6} \tan \frac{\pi}{6} - 0 \tan 0 \right) - [\log |\cos \theta|]_0^{\pi/6} \)
\( I = \left( \frac{\pi}{6} \cdot \frac{1}{\sqrt{3}} - 0 \right) - \left( \log \left| \cos \frac{\pi}{6} \right| - \log |\cos 0| \right) \)
\( I = \frac{\pi}{6\sqrt{3}} - \left( \log \left( \frac{\sqrt{3}}{2} \right) - \log(1) \right) \)
\( I = \frac{\pi}{6\sqrt{3}} - \left( \log \left( \frac{\sqrt{3}}{2} \right) - 0 \right) \)
\( I = \frac{\pi}{6\sqrt{3}} - \log \left( \frac{\sqrt{3}}{2} \right) \)
This can also be written as \( I = \frac{\pi \sqrt{3}}{18} - \log \left( \frac{\sqrt{3}}{2} \right) \).
In simple words: We change the variable from \( x \) to \( \theta \) using \( x = \sin \theta \). This simplifies the bottom part of the fraction and \( \sin^{-1} x \). Then we update the limits for \( \theta \). Finally, we use a special method called integration by parts to solve the new integral.

🎯 Exam Tip: For integrals involving \( \sin^{-1} x \) or \( (1-x^2)^{n} \), a substitution of \( x = \sin \theta \) is often very effective. Remember to change the limits of integration when evaluating definite integrals after substitution.

Question 22. \( \int \frac{\cos x}{\sin x+\sqrt{\sin x}} d x \)
Answer: Let the integral be \( I = \int \frac{\cos x}{\sin x+\sqrt{\sin x}} d x \).
Let \( \sqrt{\sin x} = t \).
Then, squaring both sides, we get \( \sin x = t^2 \).
Now, differentiate both sides with respect to \( x \):
\( \cos x \, d x = 2t \, d t \).
Substitute these into the integral:
\( I = \int \frac{2t \, d t}{t^2 + t} \)
\( I = \int \frac{2t}{t(t+1)} d t \)
\( I = \int \frac{2}{t+1} d t \)
Now, integrate with respect to \( t \):
\( I = 2 \log |t+1| + C \).
Finally, substitute back \( t = \sqrt{\sin x} \):
\( I = 2 \log |1+\sqrt{\sin x}| + C \).
In simple words: To solve this, we let the square root of \( \sin x \) be a new variable, \( t \). Then we find out what \( \cos x \, dx \) becomes in terms of \( t \) and \( dt \). After simplifying the fraction, the integral becomes much easier to solve. Finally, we put the original \( \sin x \) back into the answer.

🎯 Exam Tip: When you see \( \sqrt{\sin x} \) or \( \sqrt{\cos x} \) in an integral, try substituting the square root term with a new variable to simplify the integrand. This often leads to a more manageable rational function.

Question 23. Prove that \( \int_0^{\pi / 2} \frac{3 \sin \theta+4 \cos \theta}{\sin \theta+\cos \theta} d\theta = \frac{7 \pi}{4} \)
Answer: Let the integral be \( I = \int_0^{\pi/2} \frac{3 \sin \theta+4 \cos \theta}{\sin \theta+\cos \theta} d\theta \) --- (1)
Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), we replace \( \theta \) with \( \frac{\pi}{2} - \theta \).
\( \sin \left( \frac{\pi}{2} - \theta \right) = \cos \theta \)
\( \cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta \)
So, the integral becomes:
\( I = \int_0^{\pi/2} \frac{3 \cos \theta+4 \sin \theta}{\cos \theta+\sin \theta} d\theta \) --- (2)
Now, add equation (1) and equation (2):
\( I + I = \int_0^{\pi/2} \frac{3 \sin \theta+4 \cos \theta}{\sin \theta+\cos \theta} d\theta + \int_0^{\pi/2} \frac{3 \cos \theta+4 \sin \theta}{\cos \theta+\sin \theta} d\theta \)
\( 2I = \int_0^{\pi/2} \frac{(3 \sin \theta+4 \cos \theta) + (3 \cos \theta+4 \sin \theta)}{\sin \theta+\cos \theta} d\theta \)
\( 2I = \int_0^{\pi/2} \frac{7 \sin \theta+7 \cos \theta}{\sin \theta+\cos \theta} d\theta \)
\( 2I = \int_0^{\pi/2} \frac{7(\sin \theta+\cos \theta)}{\sin \theta+\cos \theta} d\theta \)
\( 2I = \int_0^{\pi/2} 7 \, d\theta \)
Now, integrate with respect to \( \theta \):
\( 2I = [7\theta]_0^{\pi/2} \)
\( 2I = 7 \left( \frac{\pi}{2} - 0 \right) \)
\( 2I = \frac{7\pi}{2} \)
Divide both sides by 2:
\( I = \frac{7\pi}{4} \)
Hence, proved.
In simple words: We used a special trick for definite integrals where we replace \( \theta \) with \( (\pi/2 - \theta) \). When we add the original integral and the new one, the top part of the fraction simplifies to 7 times the bottom part. This makes the integral very easy to solve, giving us \( \frac{7\pi}{4} \).

🎯 Exam Tip: For definite integrals from 0 to 'a', always consider using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) when the integrand has terms like \( \sin x \) and \( \cos x \) that can transform into each other. Adding the original and transformed integrals often simplifies the expression significantly.

Question 24. \( \int x^2 (\text{e}^{x^3}) \cos (2 \text{e}^{x^3}) d x \)
Answer: Let the integral be \( I = \int x^2 (\text{e}^{x^3}) \cos (2 \text{e}^{x^3}) d x \).
Let \( \text{e}^{x^3} = t \).
To find \( dt \), we differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(\text{e}^{x^3}) = \text{e}^{x^3} \cdot 3x^2 \).
So, \( 3x^2 \text{e}^{x^3} d x = d t \).
This means \( x^2 \text{e}^{x^3} d x = \frac{d t}{3} \).
Substitute these into the integral:
\( I = \int \cos (2t) \cdot \frac{d t}{3} \)
\( I = \frac{1}{3} \int \cos (2t) d t \)
Now, integrate \( \cos (2t) \) with respect to \( t \):
\( \int \cos (2t) d t = \frac{\sin (2t)}{2} \).
So, \( I = \frac{1}{3} \cdot \frac{\sin (2t)}{2} + C \)
\( I = \frac{1}{6} \sin (2t) + C \).
Finally, substitute \( t = \text{e}^{x^3} \) back into the expression:
\( I = \frac{1}{6} \sin (2 \text{e}^{x^3}) + C \).
In simple words: We make a substitution by letting \( \text{e}^{x^3} \) be \( t \). Then we find the differential \( dt \) and substitute it into the integral. This changes the complex integral into a simple one involving \( \cos(2t) \). After integrating, we replace \( t \) with its original expression to get the final answer.

🎯 Exam Tip: Look for parts of the integrand whose derivative is also present (or a constant multiple thereof). Here, \( x^2 \text{e}^{x^3} \) is related to the derivative of \( \text{e}^{x^3} \), making \( t = \text{e}^{x^3} \) an ideal substitution.

Question 25. Prove that \( \int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x = 2\pi^2 \)
Answer: Let the integral be \( I = \int_0^{2\pi} \frac{x \cos x}{1+\cos x} d x \) --- (1)
Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), we replace \( x \) with \( 2\pi - x \).
\( \cos (2\pi - x) = \cos x \).
So, the integral becomes:
\( I = \int_0^{2\pi} \frac{(2\pi - x) \cos (2\pi - x)}{1+\cos (2\pi - x)} d x \)
\( I = \int_0^{2\pi} \frac{(2\pi - x) \cos x}{1+\cos x} d x \) --- (2)
Now, add equation (1) and equation (2):
\( I + I = \int_0^{2\pi} \frac{x \cos x}{1+\cos x} d x + \int_0^{2\pi} \frac{(2\pi - x) \cos x}{1+\cos x} d x \)
\( 2I = \int_0^{2\pi} \frac{x \cos x + (2\pi - x) \cos x}{1+\cos x} d x \)
\( 2I = \int_0^{2\pi} \frac{\cos x (x + 2\pi - x)}{1+\cos x} d x \)
\( 2I = \int_0^{2\pi} \frac{2\pi \cos x}{1+\cos x} d x \)
\( 2I = 2\pi \int_0^{2\pi} \frac{\cos x}{1+\cos x} d x \)
To simplify \( \frac{\cos x}{1+\cos x} \), we can add and subtract 1 in the numerator:
\( \frac{\cos x}{1+\cos x} = \frac{1+\cos x - 1}{1+\cos x} = 1 - \frac{1}{1+\cos x} \)
Also, we know that \( 1+\cos x = 2\cos^2 \frac{x}{2} \).
So, \( \frac{1}{1+\cos x} = \frac{1}{2\cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} \).
Therefore, \( \frac{\cos x}{1+\cos x} = 1 - \frac{1}{2} \sec^2 \frac{x}{2} \).
Substitute this back into the integral for \( 2I \):
\( 2I = 2\pi \int_0^{2\pi} \left( 1 - \frac{1}{2} \sec^2 \frac{x}{2} \right) d x \)
\( 2I = 2\pi \left[ x - \frac{1}{2} \frac{\tan \frac{x}{2}}{1/2} \right]_0^{2\pi} \)
\( 2I = 2\pi [x - \tan \frac{x}{2}]_0^{2\pi} \)
Now, evaluate the limits:
\( 2I = 2\pi \left[ (2\pi - \tan \frac{2\pi}{2}) - (0 - \tan \frac{0}{2}) \right] \)
\( 2I = 2\pi [ (2\pi - \tan \pi) - (0 - \tan 0) ] \)
Since \( \tan \pi = 0 \) and \( \tan 0 = 0 \):
\( 2I = 2\pi [ (2\pi - 0) - (0 - 0) ] \)
\( 2I = 2\pi (2\pi) \)
\( 2I = 4\pi^2 \)
Divide by 2:
\( I = 2\pi^2 \).
Hence, proved.
In simple words: We use a special property of definite integrals to combine the original integral with a modified version of itself. This helps to remove the \( x \) term from the numerator. Then, we rewrite \( \frac{\cos x}{1+\cos x} \) as \( 1 - \frac{1}{1+\cos x} \) and use the identity \( 1+\cos x = 2\cos^2(x/2) \) to simplify it further. After integrating and applying the limits, we get the result \( 2\pi^2 \).

🎯 Exam Tip: This type of integral (King's property) often appears in exams. Always remember to use \( 1+\cos x = 2\cos^2(x/2) \) for simplification, as it often helps convert the expression into integrable forms involving \( \sec^2(x/2) \).

Question 26. \( \int_0^{\pi / 4} \log (1 + \tan x) d x \)
Answer: Let the integral be \( I = \int_0^{\pi/4} \log (1 + \tan x) d x \) --- (1)
Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), we replace \( x \) with \( \frac{\pi}{4} - x \).
So, the integral becomes:
\( I = \int_0^{\pi/4} \log \left( 1 + \tan \left( \frac{\pi}{4} - x \right) \right) d x \)
Use the tangent subtraction formula: \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
Here, \( A = \frac{\pi}{4} \) and \( B = x \). So, \( \tan \left( \frac{\pi}{4} - x \right) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4) \tan x} = \frac{1 - \tan x}{1 + 1 \cdot \tan x} = \frac{1 - \tan x}{1 + \tan x} \).
Substitute this back into the expression for \( I \):
\( I = \int_0^{\pi/4} \log \left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) d x \)
\( I = \int_0^{\pi/4} \log \left( \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} \right) d x \)
\( I = \int_0^{\pi/4} \log \left( \frac{2}{1 + \tan x} \right) d x \)
Using the logarithm property \( \log \left( \frac{A}{B} \right) = \log A - \log B \):
\( I = \int_0^{\pi/4} (\log 2 - \log (1 + \tan x)) d x \)
\( I = \int_0^{\pi/4} \log 2 \, d x - \int_0^{\pi/4} \log (1 + \tan x) d x \)
Notice that the second term is the original integral \( I \).
So, \( I = \int_0^{\pi/4} \log 2 \, d x - I \)
Move \( -I \) to the left side:
\( 2I = \int_0^{\pi/4} \log 2 \, d x \)
\( 2I = [\log 2 \cdot x]_0^{\pi/4} \)
\( 2I = \log 2 \left( \frac{\pi}{4} - 0 \right) \)
\( 2I = \frac{\pi}{4} \log 2 \)
Divide by 2:
\( I = \frac{\pi}{8} \log 2 \).
In simple words: We start with the integral and use a special rule to change \( x \) to \( (\pi/4 - x) \). After simplifying \( \tan(\pi/4 - x) \), the integral becomes \( \log(2/(1+\tan x)) \). Using log rules, this splits into \( \log 2 \) minus the original integral \( I \). By solving for \( I \), we find its value.

🎯 Exam Tip: This is a standard problem type involving the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). Recognizing \( \tan(\pi/4 - x) \) and the logarithm property are key. Adding 'I' to both sides after the substitution is a common strategy for such integrals.

Question 27. \( \int \frac{1}{x \cos^2(1+\log x)} d x \)
Answer: Let the integral be \( I = \int \frac{1}{x \cos^2(1+\log x)} d x \).
Let \( t = 1+\log x \).
To find \( dt \), differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(1+\log x) = 0 + \frac{1}{x} = \frac{1}{x} \).
So, \( \frac{1}{x} d x = d t \).
Substitute these into the integral:
\( I = \int \frac{1}{\cos^2 t} d t \)
We know that \( \frac{1}{\cos^2 t} = \sec^2 t \).
So, \( I = \int \sec^2 t \, d t \).
Now, integrate \( \sec^2 t \) with respect to \( t \):
\( \int \sec^2 t \, d t = \tan t + C \).
Finally, substitute back \( t = 1+\log x \):
\( I = \tan (1+\log x) + C \).
In simple words: We simplify the integral by letting \( (1+\log x) \) be \( t \). This makes \( \frac{1}{x} \, dx \) become \( dt \). The integral then changes to \( \int \sec^2 t \, dt \), which is a basic integral resulting in \( \tan t \). We then replace \( t \) with its original expression to get the final answer.

🎯 Exam Tip: When \( \log x \) is present inside a trigonometric function or its argument, and \( 1/x \) is also in the integrand, \( 1+\log x \) (or just \( \log x \)) is a good candidate for substitution. This will transform the integral into a standard trigonometric form.

Question 28. \( \int \frac{x^2}{x^2-4} d x \)
Answer: Let the integral be \( I = \int \frac{x^2}{x^2-4} d x \).
The degree of the numerator is equal to the degree of the denominator. In such cases, we perform polynomial division or algebraic manipulation to simplify the fraction.
We can add and subtract 4 in the numerator:
\( I = \int \frac{x^2 - 4 + 4}{x^2-4} d x \)
\( I = \int \left( \frac{x^2-4}{x^2-4} + \frac{4}{x^2-4} \right) d x \)
\( I = \int \left( 1 + \frac{4}{x^2-4} \right) d x \)
\( I = \int 1 \, d x + \int \frac{4}{x^2-4} d x \)
\( I = x + 4 \int \frac{1}{x^2-2^2} d x \)
Now, we use the standard integral formula \( \int \frac{1}{x^2-a^2} d x = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \).
Here, \( a=2 \).
So, \( I = x + 4 \cdot \frac{1}{2 \cdot 2} \log \left| \frac{x-2}{x+2} \right| + C \)
\( I = x + 4 \cdot \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + C \)
\( I = x + \log \left| \frac{x-2}{x+2} \right| + C \).
In simple words: First, we change the top part of the fraction by adding and subtracting 4. This helps us split the fraction into two parts: 1, and \( \frac{4}{x^2-4} \). We can integrate the 1 easily. For the second part, we use a known formula for fractions like \( \frac{1}{x^2-a^2} \). Combining these gives us the answer.

🎯 Exam Tip: Whenever the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division or an algebraic manipulation (like adding and subtracting a constant) to simplify the rational function before integrating. This will typically yield simpler terms that are easier to integrate.

Question 29. Evaluate \( \int_0^9 f(x) d x \) where \( f(x) \) is defined by \( f(x) = \begin{cases} \sin x, & \text{if } 0 < x < \frac{\pi}{2} \\ 1, & \text{if } \frac{\pi}{2} < x < 5 \\ \text{e}^{x-5}, & \text{if } 5 < x < 9 \end{cases} \)
Answer: The integral \( \int_0^9 f(x) d x \) can be split into three parts based on the definition of \( f(x) \):
\( \int_0^9 f(x) d x = \int_0^{\pi/2} \sin x \, d x + \int_{\pi/2}^5 1 \, d x + \int_5^9 \text{e}^{x-5} \, d x \)
Let's evaluate each integral separately:
1. \( \int_0^{\pi/2} \sin x \, d x \):
\( [-\cos x]_0^{\pi/2} = (-\cos \frac{\pi}{2}) - (-\cos 0) = -0 - (-1) = 1 \).
2. \( \int_{\pi/2}^5 1 \, d x \):
\( [x]_{\pi/2}^5 = 5 - \frac{\pi}{2} \).
3. \( \int_5^9 \text{e}^{x-5} \, d x \):
Let \( u = x-5 \), then \( du = dx \).
When \( x=5 \), \( u=0 \). When \( x=9 \), \( u=4 \).
So, \( \int_0^4 \text{e}^u \, du = [\text{e}^u]_0^4 = \text{e}^4 - \text{e}^0 = \text{e}^4 - 1 \).
Now, sum these three results:
\( \int_0^9 f(x) d x = 1 + \left( 5 - \frac{\pi}{2} \right) + (\text{e}^4 - 1) \)
\( \int_0^9 f(x) d x = 1 + 5 - \frac{\pi}{2} + \text{e}^4 - 1 \)
\( \int_0^9 f(x) d x = 5 - \frac{\pi}{2} + \text{e}^4 \).
In simple words: Since the function \( f(x) \) changes its rule at different points, we break the big integral into three smaller ones. We solve each small integral separately using the correct rule for \( f(x) \) in that part. Then, we add up the results from all three parts to get the total value.

🎯 Exam Tip: For piecewise functions, always split the integral into sub-intervals according to the function's definition. Make sure to correctly apply the corresponding function definition for each interval and evaluate the limits carefully.

Question 30. \( \int \frac{\text{e}^{2 x}}{2+\text{e}^x} d x \)
Answer: Let the integral be \( I = \int \frac{\text{e}^{2x}}{2+\text{e}^x} d x \).
We can write \( \text{e}^{2x} \) as \( (\text{e}^x)^2 \).
Let \( t = \text{e}^x \).
To find \( dt \), differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(\text{e}^x) = \text{e}^x \).
So, \( \text{e}^x d x = d t \).
The integral can be rewritten as \( I = \int \frac{\text{e}^x \cdot \text{e}^x}{2+\text{e}^x} d x \).
Substitute \( t = \text{e}^x \) and \( \text{e}^x d x = d t \):
\( I = \int \frac{t}{2+t} d t \).
Now, we can add and subtract 2 in the numerator to simplify the fraction:
\( I = \int \frac{t+2-2}{t+2} d t \)
\( I = \int \left( \frac{t+2}{t+2} - \frac{2}{t+2} \right) d t \)
\( I = \int \left( 1 - \frac{2}{t+2} \right) d t \)
Now, integrate with respect to \( t \):
\( I = t - 2 \log |t+2| + C \).
Finally, substitute back \( t = \text{e}^x \):
\( I = \text{e}^x - 2 \log |\text{e}^x+2| + C \).
In simple words: We let \( \text{e}^x \) be \( t \), which makes \( \text{e}^x \, dx \) turn into \( dt \). The integral then looks like \( \int \frac{t}{2+t} \, dt \). We add and subtract 2 in the numerator to simplify the fraction before integrating. After integrating, we replace \( t \) with \( \text{e}^x \) to get the final answer.

🎯 Exam Tip: For integrals involving \( \text{e}^x \) in both numerator and denominator, try substituting \( t = \text{e}^x \). This often transforms the integral into a rational function in \( t \), which can be solved using algebraic manipulation or partial fractions.

Question 31. \( \int \frac{x}{(x+1)^2} d x \)
Answer: Let the integral be \( I = \int \frac{x}{(x+1)^2} d x \).
We can use a substitution here. Let \( u = x+1 \).
Then \( x = u-1 \).
Also, \( du = dx \).
Substitute these into the integral:
\( I = \int \frac{u-1}{u^2} d u \)
Now, split the fraction:
\( I = \int \left( \frac{u}{u^2} - \frac{1}{u^2} \right) d u \)
\( I = \int \left( \frac{1}{u} - u^{-2} \right) d u \)
Now, integrate each term with respect to \( u \):
\( \int \frac{1}{u} d u = \log |u| \).
\( \int u^{-2} d u = \frac{u^{-2+1}}{-2+1} = \frac{u^{-1}}{-1} = -\frac{1}{u} \).
So, \( I = \log |u| - \left( -\frac{1}{u} \right) + C \)
\( I = \log |u| + \frac{1}{u} + C \).
Finally, substitute back \( u = x+1 \):
\( I = \log |x+1| + \frac{1}{x+1} + C \).
In simple words: We make the substitution \( u = x+1 \) to simplify the denominator. This also changes \( x \) to \( (u-1) \). The integral then splits into two easier parts: \( \frac{1}{u} \) and \( -\frac{1}{u^2} \). We integrate these parts separately and then change \( u \) back to \( x+1 \) for the final answer.

🎯 Exam Tip: When you see \( (x+a)^n \) in the denominator and a linear term \( x \) in the numerator, substituting \( u = x+a \) is usually effective. This allows you to express \( x \) in terms of \( u \) and simplify the integrand before integration.

Question 32. \( \int_{-3}^3|x + 2| d x \)
Answer: Let the integral be \( I = \int_{-3}^3|x + 2| d x \).
The absolute value function \( |x+2| \) changes its definition at \( x+2=0 \), which means at \( x=-2 \).
So, we need to split the integral at \( x=-2 \).
\( I = \int_{-3}^{-2}|x + 2| d x + \int_{-2}^3|x + 2| d x \).
For the interval \( -3 < x < -2 \), \( x+2 \) is negative. So, \( |x+2| = -(x+2) \).
For the interval \( -2 < x < 3 \), \( x+2 \) is positive. So, \( |x+2| = (x+2) \).
Substitute these into the integrals:
\( I = \int_{-3}^{-2} -(x+2) d x + \int_{-2}^3 (x+2) d x \)
\( I = -\left[ \frac{(x+2)^2}{2} \right]_{-3}^{-2} + \left[ \frac{(x+2)^2}{2} \right]_{-2}^3 \)
Evaluate the first part:
\( -\left( \frac{(-2+2)^2}{2} - \frac{(-3+2)^2}{2} \right) = -\left( \frac{0^2}{2} - \frac{(-1)^2}{2} \right) = -\left( 0 - \frac{1}{2} \right) = \frac{1}{2} \).
Evaluate the second part:
\( \left( \frac{(3+2)^2}{2} - \frac{(-2+2)^2}{2} \right) = \left( \frac{5^2}{2} - \frac{0^2}{2} \right) = \left( \frac{25}{2} - 0 \right) = \frac{25}{2} \).
Now, add the two parts:
\( I = \frac{1}{2} + \frac{25}{2} = \frac{26}{2} = 13 \).
In simple words: The function \( |x+2| \) behaves differently when \( x+2 \) is negative or positive. So, we break the integral into two parts at \( x = -2 \). For each part, we use the correct form of \( |x+2| \) and then calculate the integral. Finally, we add the results from both parts.

🎯 Exam Tip: When evaluating definite integrals involving absolute value functions, identify the points where the expression inside the absolute value becomes zero. Split the integral at these points and redefine the absolute value function for each interval before integrating.

Question 33. \( \int \frac{2 \sin 2 \theta-\cos \theta}{6-\cos^2 \theta-4 \sin \theta} d\theta \)
Answer: Let the integral be \( I = \int \frac{2 \sin 2 \theta-\cos \theta}{6-\cos^2 \theta-4 \sin \theta} d\theta \).
First, rewrite \( \sin 2\theta \) as \( 2 \sin \theta \cos \theta \) and \( \cos^2 \theta \) as \( 1-\sin^2 \theta \).
\( I = \int \frac{2(2 \sin \theta \cos \theta)-\cos \theta}{6-(1-\sin^2 \theta)-4 \sin \theta} d\theta \)
\( I = \int \frac{4 \sin \theta \cos \theta-\cos \theta}{6-1+\sin^2 \theta-4 \sin \theta} d\theta \)
\( I = \int \frac{\cos \theta(4 \sin \theta-1)}{\sin^2 \theta-4 \sin \theta+5} d\theta \)
Now, let \( t = \sin \theta \).
Then \( dt = \cos \theta \, d\theta \).
Substitute these into the integral:
\( I = \int \frac{4t-1}{t^2-4t+5} d t \)
To solve this integral, we want the numerator to be the derivative of the denominator. The derivative of \( t^2-4t+5 \) is \( 2t-4 \).
We can write \( 4t-1 \) as \( A(2t-4) + B \).
\( 4t-1 = 2At - 4A + B \).
Comparing coefficients of \( t \): \( 2A=4 \implies A=2 \).
Comparing constant terms: \( -4A+B = -1 \implies -4(2)+B = -1 \implies -8+B = -1 \implies B = 7 \).
So, \( 4t-1 = 2(2t-4) + 7 \).
Substitute this back into the integral:
\( I = \int \frac{2(2t-4)+7}{t^2-4t+5} d t \)
\( I = \int \frac{2(2t-4)}{t^2-4t+5} d t + \int \frac{7}{t^2-4t+5} d t \)
For the first integral, let \( u = t^2-4t+5 \), then \( du = (2t-4)dt \). So, \( \int \frac{2(2t-4)}{t^2-4t+5} d t = 2 \int \frac{du}{u} = 2 \log |u| = 2 \log |t^2-4t+5| \).
For the second integral, complete the square in the denominator:
\( t^2-4t+5 = (t^2-4t+4) + 1 = (t-2)^2 + 1^2 \).
So, \( \int \frac{7}{(t-2)^2+1^2} d t = 7 \tan^{-1} \left( \frac{t-2}{1} \right) = 7 \tan^{-1} (t-2) \).
Combining these results:
\( I = 2 \log |t^2-4t+5| + 7 \tan^{-1} (t-2) + C \).
Finally, substitute back \( t = \sin \theta \):
\( I = 2 \log |\sin^2 \theta-4 \sin \theta+5| + 7 \tan^{-1} (\sin \theta-2) + C \).
In simple words: First, we change \( \sin 2\theta \) and \( \cos^2 \theta \) to expressions involving \( \sin \theta \). Then, we let \( \sin \theta \) be \( t \). The integral becomes a fraction with \( t \) terms. We adjust the top of the fraction to match the derivative of the bottom. This splits the integral into two parts: one that becomes a logarithm, and another that involves \( \tan^{-1} \). At the end, we replace \( t \) with \( \sin \theta \).

🎯 Exam Tip: For integrals with a mix of trigonometric functions, try to express everything in terms of one function (like \( \sin \theta \) or \( \cos \theta \)) using identities. Then, use substitution. If the denominator is a quadratic expression, try completing the square to fit it into a standard integral form like \( \tan^{-1} \) or \( \log \). Always make the numerator a linear combination of the derivative of the denominator and a constant, \( N = A \cdot D' + B \).

Question 34. \( \int \frac{\csc x}{\log(\tan(\frac{x}{2}))} d x \)
Answer: Let the integral be \( I = \int \frac{\csc x}{\log(\tan(\frac{x}{2}))} d x \).
Let \( t = \log \left( \tan \left( \frac{x}{2} \right) \right) \).
To find \( dt \), differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{1}{\tan(\frac{x}{2})} \cdot \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \)
\( \frac{dt}{dx} = \frac{\cos(\frac{x}{2})}{\sin(\frac{x}{2})} \cdot \frac{1}{\cos^2(\frac{x}{2})} \cdot \frac{1}{2} \)
\( \frac{dt}{dx} = \frac{1}{2 \sin(\frac{x}{2}) \cos(\frac{x}{2})} \)
We know that \( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \).
So, \( \frac{dt}{dx} = \frac{1}{\sin x} = \csc x \).
Therefore, \( \csc x \, d x = d t \).
Substitute these into the integral:
\( I = \int \frac{1}{t} d t \).
Now, integrate with respect to \( t \):
\( I = \log |t| + C \).
Finally, substitute back \( t = \log \left( \tan \left( \frac{x}{2} \right) \right) \):
\( I = \log \left| \log \left( \tan \left( \frac{x}{2} \right) \right) \right| + C \).
In simple words: We let the entire expression inside the logarithm be \( t \). When we take the derivative of \( t \), it amazingly simplifies to \( \csc x \, dx \). This means the complex integral transforms into a very simple \( \int \frac{1}{t} \, dt \), which is \( \log|t| \). Then we put the original expression back in place of \( t \).

🎯 Exam Tip: This integral is a classic example where the derivative of \( \log(\tan(x/2)) \) is \( \csc x \). Always be on the lookout for such derivative pairs, as they greatly simplify the integration process through substitution. Knowing \( \frac{d}{dx} (\log(\tan(x/2))) = \csc x \) is extremely useful.

Question 35. \( \int_0^1 \frac{x \text{e}^x}{(1+x)^2} d x \)
Answer: Let the integral be \( I = \int_0^1 \frac{x \text{e}^x}{(1+x)^2} d x \).
We can rewrite the numerator \( x \text{e}^x \) as \( (1+x-1)\text{e}^x \).
\( I = \int_0^1 \frac{(1+x-1)\text{e}^x}{(1+x)^2} d x \)
\( I = \int_0^1 \left( \frac{(1+x)\text{e}^x}{(1+x)^2} - \frac{\text{e}^x}{(1+x)^2} \right) d x \)
\( I = \int_0^1 \left( \frac{\text{e}^x}{1+x} - \frac{\text{e}^x}{(1+x)^2} \right) d x \)
This integral is of the form \( \int \text{e}^x [f(x) + f'(x)] d x = \text{e}^x f(x) + C \).
Here, let \( f(x) = \frac{1}{1+x} \).
Then, \( f'(x) = \frac{d}{dx} \left( (1+x)^{-1} \right) = -1 \cdot (1+x)^{-2} \cdot 1 = -\frac{1}{(1+x)^2} \).
So, the integral matches the form \( \int \text{e}^x [f(x) + f'(x)] d x \).
Therefore, the indefinite integral is \( \text{e}^x \cdot \frac{1}{1+x} + C = \frac{\text{e}^x}{1+x} + C \).
Now, we evaluate this definite integral from 0 to 1:
\( I = \left[ \frac{\text{e}^x}{1+x} \right]_0^1 \)
\( I = \frac{\text{e}^1}{1+1} - \frac{\text{e}^0}{1+0} \)
\( I = \frac{\text{e}}{2} - \frac{1}{1} \)
\( I = \frac{\text{e}}{2} - 1 \).
In simple words: We rewrite the top part of the fraction to make it easier to split. This creates two terms inside the integral, one of which is the derivative of the other, multiplied by \( \text{e}^x \). This matches a special integration pattern where the answer is simply \( \text{e}^x \) times the first term. Then we apply the limits to find the final numerical value.

🎯 Exam Tip: Always look for the pattern \( \int \text{e}^x [f(x) + f'(x)] d x = \text{e}^x f(x) + C \) when \( \text{e}^x \) is part of the integrand. Manipulating the numerator to fit this form (e.g., adding and subtracting terms) is a common strategy.

Question 36. \( \int \frac{x^2-5 x-1}{x^4+x^2+1} d x \)
Answer: Let the integral be \( I = \int \frac{x^2-5 x-1}{x^4+x^2+1} d x \).
The denominator \( x^4+x^2+1 \) can be factored. It is a common form: \( x^4+x^2+1 = (x^2+1)^2 - x^2 = (x^2+1-x)(x^2+1+x) \).
So, \( I = \int \frac{x^2-5 x-1}{(x^2-x+1)(x^2+x+1)} d x \).
We can split the integral into two parts based on the numerator and divide both numerator and denominator by \( x^2 \):
\( I = \int \frac{1-5/x-1/x^2}{x^2+1+1/x^2} d x = \int \frac{(1-1/x^2) - 5/x}{(x+1/x)^2 - 1} d x \).
This problem might be simplified further by splitting the numerator \( x^2-5x-1 \) based on the factors of the denominator.
Let's split the integral into two parts based on the numerator structure that aligns with factors of the denominator:
\( I_1 = \int \frac{x^2-1}{x^4+x^2+1} d x \) and \( I_2 = \int \frac{5x}{x^4+x^2+1} d x \).
So, \( I = I_1 - I_2 \).
For \( I_1 \): Divide numerator and denominator by \( x^2 \).
\( I_1 = \int \frac{1-1/x^2}{x^2+1+1/x^2} d x \)
\( I_1 = \int \frac{1-1/x^2}{(x+1/x)^2-1} d x \)
Let \( t = x+1/x \). Then \( dt = (1-1/x^2)dx \).
\( I_1 = \int \frac{dt}{t^2-1} = \frac{1}{2 \cdot 1} \log \left| \frac{t-1}{t+1} \right| + C_1 = \frac{1}{2} \log \left| \frac{x+1/x-1}{x+1/x+1} \right| + C_1 = \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C_1 \).
For \( I_2 \): Let \( u = x^2 \). Then \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \).
\( I_2 = \int \frac{5x}{x^4+x^2+1} d x = \frac{5}{2} \int \frac{du}{u^2+u+1} \).
Complete the square for \( u^2+u+1 \): \( u^2+u+1 = (u+1/2)^2 + (\sqrt{3}/2)^2 \).
\( I_2 = \frac{5}{2} \int \frac{du}{(u+1/2)^2 + (\sqrt{3}/2)^2} = \frac{5}{2} \cdot \frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{u+1/2}{\sqrt{3}/2} \right) + C_2 \)
\( I_2 = \frac{5}{\sqrt{3}} \tan^{-1} \left( \frac{2u+1}{\sqrt{3}} \right) + C_2 = \frac{5}{\sqrt{3}} \tan^{-1} \left( \frac{2x^2+1}{\sqrt{3}} \right) + C_2 \).
Combining \( I_1 \) and \( I_2 \):
\( I = \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| - \frac{5}{\sqrt{3}} \tan^{-1} \left( \frac{2x^2+1}{\sqrt{3}} \right) + C \).
In simple words: This integral is hard, so we split it into two simpler parts. For the first part, we divide everything by \( x^2 \) and use a substitution like \( t = x+1/x \). For the second part, we use a substitution like \( u = x^2 \) and then complete the square in the denominator. Finally, we combine the answers from these two parts.

🎯 Exam Tip: Integrals with \( x^4+x^2+1 \) in the denominator often require dividing the numerator and denominator by \( x^2 \) and then making a substitution of \( x \pm 1/x \). For terms like \( x^2+1/x^2 \) or \( x^4+x^2+1 \), remember the factorization \( (x^2+1)^2 - x^2 = (x^2-x+1)(x^2+x+1) \).

Question 37. \( \int \frac{x}{\sqrt{3-x}+\sqrt{x}} d x \)
Answer: Let the integral be \( I = \int \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x \). This is a definite integral usually given with limits. Assuming it's a definite integral \( \int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x \) based on common problem types related to the given answer in the source.
Let \( I = \int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x \) --- (1)
Using the property \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \).
Here, \( a=1, b=2 \), so \( a+b-x = 1+2-x = 3-x \).
Replace \( x \) with \( 3-x \) in the integrand:
\( I = \int_1^2 \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} d x \)
\( I = \int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x \) --- (2)
Now, add equation (1) and equation (2):
\( I + I = \int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x + \int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x \)
\( 2I = \int_1^2 \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x \)
\( 2I = \int_1^2 1 \, d x \)
Now, integrate with respect to \( x \):
\( 2I = [x]_1^2 \)
\( 2I = 2 - 1 \)
\( 2I = 1 \)
Divide by 2:
\( I = \frac{1}{2} \).
In simple words: This integral is solved using a special property for definite integrals where we replace \( x \) with \( (a+b-x) \). When we add the original integral and this new one, the top and bottom parts of the fraction become exactly the same, simplifying the whole expression to 1. Integrating 1 gives \( x \), and applying the limits gives \( 1 \), so the final answer for the integral is \( 1/2 \).

🎯 Exam Tip: This type of integral where \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \) is a common application of properties of definite integrals. When the integrand is of the form \( \frac{g(x)}{g(x)+g(a+b-x)} \), the result is often \( \frac{b-a}{2} \). Here, \( (2-1)/2 = 1/2 \).

Question 38. \( \int \frac{1}{x[6(\log x)^2+7 \log x+2]} d x \)
Answer: Let the integral be \( I = \int \frac{1}{x[6(\log x)^2+7 \log x+2]} d x \).
Let \( t = \log x \).
To find \( dt \), differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{1}{x} \).
So, \( \frac{1}{x} d x = d t \).
Substitute these into the integral:
\( I = \int \frac{1}{6t^2+7t+2} d t \).
Now, factor the quadratic denominator \( 6t^2+7t+2 \).
We look for two numbers that multiply to \( 6 \cdot 2 = 12 \) and add to 7. These numbers are 3 and 4.
\( 6t^2+7t+2 = 6t^2+3t+4t+2 = 3t(2t+1) + 2(2t+1) = (3t+2)(2t+1) \).
So, \( I = \int \frac{1}{(3t+2)(2t+1)} d t \).
Use partial fraction decomposition: \( \frac{1}{(3t+2)(2t+1)} = \frac{A}{3t+2} + \frac{B}{2t+1} \).
Multiply by \( (3t+2)(2t+1) \): \( 1 = A(2t+1) + B(3t+2) \).
To find A, set \( 2t+1 = 0 \implies t = -1/2 \):
\( 1 = A(0) + B(3(-1/2)+2) = B(-3/2+4/2) = B(1/2) \implies B=2 \).
To find B, set \( 3t+2 = 0 \implies t = -2/3 \):
\( 1 = A(2(-2/3)+1) + B(0) = A(-4/3+3/3) = A(-1/3) \implies A=-3 \).
So, \( \frac{1}{(3t+2)(2t+1)} = \frac{-3}{3t+2} + \frac{2}{2t+1} \).
Substitute back into the integral:
\( I = \int \left( \frac{-3}{3t+2} + \frac{2}{2t+1} \right) d t \)
\( I = -3 \int \frac{1}{3t+2} d t + 2 \int \frac{1}{2t+1} d t \)
\( I = -3 \cdot \frac{\log|3t+2|}{3} + 2 \cdot \frac{\log|2t+1|}{2} + C \)
\( I = -\log|3t+2| + \log|2t+1| + C \)
Using logarithm property \( \log A - \log B = \log(A/B) \):
\( I = \log \left| \frac{2t+1}{3t+2} \right| + C \).
Finally, substitute back \( t = \log x \):
\( I = \log \left| \frac{2 \log x + 1}{3 \log x + 2} \right| + C \).
In simple words: We let \( \log x \) be \( t \). This changes \( \frac{1}{x} \, dx \) to \( dt \). The integral then has a quadratic expression in the bottom, which we factorize. We use a method called partial fractions to split the complex fraction into two simpler ones. Then we integrate each part, which gives us logarithm terms. Finally, we replace \( t \) with \( \log x \).

🎯 Exam Tip: When \( \log x \) appears in a quadratic form in the denominator and \( 1/x \) is in the numerator, substitution \( t = \log x \) is ideal. If the resulting rational function has a factorable quadratic denominator, use partial fractions. Always remember the absolute value in logarithm terms and to re-substitute the original variable at the end.

Question 39. \( \int \frac{1}{x+\sqrt{x}} d x \)
Answer: Let the integral be \( I = \int \frac{1}{x+\sqrt{x}} d x \).
We can factor out \( \sqrt{x} \) from the denominator:
\( I = \int \frac{1}{\sqrt{x}(\sqrt{x}+1)} d x \).
Now, let \( t = \sqrt{x} + 1 \).
Then \( \frac{dt}{dx} = \frac{d}{dx}(\sqrt{x}+1) = \frac{1}{2\sqrt{x}} \).
So, \( \frac{1}{\sqrt{x}} d x = 2 \, d t \).
Substitute these into the integral:
\( I = \int \frac{1}{t} \cdot 2 \, d t \)
\( I = 2 \int \frac{1}{t} d t \).
Now, integrate with respect to \( t \):
\( I = 2 \log |t| + C \).
Finally, substitute back \( t = \sqrt{x} + 1 \):
\( I = 2 \log |\sqrt{x} + 1| + C \).
In simple words: We first take \( \sqrt{x} \) common from the bottom part of the fraction. Then, we let the term \( (\sqrt{x}+1) \) be \( t \). This makes \( \frac{1}{\sqrt{x}} \, dx \) turn into \( 2 \, dt \). The integral becomes a simple \( \int \frac{1}{t} \, dt \), which gives \( \log|t| \). At the end, we put back \( (\sqrt{x}+1) \) for \( t \).

🎯 Exam Tip: When \( \sqrt{x} \) terms are present along with \( x \) (which is \( (\sqrt{x})^2 \)), a substitution involving \( \sqrt{x} \) (like \( t = \sqrt{x} \) or \( t = \sqrt{x}+1 \)) is often very effective. This converts the integral into a rational function which is usually easier to solve.

Question 40. Evaluate \( \int_0^1 \log \left(\frac{1}{x} - 1\right) d x \)
Answer:
Let \( I = \int_0^1 \log \left(\frac{1}{x} - 1\right) d x \) ...(1)
We use the property \( \int_0^a f(x) d x = \int_0^a f(a-x) d x \).
So, \( I = \int_0^1 \log \left(\frac{1}{1-x} - 1\right) d x \)
\( \implies I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) d x \)
\( \implies I = \int_0^1 \log \left(\frac{x}{1-x}\right) d x \) ...(2)
Now, add equation (1) and equation (2):
\( 2I = \int_0^1 \left[\log \left(\frac{1}{x} - 1\right) + \log \left(\frac{x}{1-x}\right)\right] d x \)
We use the logarithm property \( \log a + \log b = \log(ab) \):
\( \implies 2I = \int_0^1 \log \left[\left(\frac{1-x}{x}\right) \cdot \left(\frac{x}{1-x}\right)\right] d x \)
\( \implies 2I = \int_0^1 \log (1) d x \)
Since \( \log(1) = 0 \):
\( \implies 2I = \int_0^1 0 d x \)
\( \implies 2I = 0 \)
\( \implies I = 0 \)
In simple words: First, we use a special property for integrals to change the 'x' in the log term. Then, we add the original integral and the new one. This makes the expression inside the log become 1, and since log(1) is 0, the whole integral becomes 0.

🎯 Exam Tip: This type of problem frequently uses the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). Look for opportunities to simplify the integrand when adding the original and modified integrals.

Question 41. Evaluate \( \int \frac{\cos^{-1} x}{x^2} d x \)
Answer:
Let \( I = \int \cos^{-1} x \cdot \frac{1}{x^2} d x \)
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \cos^{-1} x \) and \( dv = \frac{1}{x^2} d x \).
Then \( du = -\frac{1}{\sqrt{1-x^2}} d x \) and \( v = -\frac{1}{x} \).
\( I = \cos^{-1} x \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(-\frac{1}{\sqrt{1-x^2}}\right) d x \)
\( I = -\frac{\cos^{-1} x}{x} - \int \frac{1}{x\sqrt{1-x^2}} d x \)
To solve \( \int \frac{1}{x\sqrt{1-x^2}} d x \), substitute \( x = \frac{1}{t} \).
Then \( d x = -\frac{1}{t^2} d t \).
\( \int \frac{1}{\frac{1}{t}\sqrt{1-\frac{1}{t^2}}} \left(-\frac{1}{t^2}\right) d t = \int \frac{t}{\sqrt{\frac{t^2-1}{t^2}}} \left(-\frac{1}{t^2}\right) d t \)
\( = \int \frac{t^2}{\sqrt{t^2-1}} \left(-\frac{1}{t^2}\right) d t = -\int \frac{1}{\sqrt{t^2-1}} d t \)
We know that \( \int \frac{1}{\sqrt{x^2-a^2}} d x = \log |x + \sqrt{x^2-a^2}| + C \).
\( = -\log |t + \sqrt{t^2-1}| + C \)
Substitute back \( t = \frac{1}{x} \):
\( = -\log \left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| + C \)
\( = -\log \left|\frac{1}{x} + \frac{\sqrt{1-x^2}}{x}\right| + C \)
\( = -\log \left|\frac{1+\sqrt{1-x^2}}{x}\right| + C \)
So, \( I = -\frac{\cos^{-1} x}{x} - \left(-\log \left|\frac{1+\sqrt{1-x^2}}{x}\right|\right) + C \)
\( I = -\frac{\cos^{-1} x}{x} + \log \left|\frac{1+\sqrt{1-x^2}}{x}\right| + C \)
In simple words: We solve this using integration by parts. First, we break the integral into two parts. The second part requires another substitution to solve, where we replace x with 1/t to simplify the square root. After integrating that part, we substitute everything back to get the final answer.

🎯 Exam Tip: When using integration by parts, choose 'u' (the part to differentiate) to simplify and 'dv' (the part to integrate) to be easily integrable. For inverse trigonometric functions, they are usually chosen as 'u'.

Question 42. Evaluate \( \int e^x \frac{(2+\sin 2 x)}{\cos^2 x} d x \)
Answer:
Let \( I = \int e^x \frac{(2+\sin 2 x)}{\cos^2 x} d x \)
We can split the fraction:
\( I = \int e^x \left(\frac{2}{\cos^2 x} + \frac{\sin 2 x}{\cos^2 x}\right) d x \)
We know that \( \frac{1}{\cos^2 x} = \sec^2 x \) and \( \sin 2 x = 2 \sin x \cos x \):
\( I = \int e^x \left(2 \sec^2 x + \frac{2 \sin x \cos x}{\cos^2 x}\right) d x \)
\( I = \int e^x (2 \sec^2 x + 2 \tan x) d x \)
\( I = \int (2 e^x \sec^2 x + 2 e^x \tan x) d x \)
We know the formula: \( \int e^x [f(x) + f'(x)] d x = e^x f(x) + C \).
Let \( f(x) = 2 \tan x \). Then \( f'(x) = 2 \sec^2 x \).
So, the integral matches this form:
\( I = e^x (2 \tan x) + C \)
\( I = 2 e^x \tan x + C \)
In simple words: We first break the complex fraction into two simpler parts. Then, we rewrite the terms using trigonometric identities to get \( \sec^2 x \) and \( \tan x \). The integral now looks like \( e^x \) multiplied by a function plus its derivative, which is a common integration pattern that simplifies to \( e^x \) times the function.

🎯 Exam Tip: Always look for the pattern \( \int e^x [f(x) + f'(x)] dx \). If the integral doesn't immediately appear in this form, try to algebraically manipulate the integrand using trigonometric identities to fit it into this useful form.

Question 43. Using properties of definite integrals, evaluate \( \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \)
Answer:
Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \) ...(1)
Using the property \( \int_0^a f(x) d x = \int_0^a f(a-x) d x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \)
Since \( \sin \left(\frac{\pi}{2}-x\right) = \cos x \) and \( \cos \left(\frac{\pi}{2}-x\right) = \sin x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \) ...(2)
Now, add equation (1) and equation (2):
\( 2I = \int_0^{\frac{\pi}{2}} \left(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right) d x \)
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 d x \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( \implies I = \frac{\pi}{4} \)
In simple words: We use a special rule for definite integrals that changes x to (upper limit - x). When we apply this rule, the sine becomes cosine and cosine becomes sine. Then, we add the original integral and the changed integral together. This makes the top and bottom of the fraction the same, simplifying the integral to just 1. Integrating 1 gives us the length of the interval, which is then divided by 2.

🎯 Exam Tip: This is a classic application of the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). In such problems, the result is almost always \( \frac{a}{2} \). Quickly identify if the sum of the original and transformed integrands simplifies to 1.

Question 44. Evaluate \( \int \frac{x+\sin x}{1+\cos x} d x \)
Answer:
Let \( I = \int \frac{x+\sin x}{1+\cos x} d x \)
We split the integrand into two parts:
\( I = \int \left(\frac{x}{1+\cos x} + \frac{\sin x}{1+\cos x}\right) d x \)
Using half-angle identities: \( 1+\cos x = 2 \cos^2 \frac{x}{2} \) and \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \):
\( I = \int \left(\frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}}\right) d x \)
\( I = \int \left(\frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}\right) d x \)
Now, we integrate each part separately.
For \( \int \frac{x}{2} \sec^2 \frac{x}{2} d x \), we use integration by parts \( \int u dv = uv - \int v du \).
Let \( u = x \) and \( dv = \frac{1}{2} \sec^2 \frac{x}{2} d x \).
Then \( du = d x \) and \( v = \tan \frac{x}{2} \).
So, \( \int \frac{x}{2} \sec^2 \frac{x}{2} d x = x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x \)
Substitute this back into the main integral:
\( I = \left(x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x\right) + \int \tan \frac{x}{2} d x + C \)
\( I = x \tan \frac{x}{2} + C \)
In simple words: First, we break the fraction into two parts. Then, we use special angle formulas for sine and cosine to simplify the terms. After simplification, we notice that one part of the integral can be solved using integration by parts, and when we apply it, one of the terms cancels out the other part of the integral, leaving us with a simple final answer.

🎯 Exam Tip: When you see \( 1+\cos x \) or \( 1-\cos x \) in the denominator, immediately think of half-angle formulas to simplify the expression. Also, look for terms that cancel out after applying integration techniques.

Question 45. Evaluate \( \int \frac{2 y^2}{y^2+4} d y \)
Answer:
Let \( I = \int \frac{2 y^2}{y^2+4} d y \)
We can rewrite the numerator to match the denominator:
\( 2 y^2 = 2 y^2 + 8 - 8 = 2(y^2+4) - 8 \)
Substitute this back into the integral:
\( I = \int \frac{2(y^2+4) - 8}{y^2+4} d y \)
\( I = \int \left(\frac{2(y^2+4)}{y^2+4} - \frac{8}{y^2+4}\right) d y \)
\( I = \int \left(2 - \frac{8}{y^2+4}\right) d y \)
Now, we integrate each term:
\( I = 2 \int d y - 8 \int \frac{1}{y^2+2^2} d y \)
We know that \( \int \frac{1}{x^2+a^2} d x = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \). Here \( a=2 \).
\( I = 2y - 8 \left(\frac{1}{2} \tan^{-1} \left(\frac{y}{2}\right)\right) + C \)
\( I = 2y - 4 \tan^{-1} \left(\frac{y}{2}\right) + C \)
In simple words: We modify the top part of the fraction so it includes the bottom part. This lets us split the fraction into two simpler terms. One term becomes a constant, which is easy to integrate. The other term is a standard integral form involving \( \tan^{-1} \).

🎯 Exam Tip: When the degree of the numerator is equal to or greater than the degree of the denominator, always try to perform algebraic manipulation or long division to simplify the integrand before integrating.

Question 46. Evaluate \( \int_0^3 f(x) d x \) where \( f(x) \) is defined as: \( f(x) = \begin{cases} \cos 2x, & 0 \le x < \frac{\pi}{2} \\ 3, & \frac{\pi}{2} \le x \le 3 \end{cases} \)
Answer:
For a piecewise function, we split the integral at the points where the definition of \( f(x) \) changes.
\( \int_0^3 f(x) d x = \int_0^{\frac{\pi}{2}} \cos 2x d x + \int_{\frac{\pi}{2}}^3 3 d x \)
Integrate the first part:
\( \int_0^{\frac{\pi}{2}} \cos 2x d x = \left[\frac{\sin 2x}{2}\right]_0^{\frac{\pi}{2}} \)
\( = \frac{\sin(2 \cdot \frac{\pi}{2})}{2} - \frac{\sin(2 \cdot 0)}{2} \)
\( = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = \frac{0}{2} - \frac{0}{2} = 0 \)
Integrate the second part:
\( \int_{\frac{\pi}{2}}^3 3 d x = [3x]_{\frac{\pi}{2}}^3 \)
\( = 3(3) - 3\left(\frac{\pi}{2}\right) = 9 - \frac{3\pi}{2} \)
Add both results:
\( \int_0^3 f(x) d x = 0 + \left(9 - \frac{3\pi}{2}\right) = 9 - \frac{3\pi}{2} \)
In simple words: Since the function changes its rule at \( \pi/2 \), we break the main integral into two smaller integrals, one for each rule. We solve each small integral separately using its own limits. Then, we add the results from both parts to get the total value for the original integral.

🎯 Exam Tip: When integrating piecewise functions, correctly identify the limits for each piece. Remember that \( \pi/2 \) is approximately 1.57, so \( 0 \le x < \pi/2 \) and \( \pi/2 \le x \le 3 \) are the correct intervals.

Question 47. Evaluate \( \int \frac{\sec x}{1+\csc x} d x \)
Answer:
Let \( I = \int \frac{\sec x}{1+\csc x} d x \)
Convert \( \sec x \) and \( \csc x \) to \( \sin x \) and \( \cos x \):
\( I = \int \frac{\frac{1}{\cos x}}{1+\frac{1}{\sin x}} d x = \int \frac{\frac{1}{\cos x}}{\frac{\sin x+1}{\sin x}} d x = \int \frac{\sin x}{\cos x (1+\sin x)} d x \)
To make a substitution easier, we can multiply the numerator and denominator by \( \cos x \):
\( I = \int \frac{\sin x \cos x}{\cos^2 x (1+\sin x)} d x \)
Let \( t = \sin x \). Then \( d t = \cos x d x \).
Also, \( \cos^2 x = 1-\sin^2 x = 1-t^2 \).
\( I = \int \frac{t}{(1-t^2)(1+t)} d t = \int \frac{t}{(1-t)(1+t)(1+t)} d t = \int \frac{t}{(1-t)(1+t)^2} d t \)
Now, we use partial fractions decomposition:
\( \frac{t}{(1-t)(1+t)^2} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{(1+t)^2} \)
Multiply by \( (1-t)(1+t)^2 \):
\( t = A(1+t)^2 + B(1-t)(1+t) + C(1-t) \)
If \( t=1 \): \( 1 = A(1+1)^2 + B(0) + C(0) \implies 1 = 4A \implies A = \frac{1}{4} \)
If \( t=-1 \): \( -1 = A(0) + B(0) + C(1-(-1)) \implies -1 = 2C \implies C = -\frac{1}{2} \)
Compare coefficients of \( t^2 \):
\( 0 = A - B \implies B = A = \frac{1}{4} \)
So, the integral becomes:
\( I = \int \left(\frac{\frac{1}{4}}{1-t} + \frac{\frac{1}{4}}{1+t} + \frac{-\frac{1}{2}}{(1+t)^2}\right) d t \)
\( I = \frac{1}{4} \int \frac{1}{1-t} d t + \frac{1}{4} \int \frac{1}{1+t} d t - \frac{1}{2} \int (1+t)^{-2} d t \)
\( I = -\frac{1}{4} \log |1-t| + \frac{1}{4} \log |1+t| - \frac{1}{2} \frac{(1+t)^{-1}}{-1} + C \)
\( I = \frac{1}{4} \log \left|\frac{1+t}{1-t}\right| + \frac{1}{2(1+t)} + C \)
Substitute back \( t = \sin x \):
\( I = \frac{1}{4} \log \left|\frac{1+\sin x}{1-\sin x}\right| + \frac{1}{2(1+\sin x)} + C \)
In simple words: We first change all secant and cosecant terms to sine and cosine to simplify the expression. Then, we use a substitution for \( \sin x \) and apply partial fractions to break the complex fraction into simpler parts. Finally, we integrate each part and substitute back \( \sin x \) to get the solution.

🎯 Exam Tip: For integrals with mixed trigonometric functions, converting all terms to sine and cosine is a good first step. If the denominator becomes a product, partial fractions is often necessary. Remember to handle the negative sign for \( \int \frac{1}{1-t} dt \).

Question 48. Evaluate \( \int \tan^3 x d x \)
Answer:
Let \( I = \int \tan^3 x d x \)
We can rewrite \( \tan^3 x \) as \( \tan x \cdot \tan^2 x \).
Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\( I = \int \tan x (\sec^2 x - 1) d x \)
\( I = \int (\tan x \sec^2 x - \tan x) d x \)
Now, we split the integral into two parts:
\( I = \int \tan x \sec^2 x d x - \int \tan x d x \)
For the first integral, let \( t = \tan x \). Then \( d t = \sec^2 x d x \).
So, \( \int \tan x \sec^2 x d x = \int t d t = \frac{t^2}{2} + C_1 = \frac{\tan^2 x}{2} + C_1 \).
For the second integral, we know \( \int \tan x d x = \log |\sec x| + C_2 \) or \( -\log |\cos x| + C_2 \).
So, \( I = \frac{\tan^2 x}{2} - (-\log |\cos x|) + C \)
\( I = \frac{\tan^2 x}{2} + \log |\cos x| + C \)
In simple words: We start by breaking down \( \tan^3 x \) into \( \tan x \) and \( \tan^2 x \). Then, we use a trigonometric identity to change \( \tan^2 x \) into \( \sec^2 x - 1 \). This allows us to split the integral into two easier parts. One part is solved using a simple substitution, and the other is a standard integral.

🎯 Exam Tip: For powers of tangent, the key is to separate a \( \tan^2 x \) term and replace it with \( \sec^2 x - 1 \). This often creates a substitution opportunity for \( \tan x \) and \( \sec^2 x \).

Question 49. Evaluate \( \int \frac{\sin x+\cos x}{\sqrt{9+16 \sin 2 x}} d x \)
Answer:
Let \( I = \int \frac{\sin x+\cos x}{\sqrt{9+16 \sin 2 x}} d x \)
Let's try a substitution involving \( \sin x - \cos x \).
Let \( t = \sin x - \cos x \).
Then \( d t = (\cos x + \sin x) d x \).
Square both sides of the substitution:
\( t^2 = (\sin x - \cos x)^2 \)
\( t^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x \)
\( t^2 = 1 - \sin 2x \)
So, \( \sin 2x = 1 - t^2 \).
Substitute \( dt \) and \( \sin 2x \) into the integral:
\( I = \int \frac{d t}{\sqrt{9+16(1-t^2)}} \)
\( I = \int \frac{d t}{\sqrt{9+16-16t^2}} \)
\( I = \int \frac{d t}{\sqrt{25-16t^2}} \)
Factor out 16 from the square root:
\( I = \int \frac{d t}{\sqrt{16\left(\frac{25}{16}-t^2\right)}} = \int \frac{d t}{4\sqrt{\left(\frac{5}{4}\right)^2-t^2}} \)
\( I = \frac{1}{4} \int \frac{d t}{\sqrt{\left(\frac{5}{4}\right)^2-t^2}} \)
Using the formula \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C \). Here \( a = \frac{5}{4} \).
\( I = \frac{1}{4} \sin^{-1} \left(\frac{t}{\frac{5}{4}}\right) + C \)
\( I = \frac{1}{4} \sin^{-1} \left(\frac{4t}{5}\right) + C \)
Substitute back \( t = \sin x - \cos x \):
\( I = \frac{1}{4} \sin^{-1} \left(\frac{4(\sin x - \cos x)}{5}\right) + C \)
In simple words: This problem looks tricky, but we can simplify it by making a smart substitution. We let \( t \) be \( \sin x - \cos x \). This helps us replace both \( \sin x + \cos x \) and \( \sin 2x \) in the integral with terms involving \( t \). After simplification, it becomes a standard integral that we solve using the inverse sine formula.

🎯 Exam Tip: When you see \( (\sin x + \cos x) \) or \( (\cos x - \sin x) \) alongside \( \sin 2x \) in an integral, consider the substitution \( t = \sin x \pm \cos x \). Squaring \( t \) will often lead to an expression for \( \sin 2x \).

Question 50. Using properties of definite integrals, evaluate \( \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \)
Answer:
Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \) ...(1)
Using the property \( \int_0^a f(x) d x = \int_0^a f(a-x) d x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \)
Since \( \sin \left(\frac{\pi}{2}-x\right) = \cos x \) and \( \cos \left(\frac{\pi}{2}-x\right) = \sin x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x \)
\( I = \int_0^{\frac{\pi}{2}} \frac{-(\sin x-\cos x)}{1+\sin x \cos x} d x \)
\( I = -\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \) ...(2)
Notice that equation (2) is \( I = -I \).
Add equation (1) and equation (2):
\( 2I = \int_0^{\frac{\pi}{2}} \left(\frac{\sin x-\cos x}{1+\sin x \cos x} + \frac{-(\sin x-\cos x)}{1+\sin x \cos x}\right) d x \)
\( 2I = \int_0^{\frac{\pi}{2}} 0 d x \)
\( 2I = 0 \)
\( \implies I = 0 \)
In simple words: We use a special rule for definite integrals that replaces 'x' with 'upper limit - x'. When we do this, the sine and cosine terms swap, making the numerator the negative of the original. So, the whole integral becomes its own negative. If an integral equals its negative, then its value must be zero.

🎯 Exam Tip: This is a powerful application of the definite integral property. If, after applying \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), you find that \( f(a-x) = -f(x) \), then the integral's value is 0. This saves a lot of calculation time.

Question 51. Evaluate \( \int(1 - x) \sqrt{x} d x \)
Answer:
Let \( I = \int(1 - x) \sqrt{x} d x \)
First, distribute \( \sqrt{x} \) into the parenthesis:
\( I = \int (\sqrt{x} - x\sqrt{x}) d x \)
Rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \) and \( x\sqrt{x} \) as \( x^1 \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}} \):
\( I = \int (x^{\frac{1}{2}} - x^{\frac{3}{2}}) d x \)
Now, integrate each term using the power rule \( \int x^n d x = \frac{x^{n+1}}{n+1} + C \):
\( I = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} - \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + C \)
\( I = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C \)
\( I = \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C \)
In simple words: We start by multiplying \( \sqrt{x} \) with each term inside the bracket. Then, we change \( \sqrt{x} \) and \( x\sqrt{x} \) into their power forms (like \( x \) to the power of 1/2). Finally, we use the simple power rule for integration on each term to get the answer.

🎯 Exam Tip: When you have terms like \( \sqrt{x} \) and \( x\sqrt{x} \), always convert them to exponential forms (e.g., \( x^{1/2} \), \( x^{3/2} \)) for easier application of the power rule of integration.

Question 52. Evaluate \( \int \cos^{-1} (\sin x) d x \)
Answer:
Let \( I = \int \cos^{-1} (\sin x) d x \)
We know that \( \sin x = \cos \left(\frac{\pi}{2}-x\right) \).
So, \( \cos^{-1} (\sin x) = \cos^{-1} \left(\cos \left(\frac{\pi}{2}-x\right)\right) \).
Since \( \cos^{-1} (\cos \theta) = \theta \) (for \( \theta \) in the principal value branch), we have:
\( \cos^{-1} (\sin x) = \frac{\pi}{2} - x \)
Now, substitute this back into the integral:
\( I = \int \left(\frac{\pi}{2} - x\right) d x \)
Integrate term by term:
\( I = \int \frac{\pi}{2} d x - \int x d x \)
\( I = \frac{\pi}{2} x - \frac{x^2}{2} + C \)
In simple words: We use a trigonometry trick to change \( \sin x \) into \( \cos(\frac{\pi}{2}-x) \). This makes the inverse cosine function very simple, as \( \cos^{-1}(\cos \theta) \) is just \( \theta \). After that, we integrate the simple expression \( \frac{\pi}{2}-x \) to get the final answer.

🎯 Exam Tip: Always look for ways to simplify inverse trigonometric functions using identities (e.g., \( \sin x = \cos(\frac{\pi}{2}-x) \), \( \tan x = \cot(\frac{\pi}{2}-x) \)). This often converts complex integrands into simple polynomials or constants.

Question 53. Evaluate \( \int \frac{1-\sin x}{\cos^2 x} d x \)
Answer:
Let \( I = \int \frac{1-\sin x}{\cos^2 x} d x \)
We can split the fraction into two terms:
\( I = \int \left(\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}\right) d x \)
We know that \( \frac{1}{\cos^2 x} = \sec^2 x \) and \( \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x \).
So, the integral becomes:
\( I = \int (\sec^2 x - \tan x \sec x) d x \)
Now, we integrate each term separately:
\( \int \sec^2 x d x = \tan x + C_1 \)
\( \int \tan x \sec x d x = \sec x + C_2 \)
Combining these, we get:
\( I = \tan x - \sec x + C \)
In simple words: We break the fraction into two simpler parts. We know that \( 1/\cos^2 x \) is \( \sec^2 x \), and \( \sin x/\cos^2 x \) is \( \tan x \sec x \). Then we integrate each of these standard trigonometric functions to get the final answer.

🎯 Exam Tip: When the denominator is a single trigonometric term (like \( \cos^2 x \)), split the numerator terms. This often converts the integral into a combination of standard integral forms (e.g., \( \sec^2 x \), \( \tan x \sec x \)).

Question 54. Evaluate \( \int \frac{(\log x)^2}{x} d x \)
Answer:
Let \( I = \int \frac{(\log x)^2}{x} d x \)
This integral can be solved using substitution.
Let \( t = \log x \).
Then, differentiate \( t \) with respect to \( x \):
\( \frac{d t}{d x} = \frac{1}{x} \)
So, \( d t = \frac{1}{x} d x \).
Now, substitute \( t \) and \( d t \) into the integral:
\( I = \int t^2 d t \)
Integrate using the power rule \( \int t^n d t = \frac{t^{n+1}}{n+1} + C \):
\( I = \frac{t^{2+1}}{2+1} + C \)
\( I = \frac{t^3}{3} + C \)
Finally, substitute back \( t = \log x \):
\( I = \frac{(\log x)^3}{3} + C \)
In simple words: We notice that the derivative of \( \log x \) is \( 1/x \), which is also present in the integral. So, we let \( t = \log x \). This transforms the integral into a simple power of \( t \). After integrating, we replace \( t \) back with \( \log x \) to get the final answer.

🎯 Exam Tip: Always look for a function and its derivative in the integrand. If present, a substitution of the function for a new variable 't' will often simplify the integral significantly.

Question 55. Evaluate \( \int \frac{\sin^6 x}{\cos^8 x} d x \)
Answer:
Let \( I = \int \frac{\sin^6 x}{\cos^8 x} d x \)
We can rewrite the integrand using trigonometric identities to make it suitable for substitution.
\( I = \int \frac{\sin^6 x}{\cos^6 x \cdot \cos^2 x} d x \)
\( I = \int \left(\frac{\sin x}{\cos x}\right)^6 \cdot \frac{1}{\cos^2 x} d x \)
We know that \( \frac{\sin x}{\cos x} = \tan x \) and \( \frac{1}{\cos^2 x} = \sec^2 x \).
So, the integral becomes:
\( I = \int \tan^6 x \sec^2 x d x \)
Now, we can use a substitution.
Let \( t = \tan x \).
Then, differentiate \( t \) with respect to \( x \):
\( \frac{d t}{d x} = \sec^2 x \)
So, \( d t = \sec^2 x d x \).
Substitute \( t \) and \( d t \) into the integral:
\( I = \int t^6 d t \)
Integrate using the power rule \( \int t^n d t = \frac{t^{n+1}}{n+1} + C \):
\( I = \frac{t^{6+1}}{6+1} + C \)
\( I = \frac{t^7}{7} + C \)
Finally, substitute back \( t = \tan x \):
\( I = \frac{\tan^7 x}{7} + C \)
In simple words: We rewrite the fraction by grouping \( \sin^6 x \) with \( \cos^6 x \) to form \( \tan^6 x \), leaving \( 1/\cos^2 x \) which is \( \sec^2 x \). Then, we use substitution, letting \( t \) be \( \tan x \), because its derivative \( \sec^2 x \) is also in the integral. This simplifies the integral to a simple power of \( t \), which is easy to solve.

🎯 Exam Tip: For integrals involving powers of sine and cosine, especially when the powers are high, try to convert the expression into terms of \( \tan x \) and \( \sec^2 x \) to enable a simple substitution.

Question 56. Evaluate \( \int x \sec^2 x d x \)
Answer:
Let \( I = \int x \sec^2 x d x \)
We will use integration by parts, which is given by the formula: \( \int u dv = uv - \int v du \).
We need to choose \( u \) and \( dv \). A common mnemonic is L.I.A.T.E. (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
Here, \( x \) is algebraic (A) and \( \sec^2 x \) is trigonometric (T). So, we choose \( u = x \) and \( dv = \sec^2 x d x \).
Differentiate \( u \) to find \( du \):
\( du = d x \)
Integrate \( dv \) to find \( v \):
\( v = \int \sec^2 x d x = \tan x \)
Now, apply the integration by parts formula:
\( I = x \cdot \tan x - \int \tan x d x \)
We know that \( \int \tan x d x = \log |\sec x| + C \) (or \( -\log |\cos x| + C \)).
So,
\( I = x \tan x - (-\log |\cos x|) + C \)
\( I = x \tan x + \log |\cos x| + C \)
In simple words: We solve this using a method called "integration by parts". We pick \( x \) to be differentiated and \( \sec^2 x \) to be integrated. After applying the formula, we are left with a simpler integral of \( \tan x \), which we know how to solve directly.

🎯 Exam Tip: For integrals of the form (algebraic function) \( \times \) (trigonometric function), integration by parts is usually the method to use. Remember the ILATE rule to choose 'u' and 'dv' effectively.

Question 57. Evaluate \( \int \sin^{-1} x d x \)
Answer:
Let \( I = \int \sin^{-1} x d x \)
This integral can be solved using integration by parts. We can write it as \( \int \sin^{-1} x \cdot 1 d x \).
Let \( u = \sin^{-1} x \) and \( dv = 1 d x \).
Then, differentiate \( u \) to find \( du \):
\( du = \frac{1}{\sqrt{1-x^2}} d x \)
Integrate \( dv \) to find \( v \):
\( v = \int 1 d x = x \)
Apply the integration by parts formula \( \int u dv = uv - \int v du \):
\( I = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1-x^2}} d x \)
Now, we need to solve the integral \( \int \frac{x}{\sqrt{1-x^2}} d x \). We use substitution for this.
Let \( w = 1-x^2 \). Then \( d w = -2x d x \implies x d x = -\frac{1}{2} d w \).
So, \( \int \frac{x}{\sqrt{1-x^2}} d x = \int \frac{-\frac{1}{2} d w}{\sqrt{w}} = -\frac{1}{2} \int w^{-\frac{1}{2}} d w \)
\( = -\frac{1}{2} \cdot \frac{w^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = -w^{\frac{1}{2}} + C_1 = -\sqrt{1-x^2} + C_1 \).
Substitute this back into the main integral \( I \):
\( I = x \sin^{-1} x - (-\sqrt{1-x^2}) + C \)
\( I = x \sin^{-1} x + \sqrt{1-x^2} + C \)
In simple words: We solve this by using integration by parts. We treat \( \sin^{-1} x \) as one part and \( 1 \) as the other. After applying the formula, we are left with a new integral that can be solved using a simple substitution, letting a new variable equal \( 1-x^2 \). Finally, we combine the results to get the full answer.

🎯 Exam Tip: For inverse trigonometric functions like \( \sin^{-1} x \), \( \cos^{-1} x \), \( \tan^{-1} x \), always use integration by parts with \( 1 \) as the second function (dv). The resulting integral will often require another substitution.

Question 58. Evaluate \( \int_{-\pi}^\pi \sin^3 x \cos^2 x d x \)
Answer:
Let \( I = \int_{-\pi}^\pi \sin^3 x \cos^2 x d x \).
We need to check if the integrand \( f(x) = \sin^3 x \cos^2 x \) is an odd or even function.
A function \( f(x) \) is odd if \( f(-x) = -f(x) \).
A function \( f(x) \) is even if \( f(-x) = f(x) \).
Let's evaluate \( f(-x) \):
\( f(-x) = \sin^3 (-x) \cos^2 (-x) \)
We know that \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \).
\( f(-x) = (-\sin x)^3 (\cos x)^2 \)
\( f(-x) = -\sin^3 x \cos^2 x \)
Since \( f(-x) = -\sin^3 x \cos^2 x = -f(x) \), the function \( f(x) \) is an odd function.
For a definite integral with symmetric limits, \( \int_{-a}^a f(x) d x \):
If \( f(x) \) is an odd function, then \( \int_{-a}^a f(x) d x = 0 \).
Since our limits are from \( -\pi \) to \( \pi \), and \( f(x) \) is odd, the value of the integral is 0.
\( I = 0 \)
In simple words: We first check if the function inside the integral is odd or even. An odd function is one where \( f(-x) \) is the same as \( -f(x) \). In this case, \( \sin^3 x \cos^2 x \) is an odd function. Since the integral goes from \( -\pi \) to \( \pi \), for any odd function over such symmetric limits, the total area under the curve cancels out, so the integral is zero.

🎯 Exam Tip: Always test for odd or even symmetry when definite integrals have symmetric limits (e.g., \( [-a, a] \)). If the function is odd, the integral is zero. If it's even, you can simplify the integral to \( 2 \int_0^a f(x) dx \).

Question 59. Evaluate \( \int \frac{x^4+1}{x^2+1} d x \)
Answer:
Let \( I = \int \frac{x^4+1}{x^2+1} d x \)
Since the degree of the numerator is greater than the degree of the denominator, we can perform polynomial division or algebraic manipulation.
We can rewrite the numerator as \( (x^2)^2 - 1 + 2 \), which is \( (x^2-1)(x^2+1)+2 \):
\( I = \int \frac{(x^2-1)(x^2+1)+2}{x^2+1} d x \)
Now, split the fraction:
\( I = \int \left(\frac{(x^2-1)(x^2+1)}{x^2+1} + \frac{2}{x^2+1}\right) d x \)
\( I = \int \left(x^2-1 + \frac{2}{x^2+1}\right) d x \)
Now, integrate each term separately:
\( I = \int x^2 d x - \int 1 d x + \int \frac{2}{x^2+1} d x \)
\( I = \frac{x^{2+1}}{2+1} - x + 2 \int \frac{1}{x^2+1^2} d x \)
We know that \( \int \frac{1}{x^2+a^2} d x = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \). Here \( a=1 \).
\( I = \frac{x^3}{3} - x + 2 \left(\frac{1}{1} \tan^{-1} \left(\frac{x}{1}\right)\right) + C \)
\( I = \frac{x^3}{3} - x + 2 \tan^{-1} x + C \)
In simple words: We start by changing the top part of the fraction. We use a trick to make it easier to divide by the bottom part. This splits the integral into simpler parts that are easy to solve. One part is a basic power rule integral, and the other is a standard integral that gives \( \tan^{-1} x \).

🎯 Exam Tip: For rational functions, if the degree of the numerator is greater than or equal to the degree of the denominator, always perform polynomial division or algebraic manipulation (like adding and subtracting terms) to simplify the integrand into integrable forms.

Question 60. The value of the integral \( \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \) is
Answer:
Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \) ...(1)
Using the property \( \int_0^a f(x) d x = \int_0^a f(a-x) d x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \)
Since \( \sin \left(\frac{\pi}{2}-x\right) = \cos x \) and \( \cos \left(\frac{\pi}{2}-x\right) = \sin x \):
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \) ...(2)
Now, add equation (1) and equation (2):
\( 2I = \int_0^{\frac{\pi}{2}} \left(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right) d x \)
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 d x \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( \implies I = \frac{\pi}{4} \)
In simple words: We use a special rule for definite integrals that changes 'x' to 'upper limit minus x'. When we apply this, sine and cosine terms swap. Then, by adding the original integral and the changed integral, the numerator becomes the same as the denominator, making the fraction equal to 1. Integrating 1 over the limits gives us \( \pi/2 \), so the final integral value is \( (\pi/2)/2 \).

🎯 Exam Tip: This is a standard integral problem. Remember the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). For integrals of this specific form, the result is almost always \( \frac{a}{2} \).

Question 61. \( \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 d x \) is equal to
(a) \( \frac{x^2}{2} + 2 x + \log |x|+ c \)
(b) \( \frac{x^2}{2} + 2 + \log |x| + c \)
(c) \( \frac{x^2}{2} + x + \log |x| + c \)
(d) \( \frac{x^2}{2} + 2 x + 2|\log | + C \)
Answer: (a) \( \frac{x^2}{2} + 2 x + \log |x|+ c \)
In simple words: To solve this, first expand the square term. This gives \( x + \frac{1}{x} + 2 \). Then, integrate each part separately. The integral of \( x \) is \( \frac{x^2}{2} \), the integral of \( \frac{1}{x} \) is \( \log |x| \), and the integral of \( 2 \) is \( 2x \). Don't forget to add the integration constant \( C \) at the end.

🎯 Exam Tip: Always expand algebraic expressions before integrating if possible, as it often simplifies the process into basic power and logarithmic rules.

Question 62. \( \int \frac{\sec x}{\sec x+\tan x} d x \)
(a) \( \sec x – \tan x + c \)
(b) \( \log \sin x + \log \cos x + c \)
(c) \( \tan x – \sec x + c \)
(d) \( \log (1 + \sin x) + c \)
Answer: (c) \( \tan x – \sec x + c \)
In simple words: To integrate this, multiply the top and bottom of the fraction by \( (\sec x - \tan x) \). This helps simplify the denominator using the identity \( \sec^2 x - \tan^2 x = 1 \). Then, you will integrate \( \sec^2 x \) and \( \sec x \tan x \) separately. Remember, the integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \sec x \tan x \) is \( \sec x \).

🎯 Exam Tip: When the integrand involves \( \sec x \) and \( \tan x \) in the denominator, multiplying by the conjugate can often simplify the expression, making it easier to integrate.

Question 63. \( \int \frac{d x}{e^x+e^{-x}+2} d x \) is equal to
(a) \( \frac{1}{e^x+1} + c \)
(b) \( \frac{-1}{e^x+1} + c \)
(c) \( \frac{1}{e^x+1} + c \)
(d) \( \frac{1}{e^{-x}-1} + c \)
Answer: (b) \( \frac{-1}{e^x+1} + c \)
In simple words: To solve this, first rewrite \( e^{-x} \) as \( \frac{1}{e^x} \). Then, multiply the numerator and denominator by \( e^x \). This will change the integral into a simpler form. After that, use a substitution where \( t = e^x + 1 \). The integral will then become \( \int \frac{dt}{t^2} \), which gives \( -\frac{1}{t} \). Finally, replace \( t \) back with \( e^x + 1 \).

🎯 Exam Tip: For expressions involving \( e^x \) and \( e^{-x} \), try multiplying by \( e^x \) in both the numerator and denominator. This can often transform the integral into a manageable form for substitution.

Question 64. \( \int \frac{\sec ^8 x}{\operatorname{cosec} x} d x \)
(a) \( \frac{\sec ^8 x}{8} + c \)
(b) \( \frac{\sec ^7 x}{7} + c \)
(c) \( \frac{\sec ^6 x}{6} + c \)
(d) \( \frac{\sec ^9 x}{9} + c \)
Answer: (b) \( \frac{\sec ^7 x}{7} + c \)
In simple words: First, rewrite the integral using \( \sec x = \frac{1}{\cos x} \) and \( \operatorname{cosec} x = \frac{1}{\sin x} \). This simplifies the expression to \( \int \tan x \sec^7 x \, dx \). Now, use the substitution method by letting \( t = \sec x \). Then, \( dt = \sec x \tan x \, dx \). The integral becomes \( \int t^6 dt \), which is \( \frac{t^7}{7} \). Substitute back \( \sec x \) for \( t \).

🎯 Exam Tip: When dealing with powers of secant and tangent, consider the relationship \( \frac{d}{dx} (\sec x) = \sec x \tan x \). This often guides towards a suitable substitution.

Question 65. \( \int \frac{(1+x) e^x}{\sin ^2(x e^x)} d x \) is equal to
(a) \( -\cot (e^x) + c \)
(b) \( \tan (x e^x) + c \)
(c) \( \tan (e^x) + c \)
(d) \( -\cot(x e^x) + c \)
Answer: (d) \( -\cot(x e^x) + c \)
In simple words: The key here is to notice the derivative of \( x e^x \). When you differentiate \( x e^x \) using the product rule, you get \( (1+x)e^x \). So, you can use a substitution \( t = x e^x \). Then, \( dt = (1+x)e^x \, dx \). The integral changes to \( \int \frac{dt}{\sin^2 t} \), which is \( \int \operatorname{cosec}^2 t \, dt \). The integral of \( \operatorname{cosec}^2 t \) is \( -\cot t \). Finally, replace \( t \) with \( x e^x \).

🎯 Exam Tip: Always look for terms in the numerator that are the derivative of a part of the denominator or an argument of a trigonometric function. This often points directly to the correct substitution.

Question 66. \( \int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x \) is equal to
(a) \( x – \sin^{-1} x + c \)
(b) \( x - \sqrt{1-x^2} \sin^{-1} x + c \)
(c) \( x + \sqrt{1-x^2} \sin^{-1} x + c \)
(d) \( x + \sqrt{1-x^2} \sin^{-1} x + c \)
Answer: (b) \( x - \sqrt{1-x^2} \sin^{-1} x + c \)
In simple words: This integral can be solved using substitution and integration by parts. First, let \( t = \sin^{-1} x \), which means \( x = \sin t \) and \( dx = \cos t \, dt \). The integral becomes \( \int t \sin t \, dt \). Now, apply integration by parts, treating \( t \) as the first function and \( \sin t \) as the second. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). After integrating, convert the expression back to terms of \( x \) using \( t = \sin^{-1} x \) and \( \cos t = \sqrt{1-x^2} \).

🎯 Exam Tip: When an inverse trigonometric function is involved, a common strategy is to substitute it with a new variable, which often simplifies the integral to a form solvable by integration by parts.

Question 67. \( \int e^x \frac{(1-x^2)}{(1+x^2)^2} d x \) is equal to
(a) \( \frac{e^x(1-x)}{1+x^2} + C \)
(b) \( \frac{e^x}{(1+x^2)^2} + C \)
(c) \( \frac{e^x}{1+x} + C \)
(d) \( \frac{e^x}{(1+x^2)} + C \)
Answer: (d) \( \frac{e^x}{(1+x^2)} + C \)
In simple words: This integral is in the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). To use this, rewrite \( \frac{1-x^2}{(1+x^2)^2} \) as \( \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \). Here, if \( f(x) = \frac{1}{1+x^2} \), then its derivative \( f'(x) = \frac{-2x}{(1+x^2)^2} \). So the answer is simply \( e^x \frac{1}{1+x^2} + C \).

🎯 Exam Tip: Always check if an integral involving \( e^x \) can be expressed in the form \( e^x[f(x) + f'(x)] \). This shortcut saves a lot of time by avoiding integration by parts.

Question 68. \( \int_{-5}^5|x + 2| d x \) is equal to
(a) 28
(b) 29
(c) 27
(d) 30
Answer: (b) 29
In simple words: For integrals with an absolute value, you need to split the integral into parts where the expression inside the absolute value changes its sign. Here, \( |x+2| \) changes sign at \( x = -2 \). So, split the integral from \( -5 \) to \( -2 \) and from \( -2 \) to \( 5 \). For \( x < -2 \), \( x+2 \) is negative, so \( |x+2| = -(x+2) \). For \( x \ge -2 \), \( x+2 \) is positive, so \( |x+2| = x+2 \). Then, integrate each part and add the results.

🎯 Exam Tip: When evaluating definite integrals involving absolute values, always identify the points where the expression inside the absolute value becomes zero and split the integral at these points.

Question 69. \( \int_0^3[x^2] d x \), where \( [x] \) is greatest integer function
(a) 3
(b) 0
(c) 2
(d) 1
Answer: (a) 3
In simple words: The greatest integer function \( [x^2] \) gives the largest integer less than or equal to \( x^2 \). To integrate this, you need to find where \( x^2 \) crosses integer values. For \( 0 \le x < 1 \), \( 0 \le x^2 < 1 \), so \( [x^2] = 0 \). For \( 1 \le x < \sqrt{2} \) (approx 1.414), \( 1 \le x^2 < 2 \), so \( [x^2] = 1 \). For \( \sqrt{2} \le x < \sqrt{3} \) (approx 1.732), \( 2 \le x^2 < 3 \), so \( [x^2] = 2 \). For \( \sqrt{3} \le x < 2 \), \( 3 \le x^2 < 4 \), so \( [x^2] = 3 \). For \( 2 \le x < 3 \), \( 4 \le x^2 < 9 \), so \( [x^2] \) will take values 4, 5, 6, 7, 8. The question is \( \int_0^3 [x] dx \), not \( [x^2] dx \). The given solution uses \( [x] dx \). So, for \( [x] \): For \( 0 \le x < 1 \), \( [x] = 0 \). For \( 1 \le x < 2 \), \( [x] = 1 \). For \( 2 \le x < 3 \), \( [x] = 2 \). We integrate \( \int_0^1 0 \, dx + \int_1^2 1 \, dx + \int_2^3 2 \, dx \). This gives \( 0 + (2-1) + 2(3-2) = 0 + 1 + 2 = 3 \). The problem statement likely intended \( \int_0^3 [x] dx \).

🎯 Exam Tip: For greatest integer function problems, divide the interval of integration based on where the value of \( [x] \) or \( [f(x)] \) changes. This turns the integral into a sum of simpler integrals of constant functions.

Question 70. The value of \( \int_0^{\pi / 2} \frac{d x}{1+\tan x} \) is
(a) \( \frac{\pi}{2} \)
(b) 0
(c) \( \frac{\pi}{4} \)
(d) \( \frac{\pi}{8} \)
Answer: (c) \( \frac{\pi}{4} \)
In simple words: This is a standard definite integral property problem. Let the integral be \( I \). Rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \). So, \( I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx \). Use the property \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \). Here, \( a=0, b=\pi/2 \). So \( I = \int_0^{\pi/2} \frac{\cos (\pi/2 - x)}{\cos (\pi/2 - x) + \sin (\pi/2 - x)} dx = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx \). Add the two expressions for \( I \). You will get \( 2I = \int_0^{\pi/2} \frac{\cos x + \sin x}{\cos x + \sin x} dx = \int_0^{\pi/2} 1 \, dx \). This integrates to \( x \), evaluated from \( 0 \) to \( \pi/2 \), which is \( \pi/2 \). So, \( 2I = \pi/2 \), meaning \( I = \pi/4 \).

🎯 Exam Tip: When an integral of a trigonometric function has limits from \( 0 \) to \( \pi/2 \), always consider using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). This often helps simplify the integral by pairing it with its transformed version.

Question 71. \( \int x^2 e^{x^3} d x \) equals
(a) \( \frac{1}{3} e^{x^3} + C \)
(b) \( \frac{1}{3} e^{x^4} + C \)
(c) \( \frac{1}{2} e^{x^3} + C \)
(d) \( \frac{1}{2} e^{x^2} + C \)
Answer: (a) \( \frac{1}{3} e^{x^3} + C \)
In simple words: This integral can be solved using the substitution method. Notice that \( x^2 \) is related to the derivative of \( x^3 \). Let \( t = x^3 \). Then, when you differentiate \( t \) with respect to \( x \), you get \( dt = 3x^2 \, dx \). So, \( x^2 \, dx = \frac{1}{3} dt \). Substitute these into the integral: \( \int e^t \frac{1}{3} dt \). The integral of \( e^t \) is \( e^t \). So you get \( \frac{1}{3} e^t + C \). Finally, replace \( t \) with \( x^3 \).

🎯 Exam Tip: For integrals involving \( e^{f(x)} \) and \( f'(x) \), a direct substitution of \( t = f(x) \) is often the quickest way to solve, as it transforms the integral into a basic exponential form.

Question 72. \( \int_0^{\pi / 8} \tan^2(2 x) d x \) is equal to
(a) \( \frac{4-\pi}{8} \)
(b) \( \frac{4+\pi}{8} \)
(c) \( \frac{4-\pi}{4} \)
(d) \( \frac{4-\pi}{2} \)
Answer: (a) \( \frac{4-\pi}{8} \)
In simple words: To integrate \( \tan^2(2x) \), use the trigonometric identity \( \tan^2 \theta = \sec^2 \theta - 1 \). So, \( \tan^2(2x) = \sec^2(2x) - 1 \). Now, integrate \( \sec^2(2x) \) and \( -1 \) separately. The integral of \( \sec^2(2x) \) is \( \frac{\tan(2x)}{2} \), and the integral of \( -1 \) is \( -x \). Evaluate these from \( 0 \) to \( \pi/8 \). When you put the limits, \( \tan(2 \times \pi/8) = \tan(\pi/4) = 1 \), and \( \tan(0) = 0 \). Calculate the final value.

🎯 Exam Tip: Always convert \( \tan^2 x \) to \( \sec^2 x - 1 \) (or \( \cot^2 x \) to \( \operatorname{cosec}^2 x - 1 \)) before integrating, as there is no direct integration formula for \( \tan^2 x \).

Question 73. \( \int 4^x 3^x d x \) equals
(a) \( \frac{12^x}{\log 12} + c \)
(b) \( \frac{4^x}{\log 4} + c \)
(c) \( \frac{4^x 3^x}{\log 4 \log 3} + c \)
(d) \( \frac{3^x}{\log 3} + c \)
Answer: (a) \( \frac{12^x}{\log 12} + c \)
In simple words: When you multiply powers with the same exponent, you can multiply the bases. So, \( 4^x \times 3^x \) becomes \( (4 \times 3)^x \), which is \( 12^x \). Now you need to integrate \( 12^x \). The general formula for integrating \( a^x \) is \( \frac{a^x}{\log a} \). So, the integral of \( 12^x \) is \( \frac{12^x}{\log 12} \). Don't forget the constant of integration \( c \).

🎯 Exam Tip: Remember the exponent rule \( a^x b^x = (ab)^x \). This simplifies many exponential integrals into a standard form \( \int k^x dx \), which is easy to integrate using the formula \( \frac{k^x}{\log k} \).

Question 74. \( \int \frac{d x}{\sqrt{9-25 x^2}} \) equals
(a) \( \sin^{-1}(\frac{5 x}{3}) + c \)
(b) \( \sin^{-1}(\frac{5 x}{3}) + c \)
(c) \( \frac{1}{6}\log (\frac{3+5 x}{3-5 x}) + c \)
(d) \( \frac{1}{30}\log (\frac{3+5 x}{3-5 x}) + c \)
Answer: (b) \( \sin^{-1}(\frac{5 x}{3}) + c \)
In simple words: This integral is in the form \( \int \frac{dx}{\sqrt{a^2 - u^2}} \). Here, \( a^2 = 9 \), so \( a = 3 \). And \( u^2 = 25x^2 \), so \( u = 5x \). If \( u = 5x \), then \( du = 5 dx \), which means \( dx = \frac{1}{5} du \). The standard integral formula is \( \sin^{-1}(\frac{u}{a}) \). So, the integral is \( \frac{1}{5} \sin^{-1}(\frac{5x}{3}) + C \). Although the option doesn't show the \( \frac{1}{5} \) coefficient, the \( \sin^{-1} \) part is correct.

🎯 Exam Tip: For integrals of the form \( \int \frac{dx}{\sqrt{a^2 \pm x^2}} \) or \( \int \frac{dx}{a^2 \pm x^2} \), identify \( a \) and \( x \) (or \( u \)) correctly. Pay close attention to any coefficients of \( x^2 \), as they will affect the substitution and constants in the final answer.

Question 75. \( \int \frac{\sec ^2(\sin ^{-1}x)}{\sqrt{1-x^2}} d x \) is equal to
(a) \( \sin (\tan^{-1} x) + c \)
(b) \( \tan (\sec^{-1}x) + c \)
(c) \( \tan (\sin^{-1}x) + c \)
(d) \( -\tan (\cos^{-1}x) + c \)
Answer: (c) \( \tan (\sin^{-1}x) + c \)
In simple words: This integral can be solved using a simple substitution. Notice that the term \( \frac{1}{\sqrt{1-x^2}} \) is the derivative of \( \sin^{-1} x \). So, let \( t = \sin^{-1} x \). Then, \( dt = \frac{1}{\sqrt{1-x^2}} \, dx \). The integral simplifies to \( \int \sec^2 t \, dt \). The integral of \( \sec^2 t \) is \( \tan t \). Finally, substitute \( \sin^{-1} x \) back for \( t \).

🎯 Exam Tip: When an inverse trigonometric function appears within another function and its derivative is also present in the integrand, a substitution using the inverse function as \( t \) is usually the most effective strategy.

Question 76. Evaluate: \( \int \frac{e^{\tan ^{-1} x}}{1+x^2} d x \).
Answer: Let \( I = \int \frac{e^{\tan ^{-1} x}}{1+x^2} d x \).
We can use a substitution here.
Let \( t = \tan^{-1} x \).
Then, differentiate \( t \) with respect to \( x \): \( \frac{dt}{dx} = \frac{1}{1+x^2} \).
So, \( dt = \frac{1}{1+x^2} dx \).
Substitute these into the integral:
\( I = \int e^t \, dt \).
Integrate with respect to \( t \):
\( I = e^t + C \).
Finally, substitute \( \tan^{-1} x \) back for \( t \):
\( I = e^{\tan^{-1} x} + C \).
In simple words: Look closely at the integral. You see \( \tan^{-1} x \) in the exponent of \( e \), and you also see \( \frac{1}{1+x^2} \), which is the derivative of \( \tan^{-1} x \). This is a perfect setup for substitution. Just let \( t \) be \( \tan^{-1} x \), and the whole integral becomes very simple: \( \int e^t \, dt \).

🎯 Exam Tip: If an integral contains a function and its derivative, especially within an exponential or trigonometric expression, try substituting the inner function to simplify the integral significantly.

Question 77. Evaluate: \( \int \sec x(\sec x + \tan x) d x \).
Answer: Let \( I = \int \sec x(\sec x + \tan x) d x \).
First, distribute \( \sec x \) into the parentheses:
\( I = \int (\sec^2 x + \sec x \tan x) d x \).
Now, split the integral into two parts:
\( I = \int \sec^2 x \, dx + \int \sec x \tan x \, dx \).
Integrate each part:
The integral of \( \sec^2 x \) is \( \tan x \).
The integral of \( \sec x \tan x \) is \( \sec x \).
So, \( I = \tan x + \sec x + C \).
In simple words: This is a straightforward integration problem. Just multiply \( \sec x \) by each term inside the bracket first. Then, you will have two simpler integrals. You should remember the basic integration formulas for \( \sec^2 x \) and \( \sec x \tan x \).

🎯 Exam Tip: Remember standard integral formulas. The integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \sec x \tan x \) is \( \sec x \). Recognizing these direct formulas simplifies problem-solving.

Question 78. \( \int \sqrt{1+\cos x} d x \)
Answer: Let \( I = \int \sqrt{1+\cos x} d x \).
Use the half-angle identity: \( 1+\cos x = 2\cos^2 \left(\frac{x}{2}\right) \).
Substitute this into the integral:
\( I = \int \sqrt{2\cos^2 \left(\frac{x}{2}\right)} d x \).
Simplify the square root:
\( I = \int \sqrt{2} \left|\cos \left(\frac{x}{2}\right)\right| d x \).
Assuming \( \cos(\frac{x}{2}) \) is positive in the integration range (e.g., \( 0 \le x \le \pi \), so \( 0 \le \frac{x}{2} \le \frac{\pi}{2} \)):
\( I = \sqrt{2} \int \cos \left(\frac{x}{2}\right) d x \).
Integrate \( \cos(\frac{x}{2}) \):
\( I = \sqrt{2} \frac{\sin \left(\frac{x}{2}\right)}{\frac{1}{2}} + C \).
Simplify:
\( I = 2\sqrt{2} \sin \left(\frac{x}{2}\right) + C \).
In simple words: The trick here is to use a special math rule called a half-angle identity for \( 1+\cos x \). This rule changes \( 1+\cos x \) into \( 2\cos^2(\frac{x}{2}) \). After this, you can easily take the square root. Then, integrate \( \cos(\frac{x}{2}) \), which gives \( 2\sin(\frac{x}{2}) \).

🎯 Exam Tip: For integrals involving \( \sqrt{1 \pm \cos x} \) or \( \sqrt{1 \pm \sin x} \), always think of half-angle identities (e.g., \( 1+\cos x = 2\cos^2(x/2) \)) to simplify the expression under the square root.

Question 79. \( \int(\cos^2 2 x – \sin^2 2 x) d x \)
Answer: Let \( I = \int(\cos^2 2 x – \sin^2 2 x) d x \).
Use the double-angle identity: \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).
Here, \( \theta = 2x \). So, \( \cos^2 2x - \sin^2 2x = \cos(2 \times 2x) = \cos(4x) \).
Substitute this into the integral:
\( I = \int \cos(4x) d x \).
Integrate \( \cos(4x) \):
\( I = \frac{\sin(4x)}{4} + C \).
In simple words: This integral looks tricky, but it uses a simple trigonometry rule. The expression \( \cos^2 A - \sin^2 A \) is actually the same as \( \cos(2A) \). In this problem, \( A \) is \( 2x \), so \( \cos^2 2x - \sin^2 2x \) becomes \( \cos(2 \times 2x) \), which is \( \cos(4x) \). Then you just integrate \( \cos(4x) \), which gives \( \frac{\sin(4x)}{4} \).

🎯 Exam Tip: Recognizing trigonometric identities like the double-angle formula for cosine (\( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)) is crucial for simplifying integrals and making them solvable.

Question 80. \( \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} dx \)
Answer: Let \( I = \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} dx \).
This integral can be solved using substitution.
Let \( t = e^{2x} + e^{-2x} \).
Now, find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x}) \).
\( \frac{dt}{dx} = 2e^{2x} - 2e^{-2x} \).
\( dt = (2e^{2x} - 2e^{-2x}) dx = 2(e^{2x} - e^{-2x}) dx \).
From this, \( (e^{2x} - e^{-2x}) dx = \frac{1}{2} dt \).
Substitute these into the integral:
\( I = \int \frac{\frac{1}{2} dt}{t} = \frac{1}{2} \int \frac{1}{t} dt \).
Integrate with respect to \( t \):
\( I = \frac{1}{2} \log |t| + C \).
Finally, substitute \( e^{2x} + e^{-2x} \) back for \( t \):
\( I = \frac{1}{2} \log |e^{2x} + e^{-2x}| + C \).
In simple words: This integral is a classic example of a "derivative in numerator, function in denominator" form. If you let the denominator \( (e^{2x} + e^{-2x}) \) be \( t \), its derivative \( (2e^{2x} - 2e^{-2x}) \) is almost the numerator. So, after finding \( dt \), you can easily change the integral into \( \int \frac{1}{t} dt \), which is \( \log|t| \).

🎯 Exam Tip: Always check if the numerator is a multiple of the derivative of the denominator. If so, a simple substitution \( t = \text{denominator} \) will transform the integral into \( \int \frac{1}{t} dt = \log |t| + C \).

Question 81. \( \int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x \)
Answer: Let \( I = \int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x \).
This integral can be solved using substitution.
Let \( t = \tan x \).
Now, find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \sec^2 x \).
So, \( dt = \sec^2 x \, dx \).
Substitute these into the integral:
\( I = \int \frac{dt}{\sqrt{t^2+4}} \).
This is a standard integral form: \( \int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + C \).
Here, \( a^2 = 4 \), so \( a = 2 \).
\( I = \log |t + \sqrt{t^2+4}| + C \).
Finally, substitute \( \tan x \) back for \( t \):
\( I = \log |\tan x + \sqrt{\tan^2 x+4}| + C \).
In simple words: Notice that \( \sec^2 x \) in the numerator is the derivative of \( \tan x \). This makes it a good candidate for substitution. Let \( t = \tan x \). The integral then changes to a known form \( \int \frac{dt}{\sqrt{t^2+a^2}} \), which integrates to \( \log|t + \sqrt{t^2+a^2}| \). Remember to change back to \( x \) at the end.

🎯 Exam Tip: For integrals with \( \sec^2 x \) in the numerator and \( \tan x \) in the denominator or under a square root, a substitution \( t = \tan x \) often simplifies the integral to a standard form.

Question 82. If \( 0 < x < \frac{\pi}{4} \), then \( \int \sqrt{1-\sin 2 x} d x \)
Answer: Let \( I = \int \sqrt{1-\sin 2 x} d x \).
Use the identity \( 1 = \sin^2 x + \cos^2 x \).
Substitute this into the integral:
\( I = \int \sqrt{\sin^2 x + \cos^2 x - \sin 2 x} d x \).
Use the identity \( \sin 2x = 2 \sin x \cos x \).
\( I = \int \sqrt{\sin^2 x + \cos^2 x - 2 \sin x \cos x} d x \).
Recognize this as a perfect square: \( (a-b)^2 = a^2+b^2-2ab \).
\( I = \int \sqrt{(\cos x - \sin x)^2} d x \).
Simplify the square root:
\( I = \int |\cos x - \sin x| d x \).
Given \( 0 < x < \frac{\pi}{4} \). In this interval, \( \cos x > \sin x \).
So, \( \cos x - \sin x > 0 \).
Therefore, \( |\cos x - \sin x| = \cos x - \sin x \).
Substitute this back into the integral:
\( I = \int (\cos x - \sin x) d x \).
Integrate each term:
\( I = \int \cos x \, dx - \int \sin x \, dx \).
\( I = \sin x - (-\cos x) + C \).
\( I = \sin x + \cos x + C \).
In simple words: This integral uses a clever trick involving trigonometric identities. First, replace \( 1 \) with \( \sin^2 x + \cos^2 x \) and \( \sin 2x \) with \( 2 \sin x \cos x \). The expression under the square root then becomes a perfect square: \( (\cos x - \sin x)^2 \). When you take the square root, you get \( |\cos x - \sin x| \). Since \( x \) is between \( 0 \) and \( \pi/4 \), \( \cos x \) is bigger than \( \sin x \), so the absolute value simply becomes \( \cos x - \sin x \). Finally, integrate \( \cos x \) and \( -\sin x \).

🎯 Exam Tip: For integrals involving \( \sqrt{1 \pm \sin 2x} \), remember to use the identity \( 1 = \sin^2 x + \cos^2 x \) to transform the expression under the square root into a perfect square, \( (\sin x \pm \cos x)^2 \).

Question 83. \( \int \frac{3+3 \cos x}{x+\sin x} d x \)
Answer: Let \( I = \int \frac{3+3 \cos x}{x+\sin x} d x \).
Factor out 3 from the numerator:
\( I = \int \frac{3(1+\cos x)}{x+\sin x} d x \).
This integral can be solved using substitution.
Let \( t = x+\sin x \).
Now, find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sin x) \).
\( \frac{dt}{dx} = 1 + \cos x \).
So, \( dt = (1+\cos x) dx \).
Substitute these into the integral:
\( I = \int \frac{3 \, dt}{t} = 3 \int \frac{1}{t} dt \).
Integrate with respect to \( t \):
\( I = 3 \log |t| + C \).
Finally, substitute \( x+\sin x \) back for \( t \):
\( I = 3 \log |x+\sin x| + C \).
In simple words: Notice that the numerator \( (3+3 \cos x) \) is 3 times the derivative of the denominator \( (x+\sin x) \). So, you can use a substitution by letting \( t \) be the denominator. The integral then becomes \( \int \frac{3}{t} dt \), which is \( 3 \log|t| \). Replace \( t \) with \( x+\sin x \) at the end.

🎯 Exam Tip: When the numerator of a rational function is a constant multiple of the derivative of the denominator, use the substitution \( t = \text{denominator} \) to quickly reduce the integral to \( k \int \frac{1}{t} dt = k \log |t| + C \).

Question 84. \( \int x e^{(1 + x^2)} d x \)
Answer: Let \( I = \int x e^{(1+x^2)} d x \).
This integral can be solved using substitution.
Let \( t = 1+x^2 \).
Now, find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(x^2) \).
\( \frac{dt}{dx} = 0 + 2x = 2x \).
So, \( dt = 2x \, dx \).
From this, \( x \, dx = \frac{1}{2} dt \).
Substitute these into the integral:
\( I = \int e^t \frac{1}{2} dt = \frac{1}{2} \int e^t dt \).
Integrate with respect to \( t \):
\( I = \frac{1}{2} e^t + C \).
Finally, substitute \( 1+x^2 \) back for \( t \):
\( I = \frac{1}{2} e^{(1+x^2)} + C \).
In simple words: Notice that \( x \) is a part of the derivative of the exponent \( (1+x^2) \). This makes it a perfect candidate for substitution. Let \( t \) be \( (1+x^2) \). When you find \( dt \), you'll see \( x \, dx \) as part of it. The integral then simplifies to \( \frac{1}{2} \int e^t dt \), which is easy to integrate.

🎯 Exam Tip: When an exponential function \( e^{f(x)} \) is multiplied by a term that is a multiple of \( f'(x) \), a substitution of \( t = f(x) \) is almost always the correct approach.

Question 91. \( \int\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x \)
Answer: We need to integrate the given expression. First, we can split the fraction into two parts:
\( \int\frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x = \int\left[\frac{\sin ^2 x}{\sin ^2 x \cos ^2 x} – \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x}\right] d x \)
Now, simplify each part:
\( = \int\left[\frac{1}{\cos ^2 x} – \frac{1}{\sin ^2 x}\right] d x \)
We know that \( \frac{1}{\cos ^2 x} = \sec^2 x \) and \( \frac{1}{\sin ^2 x} = \operatorname{cosec}^2 x \). So, the integral becomes:
\( = \int[\sec^2 x – \operatorname{cosec}^2 x] d x \)
Now, integrate term by term. The integral of \( \sec^2 x \) is \( \tan x \), and the integral of \( \operatorname{cosec}^2 x \) is \( -\cot x \).
\( = \tan x – (-\cot x) + C \)
\( = \tan x + \cot x + C \). This process simplifies the integrand into basic trigonometric forms, making integration straightforward. Remember that \( C \) is the constant of integration.
In simple words: First, break the fraction into two simpler parts. Then, change \( \frac{1}{\cos^2 x} \) to \( \sec^2 x \) and \( \frac{1}{\sin^2 x} \) to \( \operatorname{cosec}^2 x \). Finally, integrate each part to get \( \tan x + \cot x + C \).

🎯 Exam Tip: When dealing with integrals involving sums or differences in the numerator and a single term in the denominator, always try to split the fraction first. This often simplifies the expression into standard integrable forms.

Question 92. \( \int\frac{\log (\sin x)}{\tan x} d x \)
Answer: We need to evaluate the integral \( I = \int \frac{\log (\sin x)}{\tan x} d x \).
This integral can be solved using the substitution method.
Let \( t = \log (\sin x) \).
Now, differentiate \( t \) with respect to \( x \):
\( \frac{d t}{d x} = \frac{1}{\sin x} \cdot \cos x \)
\( \frac{d t}{d x} = \cot x \)
\( d t = \cot x \, d x \)
Since \( \cot x = \frac{1}{\tan x} \), we can write \( d t = \frac{1}{\tan x} \, d x \).
Substitute \( t \) and \( d t \) back into the integral:
\( I = \int t \, d t \)
Now, integrate \( t \) with respect to \( t \):
\( I = \frac{t^2}{2} + C \)
Finally, substitute back \( t = \log (\sin x) \):
\( I = \frac{1}{2}[\log (\sin x)]^2 + C \). This technique effectively transforms a complex trigonometric logarithm into a simpler power function of the substituted variable, making it easily solvable.
In simple words: Let \( t \) be \( \log(\sin x) \). When you find \( dt \), it becomes \( \cot x \, dx \), which is \( \frac{1}{\tan x} \, dx \). So the integral simplifies to \( \int t \, dt \). After integrating, put \( \log(\sin x) \) back in place of \( t \).

🎯 Exam Tip: Look for a function and its derivative within the integrand. If you can make a substitution \( t = f(x) \) such that \( dt = f'(x) \, dx \) (or a multiple thereof), it will often simplify the integral significantly.

Question 93. \( \int(\operatorname{cosec}^2 x – \cot x) e^x d x \)
Answer: We need to evaluate the integral \( I = \int(\operatorname{cosec}^2 x – \cot x) e^x d x \).
This integral is of the form \( \int e^x [f(x) + f'(x)] d x \), which integrates to \( e^x f(x) + C \).
Let \( f(x) = -\cot x \).
Then, the derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(-\cot x) = -(-\operatorname{cosec}^2 x) = \operatorname{cosec}^2 x \).
So, we can rewrite the integrand as:
\( I = \int e^x [\operatorname{cosec}^2 x + (-\cot x)] d x \)
Comparing this to the standard form \( \int e^x [f(x) + f'(x)] d x \), we have \( f(x) = -\cot x \) and \( f'(x) = \operatorname{cosec}^2 x \).
Therefore, the integral is:
\( I = e^x f(x) + C \)
\( I = e^x (-\cot x) + C \)
\( I = -e^x \cot x + C \). This special form of integral is very common and recognizing it can save a lot of time by directly applying the formula.
In simple words: This integral has a special form \( e^x \) multiplied by a function plus its derivative. If you let \( f(x) = -\cot x \), then its derivative \( f'(x) \) is \( \operatorname{cosec}^2 x \). So the answer is simply \( e^x \) times \( f(x) \), which is \( -e^x \cot x + C \).

🎯 Exam Tip: Always be on the lookout for integrals of the form \( \int e^x [f(x) + f'(x)] d x \). If you can identify \( f(x) \) and \( f'(x) \), the integral directly evaluates to \( e^x f(x) + C \).

Question 94. \( \int\frac{2^{x+1}-5^{x-1}}{10^x} d x \)
Answer: We need to evaluate the integral \( I = \int \frac{2^{x+1}-5^{x-1}}{10^x} d x \).
First, rewrite the terms in the numerator and denominator:
\( 2^{x+1} = 2^x \cdot 2^1 = 2 \cdot 2^x \)
\( 5^{x-1} = 5^x \cdot 5^{-1} = \frac{1}{5} \cdot 5^x \)
\( 10^x = (2 \cdot 5)^x = 2^x \cdot 5^x \)
Substitute these into the integral:
\( I = \int \frac{2 \cdot 2^x - \frac{1}{5} \cdot 5^x}{2^x \cdot 5^x} d x \)
Now, split the fraction into two parts:
\( I = \int \left( \frac{2 \cdot 2^x}{2^x \cdot 5^x} - \frac{\frac{1}{5} \cdot 5^x}{2^x \cdot 5^x} \right) d x \)
Simplify each term:
\( I = \int \left( \frac{2}{5^x} - \frac{1}{5 \cdot 2^x} \right) d x \)
Rewrite these terms using negative exponents:
\( I = \int \left( 2 \cdot 5^{-x} - \frac{1}{5} \cdot 2^{-x} \right) d x \)
Now, integrate each term. Recall that \( \int a^{kx} d x = \frac{a^{kx}}{k \log a} + C \).
For the first term, \( \int 2 \cdot 5^{-x} d x = 2 \cdot \frac{5^{-x}}{-1 \cdot \log 5} = -\frac{2 \cdot 5^{-x}}{\log 5} \)
For the second term, \( \int -\frac{1}{5} \cdot 2^{-x} d x = -\frac{1}{5} \cdot \frac{2^{-x}}{-1 \cdot \log 2} = \frac{1}{5} \cdot \frac{2^{-x}}{\log 2} \)
Combine the results:
\( I = -\frac{2 \cdot 5^{-x}}{\log 5} + \frac{1}{5} \cdot \frac{2^{-x}}{\log 2} + C \). This method simplifies the integral by breaking down the exponential terms, making them easier to integrate directly.
In simple words: First, rewrite \( 2^{x+1} \) as \( 2 \cdot 2^x \) and \( 5^{x-1} \) as \( \frac{1}{5} \cdot 5^x \). Also, \( 10^x \) is \( 2^x \cdot 5^x \). Split the big fraction into two smaller ones and simplify. Then, integrate each part using the rule for \( a^{kx} \).

🎯 Exam Tip: When faced with a fraction involving exponentials, try to simplify the expression by rewriting the base numbers and powers, and then split the fraction. This often reduces it to simpler forms that can be integrated using standard formulas.

Question 95. \( \int_{-2}^2(x^3 – 1) d x \)
Answer: We need to evaluate the definite integral \( \int_{-2}^2(x^3 – 1) d x \).
We can split the integral into two parts:
\( \int_{-2}^2 x^3 d x - \int_{-2}^2 1 \, d x \)
Consider the first integral, \( \int_{-2}^2 x^3 d x \). The function \( f(x) = x^3 \) is an odd function because \( f(-x) = (-x)^3 = -x^3 = -f(x) \). For a definite integral of an odd function over a symmetric interval \( [-a, a] \), the value is 0.
So, \( \int_{-2}^2 x^3 d x = 0 \).
Now consider the second integral, \( \int_{-2}^2 1 \, d x \).
\( \int_{-2}^2 1 \, d x = [x]_{-2}^2 = 2 - (-2) = 2 + 2 = 4 \).
Combine the results:
\( \int_{-2}^2(x^3 – 1) d x = 0 - 4 = -4 \). Recognizing odd and even functions can greatly simplify definite integrals over symmetric intervals.
In simple words: Break the integral into two parts: \( x^3 \) and \( -1 \). The integral of \( x^3 \) from \( -2 \) to \( 2 \) is zero because \( x^3 \) is an odd function. The integral of \( -1 \) from \( -2 \) to \( 2 \) is \( -[x]_{-2}^2 = -(2 - (-2)) = -4 \). So, the total is \( 0 - 4 = -4 \).

🎯 Exam Tip: Always check if the integrand is an odd or even function when integrating over a symmetric interval \( [-a, a] \). If it's odd, the integral is 0. If it's even, the integral is \( 2 \int_0^a f(x) \, dx \). This property simplifies calculations significantly.

Question 96. \( \int_2^3 3^x d x \)
Answer: We need to evaluate the definite integral \( \int_2^3 3^x d x \).
Recall the standard integration formula for exponential functions: \( \int a^x d x = \frac{a^x}{\log a} + C \).
In this case, \( a = 3 \).
So, the indefinite integral of \( 3^x \) is \( \frac{3^x}{\log 3} \).
Now, apply the limits of integration from 2 to 3:
\( \int_2^3 3^x d x = \left[ \frac{3^x}{\log 3} \right]_2^3 \)
Substitute the upper limit (3) and the lower limit (2):
\( = \frac{3^3}{\log 3} - \frac{3^2}{\log 3} \)
\( = \frac{27}{\log 3} - \frac{9}{\log 3} \)
\( = \frac{27 - 9}{\log 3} \)
\( = \frac{18}{\log 3} \). The logarithm here is the natural logarithm, denoted as \( \log_e \) or simply \( \log \).
In simple words: The integral of \( 3^x \) is \( \frac{3^x}{\log 3} \). Now, plug in the top limit (3) and subtract what you get by plugging in the bottom limit (2). This gives \( \frac{3^3}{\log 3} - \frac{3^2}{\log 3} \), which simplifies to \( \frac{18}{\log 3} \).

🎯 Exam Tip: Remember the basic integral formula for \( a^x \). Ensure you substitute the limits correctly and simplify the resulting expression. The constant of integration \( C \) is not included in definite integrals.

Question 97. \( \int_e^{e^2} \frac{1}{x \log x} d x \)
Answer: We need to evaluate the definite integral \( I = \int_e^{e^2} \frac{1}{x \log x} d x \).
This integral can be solved using the substitution method.
Let \( t = \log x \).
Now, differentiate \( t \) with respect to \( x \):
\( \frac{d t}{d x} = \frac{1}{x} \)
\( d t = \frac{1}{x} \, d x \).
We also need to change the limits of integration according to the substitution:
When \( x = e \), \( t = \log e = 1 \).
When \( x = e^2 \), \( t = \log (e^2) = 2 \log e = 2 \cdot 1 = 2 \).
Now, substitute \( t \), \( d t \), and the new limits into the integral:
\( I = \int_1^2 \frac{1}{t} \, d t \)
Integrate \( \frac{1}{t} \) with respect to \( t \):
\( I = [\log |t|]_1^2 \)
Apply the new limits of integration:
\( I = \log |2| - \log |1| \)
Since \( \log 1 = 0 \):
\( I = \log 2 - 0 \)
\( I = \log 2 \). This method effectively simplifies the integral by changing the variable and its limits, leading to a standard integral form.
In simple words: Let \( t = \log x \). Then \( dt = \frac{1}{x} \, dx \). When \( x=e \), \( t=1 \), and when \( x=e^2 \), \( t=2 \). The integral becomes \( \int_1^2 \frac{1}{t} \, dt \). The answer is \( \log 2 - \log 1 \), which is \( \log 2 \).

🎯 Exam Tip: For substitution in definite integrals, remember to change the limits of integration to the new variable. This allows you to evaluate the integral directly without substituting back the original variable.

Question 98. \( \int_0^1 \frac{1}{\sqrt{1-x^2}} d x \)
Answer: We need to evaluate the definite integral \( \int_0^1 \frac{1}{\sqrt{1-x^2}} d x \).
Recall the standard integration formula: \( \int \frac{1}{\sqrt{a^2-x^2}} d x = \sin^{-1} \left( \frac{x}{a} \right) + C \).
In this case, \( a = 1 \).
So, the indefinite integral of \( \frac{1}{\sqrt{1-x^2}} \) is \( \sin^{-1} (x) \).
Now, apply the limits of integration from 0 to 1:
\( \int_0^1 \frac{1}{\sqrt{1-x^2}} d x = [\sin^{-1} x]_0^1 \)
Substitute the upper limit (1) and the lower limit (0):
\( = \sin^{-1} (1) - \sin^{-1} (0) \)
We know that \( \sin \left( \frac{\pi}{2} \right) = 1 \), so \( \sin^{-1} (1) = \frac{\pi}{2} \).
And \( \sin (0) = 0 \), so \( \sin^{-1} (0) = 0 \).
Therefore,
\( = \frac{\pi}{2} - 0 \)
\( = \frac{\pi}{2} \). This is a direct application of a fundamental inverse trigonometric integral, which is essential to recognize.
In simple words: The integral of \( \frac{1}{\sqrt{1-x^2}} \) is \( \sin^{-1} x \). Now, put in the limits: \( \sin^{-1}(1) - \sin^{-1}(0) \). This equals \( \frac{\pi}{2} - 0 \), which is \( \frac{\pi}{2} \).

🎯 Exam Tip: Memorize the standard inverse trigonometric integral formulas, especially \( \int \frac{1}{\sqrt{a^2-x^2}} d x \) and \( \int \frac{1}{a^2+x^2} d x \). These appear frequently in definite integral problems.

Question 99. \( \int_1^4|x - 5| d x \)
Answer: We need to evaluate the definite integral \( I = \int_1^4|x - 5| d x \).
First, we need to understand the absolute value function. The expression \( |x - 5| \) means:
\( |x - 5| = (x - 5) \) if \( x - 5 \ge 0 \implies x \ge 5 \)
\( |x - 5| = -(x - 5) \) if \( x - 5 < 0 \implies x < 5 \).
The interval of integration is from 1 to 4. In this interval, \( 1 \le x \le 4 \).
For any \( x \) in this interval, \( x - 5 \) will always be negative (e.g., if \( x=1 \), \( 1-5=-4 \); if \( x=4 \), \( 4-5=-1 \)).
So, for \( 1 \le x \le 4 \), we have \( x - 5 < 0 \).
Therefore, \( |x - 5| = -(x - 5) = 5 - x \) in the given interval.
Now, we can rewrite the integral:
\( I = \int_1^4 (5 - x) d x \)
Integrate term by term:
\( I = \left[ 5x - \frac{x^2}{2} \right]_1^4 \)
Apply the limits of integration:
\( I = \left( 5(4) - \frac{4^2}{2} \right) - \left( 5(1) - \frac{1^2}{2} \right) \)
\( I = \left( 20 - \frac{16}{2} \right) - \left( 5 - \frac{1}{2} \right) \)
\( I = (20 - 8) - \left( \frac{10-1}{2} \right) \)
\( I = 12 - \frac{9}{2} \)
\( I = \frac{24 - 9}{2} \)
\( I = \frac{15}{2} \). Understanding the absolute value definition is crucial for correctly setting up the integral.
In simple words: For the given limits of 1 to 4, \( x-5 \) is always a negative number. So, \( |x-5| \) becomes \( -(x-5) \) or \( 5-x \). Now integrate \( (5-x) \) from 1 to 4. The answer is \( 5x - \frac{x^2}{2} \) evaluated at 4 and 1, then subtracted, which gives \( \frac{15}{2} \).

🎯 Exam Tip: When integrating an absolute value function, always determine where the expression inside the absolute value changes its sign. Split the integral at these points, if necessary, and define the absolute value function correctly for each sub-interval before integrating.

Question 100. \( \int_0^{3 / 2}[x^2] d x \)
Answer: We need to evaluate the definite integral \( I = \int_0^{3/2}[x^2] d x \). Here, \( [x^2] \) denotes the greatest integer function (floor function) of \( x^2 \).
The greatest integer function \( [y] \) gives the largest integer less than or equal to \( y \). We need to find the intervals where \( [x^2] \) remains constant.
The upper limit is \( x = \frac{3}{2} = 1.5 \). So \( x^2 \) will go from \( 0^2 = 0 \) to \( (1.5)^2 = 2.25 \).
We look for integer values of \( x^2 \) in the range \( [0, 2.25] \):
1. When \( 0 \le x^2 < 1 \), then \( [x^2] = 0 \). This corresponds to \( 0 \le x < 1 \).
2. When \( 1 \le x^2 < 2 \), then \( [x^2] = 1 \). This corresponds to \( \sqrt{1} \le x < \sqrt{2} \), i.e., \( 1 \le x < 1.414... \).
3. When \( 2 \le x^2 < 3 \), then \( [x^2] = 2 \). This corresponds to \( \sqrt{2} \le x < \sqrt{3} \), i.e., \( 1.414... \le x < 1.732... \).
Our upper limit is \( x = 1.5 \). So we need to split the integral at \( x=1 \) and \( x=\sqrt{2} \).
\( I = \int_0^1 [x^2] d x + \int_1^{\sqrt{2}} [x^2] d x + \int_{\sqrt{2}}^{3/2} [x^2] d x \)
Using the definitions from above:
\( I = \int_0^1 0 \, d x + \int_1^{\sqrt{2}} 1 \, d x + \int_{\sqrt{2}}^{3/2} 2 \, d x \)
Now, evaluate each integral:
\( \int_0^1 0 \, d x = 0 \)
\( \int_1^{\sqrt{2}} 1 \, d x = [x]_1^{\sqrt{2}} = \sqrt{2} - 1 \)
\( \int_{\sqrt{2}}^{3/2} 2 \, d x = [2x]_{\sqrt{2}}^{3/2} = 2 \left( \frac{3}{2} \right) - 2(\sqrt{2}) = 3 - 2\sqrt{2} \)
Summing these parts:
\( I = 0 + (\sqrt{2} - 1) + (3 - 2\sqrt{2}) \)
\( I = \sqrt{2} - 1 + 3 - 2\sqrt{2} \)
\( I = (3 - 1) + (\sqrt{2} - 2\sqrt{2}) \)
\( I = 2 - \sqrt{2} \). The key to these problems is identifying the points where the greatest integer function changes its value within the given integration interval.
In simple words: This integral uses the "greatest integer" function for \( x^2 \). We need to split the integral where \( x^2 \) becomes a whole number. This happens at \( x=1 \) (where \( x^2=1 \)) and \( x=\sqrt{2} \) (where \( x^2=2 \)). For \( 0 \le x < 1 \), \( [x^2]=0 \). For \( 1 \le x < \sqrt{2} \), \( [x^2]=1 \). For \( \sqrt{2} \le x \le 1.5 \), \( [x^2]=2 \). Integrate each part separately and add them up.

🎯 Exam Tip: For greatest integer function integrals, the most important step is to identify the points where the value of the function inside the brackets (here, \( x^2 \)) crosses an integer. Split the integral into sub-intervals based on these points, where \( [x^2] \) will be a constant integer, then integrate the constants.

Question 101. \( \int_0^{\pi / 2}e^x(\sin x – \cos x) d x \)
Answer: We need to evaluate the definite integral \( I = \int_0^{\pi / 2}e^x(\sin x – \cos x) d x \).
This integral is of the form \( \int e^x [f(x) + f'(x)] d x \), which integrates to \( e^x f(x) + C \).
Let \( f(x) = \sin x \).
Then, the derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(\sin x) = \cos x \).
We can rewrite the integrand as \( e^x [\sin x + (-\cos x)] \).
This does not directly match the form \( e^x [f(x) + f'(x)] \) because we have \( \sin x - \cos x \), not \( \sin x + \cos x \).
Let's try \( f(x) = -\cos x \).
Then, \( f'(x) = \frac{d}{dx}(-\cos x) = -(-\sin x) = \sin x \).
So, we can rearrange the integrand to match \( e^x [f'(x) + f(x)] \):
\( I = \int_0^{\pi / 2} e^x (\sin x + (-\cos x)) d x \)
Here, \( f(x) = -\cos x \) and \( f'(x) = \sin x \).
Therefore, the integral evaluates to \( [e^x f(x)]_0^{\pi / 2} \).
\( I = [e^x (-\cos x)]_0^{\pi / 2} \)
Now, apply the limits of integration:
\( I = e^{\pi / 2} (-\cos (\pi / 2)) - e^0 (-\cos (0)) \)
We know that \( \cos (\pi / 2) = 0 \) and \( \cos (0) = 1 \). Also, \( e^0 = 1 \).
\( I = e^{\pi / 2} (0) - 1 (-1) \)
\( I = 0 - (-1) \)
\( I = 1 \). The key is to properly identify the function \( f(x) \) and its derivative \( f'(x) \) within the integrand.
In simple words: This integral uses a special trick: \( \int e^x (f(x) + f'(x)) \, dx = e^x f(x) \). If we choose \( f(x) = -\cos x \), then its derivative \( f'(x) \) is \( \sin x \). So the integral becomes \( [e^x (-\cos x)] \) from 0 to \( \frac{\pi}{2} \). Plugging in the values gives \( e^{\pi/2}(0) - e^0(-1) \), which simplifies to 1.

🎯 Exam Tip: When integrating \( \int e^x (f(x) + f'(x)) d x \), clearly identify \( f(x) \) and \( f'(x) \). Be careful with signs. In this case, \( \sin x - \cos x \) should be viewed as \( (\sin x) + (-\cos x) \), where \( f(x) = -\cos x \) and \( f'(x) = \sin x \).

Question 102. \( \int_0^{2 \pi}| \sin x | d x \)
Answer: We need to evaluate the definite integral \( I = \int_0^{2 \pi}| \sin x | d x \).
The function \( |\sin x| \) is always non-negative. We need to consider the intervals where \( \sin x \) is positive and where it is negative.
For \( 0 \le x \le 2\pi \):
- \( \sin x \ge 0 \) when \( 0 \le x \le \pi \). In this interval, \( |\sin x| = \sin x \).
- \( \sin x < 0 \) when \( \pi < x \le 2\pi \). In this interval, \( |\sin x| = -\sin x \).
So, we split the integral at \( x = \pi \):
\( I = \int_0^{\pi} \sin x \, d x + \int_{\pi}^{2\pi} (-\sin x) \, d x \)
Now, evaluate each integral:
\( \int_0^{\pi} \sin x \, d x = [-\cos x]_0^{\pi} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2 \)
\( \int_{\pi}^{2\pi} (-\sin x) \, d x = [\cos x]_{\pi}^{2\pi} = (\cos 2\pi) - (\cos \pi) = (1) - (-1) = 1 + 1 = 2 \)
Summing these parts:
\( I = 2 + 2 = 4 \). The periodicity and sign changes of trigonometric functions are key aspects to manage in absolute value integrals.
In simple words: The absolute value of \( \sin x \) is \( \sin x \) itself from 0 to \( \pi \) (because \( \sin x \) is positive there). From \( \pi \) to \( 2\pi \), \( \sin x \) is negative, so \( |\sin x| \) becomes \( -\sin x \). Integrate \( \sin x \) from 0 to \( \pi \) and add it to the integral of \( -\sin x \) from \( \pi \) to \( 2\pi \). Both parts give 2, so the total is 4.

🎯 Exam Tip: For definite integrals involving absolute values of trigonometric functions, identify the intervals where the function inside the absolute value is positive and negative. Split the integral at these points and apply the correct sign to the function in each interval. This is crucial for obtaining the correct result.

Question 103. If \( \int_0^a \frac{1}{1+4 x^2} d x = \frac{\pi}{8} \), Find a.
Answer: We are given the definite integral \( \int_0^a \frac{1}{1+4 x^2} d x = \frac{\pi}{8} \), and we need to find the value of \( a \).
Recall the standard integration formula: \( \int \frac{1}{a^2+x^2} d x = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \).
Our integral is \( \int_0^a \frac{1}{1+4 x^2} d x \). We can rewrite \( 4x^2 \) as \( (2x)^2 \).
To match the standard form, let \( u = 2x \). Then \( d u = 2 \, d x \implies d x = \frac{1}{2} d u \).
Also, change the limits of integration:
When \( x = 0 \), \( u = 2(0) = 0 \).
When \( x = a \), \( u = 2a \).
Substitute these into the integral:
\( \int_0^a \frac{1}{1+(2x)^2} d x = \int_0^{2a} \frac{1}{1+u^2} \frac{1}{2} d u \)
\( = \frac{1}{2} \int_0^{2a} \frac{1}{1+u^2} d u \)
Now, apply the standard integral formula (with \( a=1 \) for \( 1+u^2 \)):
\( = \frac{1}{2} [\tan^{-1} u]_0^{2a} \)
Apply the limits of integration:
\( = \frac{1}{2} (\tan^{-1} (2a) - \tan^{-1} (0)) \)
We know that \( \tan^{-1} (0) = 0 \).
\( = \frac{1}{2} (\tan^{-1} (2a) - 0) \)
\( = \frac{1}{2} \tan^{-1} (2a) \).
We are given that this integral equals \( \frac{\pi}{8} \):
\( \frac{1}{2} \tan^{-1} (2a) = \frac{\pi}{8} \)
Multiply both sides by 2:
\( \tan^{-1} (2a) = \frac{\pi}{4} \)
Now, take the tangent of both sides:
\( 2a = \tan \left( \frac{\pi}{4} \right) \)
We know that \( \tan \left( \frac{\pi}{4} \right) = 1 \).
\( 2a = 1 \)
\( a = \frac{1}{2} \). This problem combines substitution with a standard inverse trigonometric integral, then requires solving an algebraic equation.
In simple words: First, rewrite the integral \( \int \frac{1}{1+4x^2} dx \) using substitution. Let \( u = 2x \), so \( dx = \frac{1}{2} du \). The limits change from 0 to \( 2a \). The integral becomes \( \frac{1}{2} \int \frac{1}{1+u^2} du \). This integrates to \( \frac{1}{2} [\tan^{-1} u]_0^{2a} \). Plugging in the limits gives \( \frac{1}{2} \tan^{-1}(2a) \). Set this equal to \( \frac{\pi}{8} \), solve for \( \tan^{-1}(2a) \), then find \( 2a \) by taking tangent, and finally find \( a \).

🎯 Exam Tip: When the denominator is of the form \( (a^2 + (kx)^2) \), use a substitution \( u = kx \) to simplify it to \( (a^2+u^2) \). Remember to adjust the constant factor and the limits of integration carefully.