CBSE Class 12 Mathematics Integrals VBQs Set A

BASIC CONCEPTS

Antiderivative (or Primitive): A function \( \phi(x) \) is said to be antiderivative or primitive of a function \( f(x) \) if \( \phi'(x) = f(x) \) i.e., \( \frac{d}{dx} \{ \phi(x) \} = f(x) \).
For example, \( \frac{x^2}{2} \) is primitive or antiderivative of \( x \) because
\[ \frac{d}{dx} \left( \frac{x^2}{2} \right) = \frac{1}{2} \cdot 2x = x \]
Similarly, \( \frac{d}{dx} \left( \frac{x^2}{2} + 1 \right) = \frac{1}{2} \cdot 2x + 0 = x \)
Similarly, \( \frac{d}{dx} \left( \frac{x^2}{2} + C \right) = \frac{1}{2} \cdot 2x + 0 = x \)
In this way, we see that a function has infinitely many antiderivatives or primitives. i.e., if \( \phi(x) \) be an antiderivative of \( f(x) \), then \( \phi(x) + C \) is also antiderivative of \( f(x) \), where \( C \) is any constant. Because, \( \frac{d}{dx} \{ \phi(x) + C \} = \phi'(x) + 0 = \phi'(x) = f(x) \)

Indefinite Integrals: If \( f(x) \) is a function, then the family of all its antiderivatives is called Indefinite Integral of \( f(x) \). It is represented by \( \int f(x) dx \) (read as indefinite integral of \( f(x) \) with respect to \( x \)).
For example, \( \int x^2 dx = \frac{x^3}{3} + C; \quad \int x^3 dx = \frac{x^4}{4} + C \)

Why is it called Indefinite Integral?

It is called indefinite because it is not unique. Actually there exist infinitely many integrals of each function, which can be obtained by choosing \( C \) arbitrarily from the set of real numbers.

Some Standard Integrals:
(i) \( \int x^n dx = \frac{x^{n+1}}{n+1} + C (n \neq -1) \)
(ii) \( \int \frac{dx}{x} = \log |x| + C \)
(iii) \( \int dx = x + C \)
(iv) \( \int \cos x dx = \sin x + C \)
(v) \( \int \sin x dx = -\cos x + C \)
(vi) \( \int \sec^2 x dx = \tan x + C \)
(vii) \( \int \csc^2 x dx = -\cot x + C \)
(viii) \( \int \sec x \tan x dx = \sec x + C \)
(ix) \( \int \csc x \cot x dx = -\csc x + C \)
(x) \( \int e^x dx = e^x + C \)
(xi) \( \int a^x dx = \frac{a^x}{\log a} + C \)
(xii) (a) \( \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C \)
(b) \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C \)
(xiii) (a) \( \int \frac{1}{1+x^2} dx = \tan^{-1} x + C \)
(b) \( \int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \)
(xiv) \( \int \frac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x + C \)
(xv) \( \int -\frac{1}{x\sqrt{x^2-1}} dx = \csc^{-1} x + C \)
(xvi) \( \int -\frac{1}{a^2+x^2} dx = \frac{1}{a} \cot^{-1} \left( \frac{x}{a} \right) + C \)
(xvii) \( \int \frac{1}{x\sqrt{x^2-a^2}} dx = \frac{1}{a} \sec^{-1} \left( \frac{x}{a} \right) + C \)
(xviii) \( \int -\frac{1}{x\sqrt{x^2-a^2}} dx = \frac{1}{a} \csc^{-1} \left( \frac{x}{a} \right) + C \)

Methods of Integration:

It is not possible to integrate each integral with the help of following methods but a large number of various problems can be solved by these methods. So, we have the following methods of integration:
(i) Integration by Substitution.
(ii) Integration by Parts.
(iii) Integration of Rational Algebraic Functions by Using Partial Fractions.

Integration by Substitution: The method of evaluating integrals of a function by suitable substitution is called Integration by substitution.
We therefore give some of the fundamental integrals when \( x \) is replaced by \( ax + b \).
(i) \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C, n \neq -1 \)
(ii) \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log |(ax+b)| + C \)
(iii) \( \int e^{ax+b} dx = \frac{1}{a} e^{ax+b} + C \)
(iv) \( \int a^{bx+c} dx = \frac{1}{b} \cdot \frac{a^{bx+c}}{\log a} + C, a > 0 \text{ and } a \neq 1 \)
(v) \( \int \sin(ax+b) dx = -\frac{1}{a} \cos(ax+b) + C \)
(vi) \( \int \cos(ax+b) dx = \frac{1}{a} \sin(ax+b) + C \)
(vii) \( \int \sec^2(ax+b) dx = \frac{1}{a} \tan(ax+b) + C \)
(viii) \( \int \csc^2(ax+b) dx = -\frac{1}{a} \cot(ax+b) + C \)
(ix) \( \int \sec(ax+b) \tan(ax+b) dx = \frac{1}{a} \sec(ax+b) + C \)
(x) \( \int \csc(ax+b) \cot(ax+b) dx = -\frac{1}{a} \csc(ax+b) + C \)
(xi) \( \int \tan(ax+b) dx = -\frac{1}{a} \log |\cos(ax+b)| + C \)
(xii) \( \int \cot(ax+b) dx = \frac{1}{a} \log |\sin(ax+b)| + C \)

More standard results:
\( \int \tan x dx = -\log |\cos x| + C = \log |\sec x| + C \), provided \( x \) is not an odd multiple of \( \frac{\pi}{2} \)
\( \int \cot x dx = \log |\sin x| + C \)
\( \int \sec x dx = \log |\sec x + \tan x| + C = \log |\tan(\frac{\pi}{4} + \frac{x}{2})| + C \)
\( \int \csc x dx = \log |\csc x - \cot x| + C = \log |\tan \frac{x}{2}| + C \)

Integration by Parts: To integrate the product of two functions, we use integration by parts. The method is as given below:
Let \( u \) and \( v \) be two functions of \( x \) then
\[ \int u \cdot v \, dx = u \int v \, dx - \int \left\{ \frac{du}{dx} \cdot \int v \, dx \right\} dx \]

Note:
(i) To integrate the product of two functions we choose the 1st function according to word ILATE, where I stands for inverse function, L stands for logarithmic function, A stands for the algebraic functions, T stands for trigonometrical function and E stands for exponential function.
(ii) If the integrand has only one function then unity, i.e., 1 is taken to be the second function.
(iii) Integration by parts is not applicable to product of functions in all cases. For example, the method does not work for \( \int \frac{1}{x} \cdot \sin x \, dx \). The reason is that there does not exist any function whose derivative is \( \frac{1}{x} \cdot \sin x \).
(iv) Observe that while finding the integral of the second function, we do not add a constant of integration on both the sides.

Results of Some Special Integrals:
(i) \( \int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \)
(ii) (a) \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \);
(b) \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \)
(iii) \( \int \frac{1}{\sqrt{a^2+x^2}} dx = \log \left| \frac{x + \sqrt{x^2+a^2}}{a} \right| + C \) or \( \log |x + \sqrt{x^2+a^2}| + C \)
(iv) \( \int \frac{1}{\sqrt{x^2-a^2}} dx = \log \left| \frac{x + \sqrt{x^2-a^2}}{a} \right| + C \) or \( \log |x + \sqrt{x^2-a^2}| + C \)
(v) (a) \( \int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1} \frac{x}{a} + C \);
(b) \( \int \sqrt{a^2-x^2} dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C \)
(vi) \( \int \sqrt{x^2-a^2} dx = \frac{x}{2} \sqrt{x^2-a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2-a^2}| + C \)
(vii) \( \int \sqrt{a^2+x^2} dx = \frac{x}{2} \sqrt{a^2+x^2} + \frac{a^2}{2} \log |x + \sqrt{x^2+a^2}| + C \)

Theorem 1. The indefinite integral of an algebraic sum of two or more functions is equal to the algebraic sum of their integrals, i.e., \( \int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx \)

Theorem 2. A constant term may be taken outside from the integral sign i.e., if \( k \) is a constant then \( \int k f(x) dx = k \int f(x) dx \)

Theorem 3. If the numerator in an integral is the exact derivative of denominator, then its integral is logarithmic of denominator, i.e., \( \int \frac{f'(x) dx}{f(x)} = \log |f(x)| + C \)

Theorem 4. To integrate a function whose numerator is unity and denominator is a homogeneous function of 1st degree in \( \cos x \) and \( \sin x \) i.e., the integrals of these forms:
\( \int \frac{dx}{a + b\cos x}, \int \frac{dx}{a \sin x + b}, \int \frac{dx}{a + b\sin x}, \int \frac{dx}{a\cos x + b\sin x}, \int \frac{dx}{a \sin x + b\cos x} \)
i.e., when integrand is a rational function of \( \sin x \) and \( \cos x \). To find these, we can use following substitution.
(i) By putting \( a = r \cos \alpha, b = r \sin \alpha \) respectively according to question OR
(ii) By putting \( \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}, \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \) and putting \( \tan \frac{x}{2} = t \) and then simplify.

Theorem 5. To integrate a function whose numerator is 1 and denominator is a homogeneous function of the second degree in \( \cos x \) and \( \sin x \) or both, i.e.,
\( \int \frac{dx}{a + b\sin^2 x}, \int \frac{dx}{a \sin^2 x + b}, \int \frac{dx}{a \cos^2 x + b}, \int \frac{dx}{a + b\cos^2 x}, \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x} \)
To evaluate such type of integrals we proceed as follows:
(i) Divide the numerator and denominator by \( \cos^2 x \) and then
(ii) Putting \( \tan x = z \) or \( \cot x = z \) and then simplify.

Theorem 6. Integrals of the type \( \int e^{f(x)} \cdot f'(x) dx, \int f'(x) \cos[f(x)] dx, \int \sin[f(x)]f'(x) dx, \int \log[f(x)]f'(x) dx \).
To evaluate these type of integrals, put \( f(x) = t \) so that \( f'(x) dx = dt \) and then integral converts to the standard forms for which the integrals are known.

Note: If the integrand is a rational function of \( e^x \), then it always needs a replacement as the differentiation and integration of \( e^x \) is the same.
Thus, if on substituting denominator = \( t \), the derivative of denominator is not present in the problem, then we need to generate it by multiplying and dividing by a suitable term containing the exponential function in numerator and denominator.

Integration by Partial Fractions

Rational function: Rational function is defined as the ratio of two polynomials in the form of \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials in \( x \). If the degree of \( P(x) \) is less than degree of \( Q(x) \) then it is said to be Proper, otherwise it is called an Improper Rational Function. Thus if \( \frac{P(x)}{Q(x)} \) is improper, then by long division method it can be reduced to proper function i.e., \( \frac{P(x)}{Q(x)} = T(x) + \frac{P_1(x)}{Q(x)} \), where \( T(x) \) is a function of \( x \) and \( \frac{P_1(x)}{Q(x)} \) is a proper rational function. Such fractions can be evaluated by breaking in factors given as follows:

1. Form of the rational function: \( \frac{px+q}{(x-a)(x-b)}, a \neq b \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{B}{(x-b)} \)

2. Form of the rational function: \( \frac{px+q}{(x-a)^2} \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{B}{(x-a)^2} \)

3. Form of the rational function: \( \frac{px^2+qx+r}{(x-a)(x-b)(x-c)} \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{B}{(x-b)} + \frac{C}{(x-c)} \)

4. Form of the rational function: \( \frac{px^2+qx+r}{(x-a)^2(x-b)} \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{B}{(x-a)^2} + \frac{C}{(x-b)} \)

5. Form of the rational function: \( \frac{px^2+qx+r}{(x-a)^3(x-b)} \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3} + \frac{D}{(x-b)} \)

6. Form of the rational function: \( \frac{px^2+qx+r}{(x-a)(x^2+bx+c)} \). Form of the partial fraction: \( \frac{A}{(x-a)} + \frac{Bx+C}{x^2+bx+c} \), where \( x^2 + bx + c \) cannot be factored further.

The constants \( A, B, C, \) etc. are obtained by equating coefficient of like terms from both sides or by substituting any value for \( x \) on both sides.

To find the integral of the form \( \int \frac{dx}{ax^2+bx+c} \), we write \( ax^2+bx+c = a \left[ x^2 + \frac{b}{a}x + \frac{c}{a} \right] = a \left[ \left( x + \frac{b}{2a} \right)^2 + \left( \frac{c}{a} - \frac{b^2}{4a^2} \right) \right] \). Now putting \( x + \frac{b}{2a} = t \) so that \( dx = dt \). Therefore, writing \( \frac{c}{a} - \frac{b^2}{4a^2} = k \), and find the integral of reduced form \( \frac{1}{|a|} \int \frac{dt}{(t^2 \pm k)} \).

Integrals of the form \( \int \frac{px+q}{ax^2+bx+c} dx \)
Step I. The numerator \( px + q \) is written in the form \( px + q = A \cdot \frac{d}{dx} (ax^2+bx+c) + B \Rightarrow px + q = A(2ax + b) + B \)
Step II. The value of \( A \) and \( B \) is obtained by equating the coefficients in the above equation.
Step III. \( (px + q) \) is replaced by \( A(2ax + b) + B \) and we write the given integral as \( \int \frac{px+q}{ax^2+bx+c} dx = \int \frac{A(2ax+b)+B}{ax^2+bx+c} dx \) and then solved.

Integrals of the form \( \int \frac{px+q}{\sqrt{ax^2+bx+c}} dx \)
Step I. The numerator \( px + q \) is written in the form \( px + q = A \frac{d}{dx} (ax^2+bx+c) + B \Rightarrow px + q = A(2ax + b) + B \)
Step II. The values of \( A \) and \( B \) are obtained by equating the coefficients in the above equation.
Step III. \( (px + q) \) is replaced by \( A(2ax + b) + B \) in given integration as \( \int \frac{(px+q)}{\sqrt{ax^2+bx+c}} dx = \int \frac{A(2ax+b)+B}{\sqrt{ax^2+bx+c}} dx \) and then solved.

Integration of the form \( \int \frac{p(x)}{q(x)} dx \), where \( p(x) \) and \( q(x) \) are polynomials such that degree of \( p(x) \geq \) degree \( q(x) \).
Step I. \( p(x) \) is divided by \( q(x) \) and it is written as \( \frac{p(x)}{q(x)} = Q(x) + \frac{R(x)}{q(x)} \), where \( Q(x) \) is quotient polynomial and \( R(x) \) is remainder polynomial.
Step II. \( \frac{p(x)}{q(x)} \) is replaced by \( \left( Q(x) + \frac{R(x)}{q(x)} \right) \) as \( \int \frac{p(x)}{q(x)} dx = \int \left( Q(x) + \frac{R(x)}{q(x)} \right) dx \) and then solved.

Integral of the form \( \int \sin^m x \cdot \cos^n x \, dx \)
(i) If the exponent of \( \sin x \) is an odd positive integer, then put \( \cos x = t \).
(ii) If the exponent of \( \cos x \) is an odd integer, then put \( \sin x = t \).

\( \int e^x (f(x) + f'(x)) dx = f(x) \cdot e^x + C \)

Definite Integrals

Definition: If \( F(x) \) is the integral of \( f(x) \) over the interval \( [a, b] \), i.e., \( \int f(x) dx = F(x) \) then the definite integral of \( f(x) \) over the interval \( [a, b] \) is denoted by \( \int_a^b f(x) dx \) is defined as \( \int_a^b f(x) dx = F(b) - F(a) \). where \( a \) is called the lower limit and \( b \) is called the upper limit of integration and the interval \( [a, b] \) is called the interval of integration.

Integration as a Limit of Sum: If a function \( f(x) \) is continuous in an interval \( [a, b] \) then it is integrable on that interval. Therefore, we have \( \int_a^b f(x) dx = \lim_{n \to \infty} S_n \). Or, \( \int_a^b f(x) dx = \lim_{h \to 0} h [f(a) + f(a+h) + f(a+2h) + \dots + f(a+(n-1)h)] \). ∴ \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} h [f(a) + f(a+h) + f(a+2h) + \dots + f(a+(n-1)h)] \). Since when \( n \to \infty \), i.e., number of intervals is very large, then the width of the interval is very small which implies that \( h \to 0 \), so that \( nh = b - a \) is a constant.

Some Useful Results: The following results will be useful in evaluating the definite integrals as the limit of sum.
(i) \( \sum (n-1) = 1 + 2 + 3 + \dots + (n-1) = \frac{n(n-1)}{2} \)
(ii) \( \sum (n-1)^2 = 1^2 + 2^2 + 3^2 + \dots + (n-1)^2 = \frac{n(n-1)(2n-1)}{6} \)
(iii) \( \sum (n-1)^3 = 1^3 + 2^3 + 3^3 + \dots + (n-1)^3 = \left[ \frac{n(n-1)}{2} \right]^2 \)
(iv) \( a + ar + ar^2 + \dots + ar^{n-1} = a \left( \frac{r^n-1}{r-1} \right) (\text{if } r > 1) \text{ or } a \left( \frac{1-r^n}{1-r} \right) (\text{if } r < 1) \)
(v) \( \sin a + \sin(a+h) + \sin(a+2h) + \dots + \sin\{a+(n-1)h\} = \frac{\sin \{a + \frac{(n-1)h}{2}\} \sin \frac{nh}{2}}{\sin \frac{h}{2}} \)
(vi) \( \cos a + \cos(a+h) + \cos(a+2h) + \dots + \cos\{a+(n-1)h\} = \frac{\cos \{a + \frac{(n-1)h}{2}\} \sin \frac{nh}{2}}{\sin \frac{h}{2}} \)

Fundamental Properties of Definite Integrals: There are certain properties of definite integrals which can be used while solving the definite integral.
(i) \( \int_a^b f(x) dx = \int_a^b f(z) dz \) (Change of variable)
(ii) \( \int_a^b f(x) dx = -\int_b^a f(x) dx \) (Inter change the limits)
(iii) \( \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx \), where \( a < c < b \) (Change the limits)
(iv) (a) \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \)
(b) \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \)
(v) \( \int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_0^a f(2a-x) dx \), then following cases will occur:
(a) \( \int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx \), if \( f(2a-x) = f(x) \)
(b) \( \int_0^{2a} f(x) dx = 0 \), if \( f(2a-x) = -f(x) \)
(vi) \( \int_0^{na} f(x) dx = n \int_0^a f(x) dx \), if \( f(x) = f(a+x) \)
(vii) (a) \( \int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx \), if \( f \) is an even function, i.e., \( f(-x) = f(x) \)
(b) \( \int_{-a}^a f(x) dx = 0 \), if \( f \) is an odd function, i.e., \( f(-x) = -f(x) \)

Questions-

Question. Find \( \int \frac{dx}{x - \sqrt{x}} \).
Answer: \( I = \int \frac{dx}{x - \sqrt{x}} = \int \frac{dx}{\sqrt{x}(\sqrt{x}-1)} \)
Let \( \sqrt{x} - 1 = t \Rightarrow \frac{1}{2\sqrt{x}} dx = dt \)
\( \Rightarrow \frac{dx}{\sqrt{x}} = 2dt \)
\( \therefore I = 2 \int \frac{dt}{t} = 2 \log |t| + C = 2 \log |\sqrt{x}-1| + C \)

Question. Find : \( \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx \)
Answer: Put \( (e^{2x} + e^{-2x}) = t \Rightarrow (2e^{2x} - 2e^{-2x}) dx = dt \Rightarrow (e^{2x} - e^{-2x}) dx = \frac{dt}{2} \)
\( \therefore \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log |t| + C = \frac{1}{2} \log |e^{2x} + e^{-2x}| + C \)

Question. Find : \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx \)
Answer: \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx \)
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt. \)
\( \therefore \int \frac{\sec^2 x}{\sqrt{\tan x}} dx = \int \frac{1}{\sqrt{t}} dt = \int t^{-1/2} dt = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C = 2\sqrt{\tan x} + C \)

Question. Find : \( \int \frac{(x+1)(x+\log x)^2}{x} dx \)
Answer: \( \int \frac{(x+1)(x+\log x)^2}{x} dx = \int \left( \frac{x+1}{x} \right) (x+\log x)^2 dx = \int \left( 1 + \frac{1}{x} \right) (x+\log x)^2 dx \)
Put \( (x + \log x) = t \Rightarrow \left( 1 + \frac{1}{x} \right) dx = dt \)
\( \therefore \int \left( 1 + \frac{1}{x} \right) (x+\log x)^2 dx = \int t^2 dt = \frac{t^3}{3} + C = \frac{1}{3}(x + \log x)^3 + C \)

Question. Find : \( \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} dx \)
Answer: \( \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} dx, \text{ put } \tan^{-1} (x^4) = t \Rightarrow \frac{4x^3}{1+x^8} dx = dt \Rightarrow \frac{x^3}{1+x^8} dx = \frac{dt}{4} \)
\( \therefore \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} dx = \frac{1}{4} \int \sin t \, dt = \frac{1}{4} (-\cos t) + C = \frac{-1}{4} \cos(\tan^{-1} x^4) + C \)

Question. Find : \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \)
Answer: Let \( I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx = \int \frac{(2 \cos^2 x - 1) - (2 \cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx = 2 \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} dx \)
\( = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{(\cos x - \cos \alpha)} dx = 2 \int (\cos x + \cos \alpha) dx \)
\( = 2 \int \cos x \, dx + 2 \cos \alpha \int 1 \, dx = 2 \sin x + 2x \cos \alpha + C \)

Question. Find : \( \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx \)
Answer: \( \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx = \int \frac{1}{\sqrt{(x^2 + 2x + 1) + 1}} dx = \int \frac{1}{\sqrt{(x+1)^2 + 1}} dx \)
Putting \( x + 1 = t \Rightarrow dx = dt \)
\( \therefore \int \frac{1}{\sqrt{(x+1)^2 + 1}} dx = \int \frac{1}{\sqrt{t^2 + 1}} dt \)
\( = \log |t + \sqrt{t^2+1}| + C = \log |(x+1) + \sqrt{(x^2+2x+1)+1}| + C \)
\( = \log |(x+1) + \sqrt{x^2+2x+2}| + C \)

Question. Find : \( \int \frac{1}{9x^2 + 6x + 5} dx \)
Answer: \( \int \frac{1}{9x^2 + 6x + 5} dx = \frac{1}{9} \int \frac{1}{x^2 + \frac{6}{9}x + \frac{5}{9}} dx \)
\( = \frac{1}{9} \int \frac{1}{x^2 + \frac{2}{3}x + \frac{5}{9} + \left( \frac{1}{3} \right)^2 - \left( \frac{1}{3} \right)^2} dx = \frac{1}{9} \int \frac{1}{\left( x + \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2} dx \), putting \( x + \frac{1}{3} = t \Rightarrow dx = dt \)
\( = \frac{1}{9} \int \frac{1}{t^2 + \left( \frac{2}{3} \right)^2} dt = \frac{1}{9} \cdot \frac{1}{2/3} \tan^{-1} \left( \frac{t}{2/3} \right) + C = \frac{1}{6} \tan^{-1} \left[ \frac{3(x+1/3)}{2} \right] + C = \frac{1}{6} \tan^{-1} \left( \frac{3x+1}{2} \right) + C \)

Question. Evaluate: \( \int \frac{5x+3}{\sqrt{x^2+4x+10}} dx \)
Answer: We can express the Nr as \( 5x + 3 = A \frac{d}{dx}(x^2 + 4x + 10) + B \)
\( \Rightarrow 5x + 3 = A (2x + 4) + B \Rightarrow 5x + 3 = 2Ax + (4A + B) \)
Equating the coefficients, we get
\( 2A = 5 \) and \( 4A + B = 3 \)
\( A = \frac{5}{2} \Rightarrow 4 \times \frac{5}{2} + B = 3 \Rightarrow B = 3 - 10 = -7 \)
\( \therefore 5x + 3 = \frac{5}{2}(2x + 4) + (-7) \)
\( \therefore I = \int \frac{\frac{5}{2}(2x+4) - 7}{\sqrt{x^2+4x+10}} dx = \frac{5}{2} \int \frac{(2x+4)}{\sqrt{x^2+4x+10}} dx - 7 \int \frac{dx}{\sqrt{x^2+4x+10}} \)
\( I = \frac{5}{2} I_1 - 7 I_2 \) ...(i)
where \( I_1 = \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx \) and \( I_2 = \int \frac{dx}{\sqrt{x^2+4x+10}} \)
Now, \( I_1 = \int \frac{(2x+4)}{\sqrt{x^2+4x+10}} dx \)
Let \( x^2 + 4x + 10 = t \Rightarrow (2x + 4)dx = dt \)
\( \therefore I_1 = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt = \frac{t^{1/2+1}}{-\frac{1}{2}+1} + C_1 = 2\sqrt{t} + C_1 \)
\( I_1 = 2\sqrt{x^2 + 4x + 10} + C_1 \)
Again, \( I_2 = \int \frac{dx}{\sqrt{x^2+2.x.2 + 2^2 - 4 + 10}} = \int \frac{dx}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} \)
\( = \log |(x+2) + \sqrt{x^2+4x+10}| + C_2 \)
Putting the value of \( I_1 \) and \( I_2 \) in (i), we get
\( I = \frac{5}{2} \times 2\sqrt{x^2+4x+10} - 7 \log |(x+2) + \sqrt{x^2+4x+10}| + \left( \frac{5}{2}C_1 - 7C_2 \right) \)
\( = 5\sqrt{x^2+4x+10} - 7 \log |(x+2) + \sqrt{x^2+4x+10}| + C \)
where \( C = \left( \frac{5}{2}C_1 - 7C_2 \right) \)

Question. Find : \( \int \frac{1-x^2}{x(1-2x)} dx \)
Answer: \( \int \frac{1-x^2}{x(1-2x)} dx = \int \left[ \frac{1}{2} + \frac{-\frac{1}{2}x + 1}{x(1-2x)} \right] dx = \frac{1}{2} \int dx - \frac{1}{2} \int \frac{x-2}{x(1-2x)} dx = \frac{x}{2} - \frac{1}{2} I_1 \) ...(i)
Now, \( I_1 = \int \frac{x-2}{x(1-2x)} dx \)
\( \frac{x-2}{x(1-2x)} \) is a proper rational function
\( \therefore \frac{x-2}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x} \) ...(ii)
\( \Rightarrow x - 2 = A(1-2x) + Bx \Rightarrow x - 2 = (-2A + B)x + A \)
\( \Rightarrow A = -2 \) and \( -2A + B = 1 \Rightarrow B = 1 + 2A = 1 + 2(-2) = -3 \)
Putting values of \( A \) and \( B \) in (ii), we have
\( \frac{x-2}{x(1-2x)} = \frac{-2}{x} - \frac{3}{1-2x} \)
\( \therefore \int \frac{x-2}{x(1-2x)} dx = \int \left[ \frac{-2}{x} - \frac{3}{1-2x} \right] dx \)
\( = -2 \int \frac{1}{x} dx - 3 \int \frac{1}{1-2x} dx = -2 \log |x| - 3 \frac{\log |1-2x|}{-2} + C_1 \)
\( = -2 \log |x| + \frac{3}{2} \log |1-2x| + C_1 \)
Putting the value \( I_1 \) in (i), we have
\( \int \frac{1-x^2}{x(1-2x)} dx = \frac{x}{2} - \frac{1}{2} \left[ -2 \log |x| + \frac{3}{2} \log |1-2x| + C_1 \right] \)
\( = \frac{x}{2} + \log |x| - \frac{3}{4} \log |1-2x| - \frac{C_1}{2} \)
\( = \frac{x}{2} + \log |x| - \frac{3}{4} \log |1-2x| + C \), where \( C = - \frac{C_1}{2} \)

Question. Find : \( \int \frac{x^3 + x + 1}{x^2 - 1} dx \)
Answer: \( \int \frac{x^3 + x + 1}{x^2 - 1} dx = \int \left[ x + \frac{2x+1}{x^2 - 1} \right] dx = \int x \, dx + \int \frac{2x+1}{x^2 - 1} dx = \frac{x^2}{2} + \int \frac{2x}{x^2-1} dx + \int \frac{1}{x^2-1} dx \)
Putting \( x^2 - 1 = t \Rightarrow 2x dx = dt \) in second integral, we get
\( = \frac{x^2}{2} + \int \frac{1}{t} dt + \int \frac{1}{x^2 - (1)^2} dx = \frac{x^2}{2} + \log |t| + \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + C \)
\( = \frac{x^2}{2} + \log |x^2 - 1| + \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + C \)

Question. Evaluate: \( \int \frac{\cos^{-1} x}{\sqrt{1 - x^2}} dx \)
Answer: Let \( I = \int \frac{\cos^{-1} x}{\sqrt{1 - x^2}} dx \)
Put \( \cos^{-1} x = z \Rightarrow \frac{-1}{\sqrt{1-x^2}} dx = dz \Rightarrow \frac{1}{\sqrt{1-x^2}} dx = -dz \)
\( I = - \int z \, dz = -\frac{z^2}{2} + C = -\frac{1}{2} (\cos^{-1} x)^2 + C \)
\( I = -\frac{1}{2} (\cos^{-1} x)^2 + C \)

Question. Find : \( \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx \)
Answer: Let \( I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = \int e^x \left[ \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right] dx \)
\( = \int e^x \left[ \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right] dx = \int e^x \left[ \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right] dx \)
\( = \int e^x \tan \frac{x}{2} dx + \frac{1}{2} \int e^x \sec^2 \frac{x}{2} dx \)
Integrating by parts the first term (I):
\( = \left( \tan \frac{x}{2} \right) e^x - \int \left( \frac{1}{2} \sec^2 \frac{x}{2} \right) e^x dx + \frac{1}{2} \int e^x \sec^2 \frac{x}{2} dx \Rightarrow I = e^x \tan \frac{x}{2} + C \)

Question. Evaluate: \( \int_1^4 (x^2 - x) dx \) as limit of sums.
Answer: \( \int_1^4 (x^2 - x) dx \)
We have to solve it by using limit of sums.
Here, \( a = 1, b = 4, h = \frac{b-a}{n} = \frac{4-1}{n} \) i.e., \( nh = 3 \)
Limit of sum for \( \int_1^4 (x^2 - x) dx \) is
\( = \lim_{h \to 0} h [f(1) + f(1+h) + f(1+2h) + \dots + f(1+(n-1)h)] \)
Now, \( f(1) = 1 - 1 = 0 \)
\( f(1+h) = (1+h)^2 - (1+h) = h^2 + h \)
\( f(1+2h) = (1+2h)^2 - (1+2h) = 4h^2 + 2h \)

\( f[1+(n-1)h] = \{1+(n-1)h\}^2 - \{1+(n-1)h\} = (n-1)^2 h^2 + (n-1)h \)
\( \therefore \int_1^4 (x^2 - x) dx = \lim_{h \to 0} h [0 + h^2 + h + 4h^2 + 2h + \dots + (n-1)^2 h^2 + (n-1)h] \)
\( = \lim_{h \to 0} h [h^2 \{1 + 4 + \dots + (n-1)^2\} + h \{1 + 2 + \dots + (n-1)\}] \)
\( = \lim_{h \to 0} h \left[ h^2 \frac{(n)(n-1)(2n-1)}{6} + h \frac{n(n-1)}{2} \right] \)
\( [\because 1 + 4 + \dots + (n-1)^2 = \frac{n(n-1)(2n-1)}{6} \text{ and } 1 + 2 + \dots + (n-1) = \frac{n(n-1)}{2}] \)
\( = \lim_{h \to 0} \left[ \frac{nh(nh-h)(2nh-h)}{6} + \frac{nh(nh-h)}{2} \right] \)
\( = \lim_{h \to 0} \left[ \frac{(3-h)(3)(6-h)}{6} + \frac{(3-h)(3)}{2} \right] = \left( \frac{3 \times 3 \times 6}{6} \right) + \left( \frac{3 \times 3}{2} \right) = 9 + \frac{9}{2} = \frac{27}{2} \)

Question. Evaluate: \( \int_{-5}^5 |x + 2| dx \)
Answer: Here, function is \( |x + 2| \) which is defined as
\( |x + 2| = \begin{cases} (x + 2), & \text{if } x \ge -2 \\ -(x + 2), & \text{if } x < -2 \end{cases} \)
So, we have
\( \int_{-5}^5 |x + 2| dx = \int_{-5}^{-2} -(x + 2) dx + \int_{-2}^5 (x + 2) dx \) \( \left[ \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx, a < c < b \right] \)
\( = \left[ -\frac{x^2}{2} - 2x \right]_{-5}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^5 \)
\( = \left( -\frac{(-2)^2}{2} - 2(-2) \right) - \left( -\frac{(-5)^2}{2} - 2(-5) \right) + \left( \frac{(5)^2}{2} + 2(5) \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right) \)
\( = (-2 + 4) + \frac{25}{2} - 10 + \frac{25}{2} + 10 - 2 + 4 = 29 \)

Question. Evaluate: \( \int_0^{\pi/4} \log(1 + \tan x) dx \)
Answer: Let \( I = \int_0^{\pi/4} \log(1 + \tan x) dx \) ...(i)
\( \therefore I = \int_0^{\pi/4} \log[1 + \tan(\frac{\pi}{4} - x)] dx \) \( [\text{By using property } \int_0^a f(x) dx = \int_0^a f(a - x) dx] \)
\( = \int_0^{\pi/4} \log \left[ 1 + \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \cdot \tan x} \right] dx = \int_0^{\pi/4} \log \left[ 1 + \frac{1 - \tan x}{1 + \tan x} \right] dx \)
\( = \int_0^{\pi/4} \log \left[ \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right] dx = \int_0^{\pi/4} \log \left[ \frac{2}{1 + \tan x} \right] dx \)
\( I = \int_0^{\pi/4} [\log 2 - \log(1 + \tan x)] dx \) ...(ii)
Adding (i) and (ii), we get
\( 2I = \int_0^{\pi/4} \log 2 \, dx = \log 2 [x]_0^{\pi/4} = \frac{\pi}{4} \log 2 \)
\( \Rightarrow I = \frac{\pi}{8} \log 2 \)

Question. Evaluate : \( \int_{-1}^{3/2} |x \sin \pi x| dx \)
Answer: \( \sin \theta = 0 \Rightarrow \theta = n\pi, n \in Z \)
\( \therefore \sin \pi x = 0 \Rightarrow x = 0, 1, 2, \dots \)
For \( -1 < x < 0, x < 0, \sin \pi x < 0 \Rightarrow x \sin \pi x > 0 \)
For \( 0 < x < 1, x > 0, \sin \pi x > 0 \Rightarrow x \sin \pi x > 0 \)
For \( 1 < x < \frac{3}{2}, x > 0, \sin \pi x < 0 \Rightarrow x \sin \pi x < 0 \)
\( \therefore \int_{-1}^{3/2} |x \sin \pi x| dx = \int_{-1}^1 x \sin \pi x \, dx + \int_1^{3/2} (-x \sin \pi x) dx \)
\( = \left[ x \cdot \left( \frac{-\cos \pi x}{\pi} \right) \right]_{-1}^1 - \int_{-1}^1 1 \cdot \left( \frac{-\cos \pi x}{\pi} \right) dx - \left[ x \cdot \left( \frac{-\cos \pi x}{\pi} \right) \right]_1^{3/2} + \int_1^{3/2} 1 \cdot \left( \frac{-\cos \pi x}{\pi} \right) dx \)
\( = \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_{-1}^1 - \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_1^{3/2} \)
\( = \left[ \frac{1}{\pi} + 0 + \frac{1}{\pi} - 0 \right] - \left[ 0 - \frac{1}{\pi^2} - \frac{1}{\pi} - 0 \right] = \left[ \frac{1}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} \right] \)
\( = \frac{1}{\pi^2} + \frac{3}{\pi} = \frac{1 + 3\pi}{\pi^2} \)

Question. Find \( \int e^x (\cos x - \sin x) \csc^2 x \, dx \).
Answer: Let \( I = \int e^x (\cos x - \sin x) \csc^2 x \, dx = \int e^x (\cot x \cdot \csc x - \csc x) dx \)
\( = \int e^x \csc x \cot x \, dx - \int e^x \csc x \, dx \)
Integrating by parts for 2nd integral:
\( = \int e^x \csc x \cot x \, dx - [ \csc x \cdot e^x + C + \int \csc x \cot x \cdot e^x dx ] \)
\( = \int e^x \csc x \cot x \, dx - e^x \csc x + C - \int e^x \csc x \cot x \, dx \)
\( = -e^x \csc x + C \).

Question. Evaluate: \( \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \)
Answer: We have, \( I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \Rightarrow I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{1 - (\cos x - \sin x)^2}} dx \)
Let \( t = (\cos x - \sin x) \Rightarrow dt = -(\sin x + \cos x) dx \)
The limits are, when \( x = \pi/6 \Rightarrow t = \cos \frac{\pi}{6} - \sin \frac{\pi}{6} = \frac{\sqrt{3}-1}{2} \)
and \( x = \pi/3 \Rightarrow t = \cos \frac{\pi}{3} - \sin \frac{\pi}{3} = \frac{1-\sqrt{3}}{2} \)
\( \therefore I = - \int_{\frac{\sqrt{3}-1}{2}}^{\frac{1-\sqrt{3}}{2}} \frac{1}{\sqrt{1-t^2}} dt \)
\( = - [\sin^{-1} t]_{\frac{\sqrt{3}-1}{2}}^{\frac{1-\sqrt{3}}{2}} = - \left[ \sin^{-1} \frac{1-\sqrt{3}}{2} - \sin^{-1} \frac{\sqrt{3}-1}{2} \right] = - \left[ -\sin^{-1} \frac{\sqrt{3}-1}{2} - \sin^{-1} \frac{\sqrt{3}-1}{2} \right] \)
\( \Rightarrow I = 2 \sin^{-1} \frac{\sqrt{3}-1}{2} \) [\( \because \sin^{-1} (-x) = -\sin^{-1} x \)]

Question. Evaluate: \( \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
Answer: Let \( I = \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
Here, we express denominator in terms of \( \sin x - \cos x \) which is integral of the numerator.
We have, \( (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x \)
\( \Rightarrow \sin 2x = 1 - (\sin x - \cos x)^2 \)
\( \therefore I = \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 [1 - (\sin x - \cos x)^2]} dx \)
\( \Rightarrow I = \int_0^{\pi/4} \frac{\sin x + \cos x}{25 - 16(\sin x - \cos x)^2} dx \)
Let \( \sin x - \cos x = t \Rightarrow (\cos x + \sin x) dx = dt \)
The limits are, when \( x = 0 \Rightarrow t = \sin 0 - \cos 0 = -1 \) and \( x = \frac{\pi}{4} \Rightarrow t = \sin \frac{\pi}{4} - \cos \frac{\pi}{4} = 0 \)
\( \therefore I = \int_{-1}^0 \frac{dt}{25 - 16t^2} \)
\( \Rightarrow I = \frac{1}{16} \int_{-1}^0 \frac{dt}{\frac{25}{16} - t^2} = \frac{1}{16} \int_{-1}^0 \frac{dt}{\left( \frac{5}{4} \right)^2 - t^2} \Rightarrow I = \frac{1}{16} \cdot \frac{1}{2 \left( \frac{5}{4} \right)} \left[ \log \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^0 \)
\( \Rightarrow I = \frac{1}{40} \left[ \log 1 - \log \left( \frac{1/4}{9/4} \right) \right] \Rightarrow I = \frac{1}{40} \left[ 0 - \log \left( \frac{1}{9} \right) \right] = \frac{1}{40} \log 9 \)

Question. Evaluate: \( \int_0^\pi \frac{x \tan x}{\sec x + \tan x} dx \)
Answer: Let \( I = \int_0^\pi \frac{x \tan x}{\sec x + \tan x} dx \) ...(i)
\( = \int_0^\pi \frac{(\pi - x) \tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} dx \) [\( \because \int_0^a f(x) dx = \int_0^a f(a - x) dx \)]
\( = \int_0^\pi \frac{(\pi - x) \tan x}{\sec x + \tan x} dx \) ...(ii)
By adding equations (i) and (ii), we get
\( 2I = \pi \int_0^\pi \frac{\tan x}{\sec x + \tan x} dx \)
Multiplying and dividing by \( (\sec x - \tan x) \), we get
\( 2I = \pi \int_0^\pi \frac{\tan x (\sec x - \tan x)}{\sec^2 x - \tan^2 x} dx = \pi \int_0^\pi (\sec x \tan x - \tan^2 x) dx \)
\( = \pi \int_0^\pi \sec x \tan x \, dx - \pi \int_0^\pi \sec^2 x \, dx + \pi \int_0^\pi dx \)
\( = \pi [\sec x]_0^\pi - \pi [\tan x]_0^\pi + \pi [x]_0^\pi = \pi (-1 - 1) - 0 + \pi(\pi - 0) = \pi(\pi - 2) \)
\( \Rightarrow 2I = \pi (\pi - 2) \Rightarrow I = \frac{\pi}{2} (\pi - 2) \)

Long Answer Questions 

Question. Evaluate the following integral as the limit of sums: \( \int_1^4 (x^2 - x) dx \)
Answer: Let \( I = \int_1^4 (x^2 - x) dx \)
\( \therefore f(x) = x^2 - x, a = 1, b = 4 \)
\( h = \frac{b - a}{n} = \frac{4 - 1}{n} = \frac{3}{n} \Rightarrow nh = 3 \)
As \( n \to \infty, h \to 0 \)
We know that
\( \int_a^b f(x) dx = \lim_{h \to 0} h [f(a) + f(a + h) + \dots + f(a + (n-1)h)] \)
\( \therefore \int_1^4 f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh) \) ...(i)
\( \therefore f(a + rh) = (a + rh)^2 - (a + rh) \)
\( \Rightarrow f(1 + rh) = (1 + rh)^2 - (1 + rh) = r^2 h^2 + rh \)
Using (i), we have
\( \int_1^4 (x^2 - x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} (r^2 h^2 + rh) \)
\( \therefore I = \lim_{h \to 0} h \{ h^2 \sum_{r=0}^{n-1} r^2 + h \sum_{r=0}^{n-1} r \} \)
\( = \lim_{h \to 0} h \left\{ h^2 \times \frac{n(n-1)(2n-1)}{6} + h \times \frac{n(n-1)}{2} \right\} \)
\( = \lim_{h \to 0} \left[ \frac{n^3 h^3(1 - \frac{1}{n})(2 - \frac{1}{n})}{6} + \frac{n^2 h^2(1 - \frac{1}{n})}{2} \right] \)
\( = \frac{(3)^3(1-0)(2-0)}{6} + \frac{(3)^2(1-0)}{2} \)
\( = 9 + \frac{9}{2} = \frac{27}{2} \)

Question. Evaluate: \( \int (\sqrt{\cot x} + \sqrt{\tan x}) dx \)
Answer: Let \( I = \int (\sqrt{\cot x} + \sqrt{\tan x}) dx \)
\( I = \int \left( \frac{\sqrt{\cos x}}{\sqrt{\sin x}} + \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \right) dx = \int \frac{(\cos x + \sin x)}{\sqrt{\sin x \cos x}} dx \)
Let \( (\sin x - \cos x) = t \Rightarrow (\cos x + \sin x) dx = dt \)
Also \( (\sin x - \cos x)^2 = t^2 \Rightarrow \sin^2 x + \cos^2 x - 2 \sin x \cdot \cos x = t^2 \)
\( \Rightarrow 1 - 2 \sin x \cdot \cos x = t^2 \Rightarrow \sin x \cos x = \frac{1 - t^2}{2} \)
Therefore, \( I = \int \frac{dt}{\sqrt{\frac{1 - t^2}{2}}} = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} \)
\( = \sqrt{2} \sin^{-1} t + C = \sqrt{2} \sin^{-1} (\sin x - \cos x) + C \)

Question. Evaluate: \( \int \frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x} dx \)
Answer: Let \( I = \int \frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x} dx \)
Dividing Nr and Dr by \( \cos^4 x \), we get
\( I = \int \frac{\sec^4 x}{\tan^4 x + \tan^2 x + 1} dx \)
Put \( z = \tan x \Rightarrow dz = \sec^2 x \, dx \)
\( \therefore I = \int \frac{(1 + z^2) dz}{z^4 + z^2 + 1} = \int \frac{z^2(1 + \frac{1}{z^2})}{z^2 \{ z^2 + \frac{1}{z^2} + 1 \}} dz = \int \frac{(1 + \frac{1}{z^2})}{(z - \frac{1}{z})^2 + 3} dz \)
Again, let \( z - \frac{1}{z} = t \Rightarrow (1 + \frac{1}{z^2}) dz = dt \)
\( \Rightarrow I = \int \frac{dt}{t^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + C \)
\( = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{z^2 - 1}{\sqrt{3} z} \right) + C = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\tan^2 x - 1}{\sqrt{3} \tan x} \right) + C \)

Question. Evaluate: \( \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \)
Answer: Let \( I = \int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx \) ..(i)
\( I = \int_0^{\pi/2} \frac{\sin^2(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} dx \) \( [\because \int_0^a f(x) dx = \int_0^a f(a - x) dx] \)
\( I = \int_0^{\pi/2} \frac{\cos^2 x}{\cos x + \sin x} dx \) ...(ii)
Adding (i) and (ii), we get
\( 2I = \int_0^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\pi/2} \frac{dx}{\sin x + \cos x} \)
\( = \int_0^{\pi/2} \frac{dx}{\sqrt{2} ( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x )} = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{dx}{\cos x \cos \frac{\pi}{4} + \sin x \sin \frac{\pi}{4}} \)
\( = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{dx}{\cos(x - \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \sec(x - \frac{\pi}{4}) dx \) [\( \because \cos(A - B) = \cos A \cos B + \sin A \sin B \)]
\( = \frac{1}{\sqrt{2}} [ \log | \sec(x - \frac{\pi}{4}) + \tan(x - \frac{\pi}{4}) | ]_0^{\pi/2} \) [\( \because \int \sec x dx = \log(\sec x + \tan x) \)]
\( = \frac{1}{\sqrt{2}} [ \log | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} | - \log | \sec(-\frac{\pi}{4}) + \tan(-\frac{\pi}{4}) | ] \)
\( = \frac{1}{\sqrt{2}} [ \log(\sqrt{2} + 1) - \log(\sqrt{2} - 1) ] = \frac{1}{\sqrt{2}} \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \)
\( = \frac{1}{\sqrt{2}} \log \left| \frac{(\sqrt{2} + 1)^2}{2 - 1} \right| = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)^2 = \frac{2}{\sqrt{2}} \log(\sqrt{2} + 1) \)
Hence, \( I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1) \)

Questions

Question. Find the scalar components of the vector \(\vec{AB}\) with initial point \(A(2, 1)\) and terminal point \(B(-5, 7)\). 

Answer: Let \(\vec{AB} = (-5 - 2)\hat{i} + (7 - 1)\hat{j} = -7\hat{i} + 6\hat{j}\)
Hence, scalar components are \(-7, 6\).
[Note: If \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) then \(x, y, z\) are called scalar components and \(x\hat{i}, y\hat{j}, z\hat{k}\) are called vector components.]

Question. Find a vector in the direction of vector \(5\hat{i} - \hat{j} + 2\hat{k}\) which has a magnitude of 8 units.

Answer: \(|\vec{a}| = \sqrt{(5)^2 + (-1)^2 + (2)^2} = \sqrt{25 + 1 + 4} = \sqrt{30}\)
The unit vector in the direction of vector \(\vec{a}\) is
\(\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k})\)
Now, vector in the direction of \(\vec{a}\) having magnitude 8 units is
\(8\hat{a} = \frac{8}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}\)

Question. Find the position vector of a point \(R\) which divides the line joining two points \(P\) and \(Q\) whose position vectors are \(\hat{i} + 2\hat{j} - \hat{k}\) and \(-\hat{i} + \hat{j} + \hat{k}\) respectively, in the ratio 2 : 1
(i) internally (ii) externally

Answer: (i) Let \(R\) be the point which divides the line joining the point \(P\) and \(Q\) internally in the ratio 2 : 1.
The position vector of point \(R = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}\).
(ii) Let \(R\) be the point which divides the line joining the points \(P\) and \(Q\) externally in the ratio 2 : 1.
\(R = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 3\hat{k}\)

Question. Find \(|\vec{x}|\), if for a unit vector \(\vec{a}, (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\).

Answer: Here \(|\vec{a}| = 1\) and \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\)
Now, \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \Rightarrow \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12\)
\(\Rightarrow |\vec{x}|^2 + \vec{x} \cdot \vec{a} - \vec{x} \cdot \vec{a} - |\vec{a}|^2 = 12\)
\(\Rightarrow |\vec{x}|^2 - |\vec{a}|^2 = 12 \Rightarrow |\vec{x}|^2 - (1)^2 = 12\)
\(\Rightarrow |\vec{x}|^2 = 12 + 1 \Rightarrow |\vec{x}|^2 = 13 \Rightarrow |\vec{x}| = \sqrt{13}\)

Question. Using vectors, prove that the points (2, -1, 3), (3, -5, 1) and (-1, 11, 9) are collinear. 

Answer: Let \(A (2, -1, 3)\), \(B (3, -5, 1)\) and \(C (-1, 11, 9)\) are three points.
To show that \(A, B, C\) are collinear.
∴ \(\vec{AB} = (3 - 2)\hat{i} + (-5 + 1)\hat{j} + (1 - 3)\hat{k} = \hat{i} - 4\hat{j} - 2\hat{k} = \sqrt{1^2 + (-4)^2 + (-2)^2} = \sqrt{21}\)
\(\vec{BC} = (-1 - 3)\hat{i} + (11 + 5)\hat{j} + (9 - 1)\hat{k} = -4\hat{i} + 16\hat{j} + 8\hat{k} = \sqrt{(-4)^2 + (16)^2 + (8)^2} = 4\sqrt{21}\)
\(\vec{AC} = (-1 - 2)\hat{i} + (11 + 1)\hat{j} + (9 - 3)\hat{k} = -3\hat{i} + 12\hat{j} + 6\hat{k} = \sqrt{(-3)^2 + (12)^2 + (6)^2} = 3\sqrt{21}\)
∵ \(|\vec{AC}| + |\vec{AB}| = |\vec{BC}|\)
\(\Rightarrow A, B, C\) are collinear.

Question. Find \(\lambda\) and \(\mu\) if \((2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}\).

Answer: Here \((2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}\)
∴ \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0}\)
\(\Rightarrow (6\mu - 27\lambda)\hat{i} - (2\mu - 27)\hat{j} + (2\lambda - 6)\hat{k} = \vec{0}\)
\(\Rightarrow 6\mu - 27\lambda = 0, \quad 2\mu - 27 = 0 \text{ and } 2\lambda - 6 = 0\)
\(\Rightarrow \mu = \frac{27}{2} \text{ and } \lambda = 3\)

Question. Find a unit vector perpendicular to each of the vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\), where \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\). 
Answer: Given, \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\)
∴ \(\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k}\) and \(\vec{a} - \vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k}\)
Now, vector perpendicular to \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is
\((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = (16 - 0)\hat{i} - (16 - 0)\hat{j} + (0 - 8)\hat{k} = 16\hat{i} - 16\hat{j} - 8\hat{k}\)
∴ Unit vector perpendicular to \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is given by
\(\pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{\sqrt{16^2 + (-16)^2 + (-8)^2}} = \pm \frac{8(2\hat{i} - 2\hat{j} - \hat{k})}{8\sqrt{2^2 + 2^2 + 1^2}}\)
\(= \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{\sqrt{9}} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) = \pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k}\)

Question. The two adjacent sides of a parallelogram are \(2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\hat{i} - 2\hat{j} - 3\hat{k}\). Find the unit vector parallel to one of its diagonals. Also, find its area. 

Answer: Let two adjacent sides \(\vec{AB}\) and \(\vec{AC}\) of a parallelogram \(ABDC\) be represented by \(2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\hat{i} - 2\hat{j} - 3\hat{k}\) in magnitude and direction respectively.
i.e., \(\vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\vec{AC} = \hat{i} - 2\hat{j} - 3\hat{k}\)
By parallelogram law of addition
\(\vec{AD} = \vec{AB} + \vec{BD} \Rightarrow \vec{AD} = \vec{AB} + \vec{AC}\) [\(\because \vec{BD} = \vec{AC}\)]
\(\vec{AD} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\hat{i} - 2\hat{j} - 3\hat{k}) = 3\hat{i} - 6\hat{j} + 2\hat{k}\)
\(|\vec{AD}| = |3\hat{i} - 6\hat{j} + 2\hat{k}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\)
Therefore, unit vector parallel to diagonal \(\vec{AD} = \frac{1}{|\vec{AD}|}\vec{AD} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}\)
Again, \(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = (12 + 10)\hat{i} - (-6 - 5)\hat{j} + (-4 + 4)\hat{k} = 22\hat{i} + 11\hat{j}\)
\(|\vec{AB} \times \vec{AC}| = |22\hat{i} + 11\hat{j}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5}\)
Now the area of parallelogram \(ABDC\) whose adjacent sides are \(\vec{AB}\) and \(\vec{AC}\) is \(= |\vec{AB} \times \vec{AC}| = 11\sqrt{5}\) sq units.

Question. Let \(\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}, \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\). Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{c} \cdot \vec{d} = 15\). 

Answer: The vector \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), so we have \(\vec{d} = \lambda(\vec{a} \times \vec{b})\).
Now \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = (28 + 4)\hat{i} - (7 - 6)\hat{j} + (-2 - 12)\hat{k} = 32\hat{i} - \hat{j} - 14\hat{k}\)
So, \(\vec{d} = 32\lambda\hat{i} - \lambda\hat{j} - 14\lambda\hat{k}\)
We have \(\vec{c} \cdot \vec{d} = 15 \Rightarrow (2\hat{i} - \hat{j} + 4\hat{k}) \cdot (32\lambda\hat{i} - \lambda\hat{j} - 14\lambda\hat{k}) = 15\)
\(\Rightarrow 64\lambda + \lambda - 56\lambda = 15 \Rightarrow 9\lambda = 15 \Rightarrow \lambda = \frac{15}{9} = \frac{5}{3}\)
∴ Required vector \(\vec{d} = \frac{5}{3}(32\hat{i} - \hat{j} - 14\hat{k}) = \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}\).

Question. The scalar product of the vector \(\hat{i} + \hat{j} + \hat{k}\) with the unit vector along the sum of vectors \(2\hat{i} + 4\hat{j} - 5\hat{k}\) and \(\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) is equal to one. Find the value of \(\lambda\). 

Answer: Let sum of vectors \(2\hat{i} + 4\hat{j} - 5\hat{k}\) and \(\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) equal to \(\vec{a}\) then \(\vec{a} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}\)
The unit vector in the direction of \(\vec{a} = \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 36 + 4}}\)
Here, \((\hat{i} + \hat{j} + \hat{k}) \cdot \hat{a} = 1 \Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}} = 1\)
\(\Rightarrow \frac{(2 + \lambda) + 6 - 2}{\sqrt{(2 + \lambda)^2 + 40}} = 1 \Rightarrow (\lambda + 6)^2 = (2 + \lambda)^2 + 40\)
\(\Rightarrow \lambda^2 + 36 + 12\lambda = 4 + \lambda^2 + 4\lambda + 40 \Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1\).

Question. If with reference to the right handed system of mutually perpendicular unit vectors \(\hat{i}, \hat{j}\) and \(\hat{k}\), \(\vec{\alpha} = 3\hat{i} - \hat{j}\), \(\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}\), then express \(\vec{\beta}\) in the form \(\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\).

Answer: Let \(\vec{\beta}_1 = \lambda\vec{\alpha}\), \(\lambda\) is a scalar, i.e., \(\vec{\beta}_1 = 3\lambda\hat{i} - \lambda\hat{j}\)
and \(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k}\)
Now, since \(\vec{\beta}_2\) is to be perpendicular to \(\vec{\alpha}\), we should have \(\vec{\alpha} \cdot \vec{\beta}_2 = 0\) i.e.,
\(3(2 - 3\lambda) - (1 + \lambda) = 0 \Rightarrow 6 - 9\lambda - 1 - \lambda = 0 \Rightarrow 10\lambda = 5 \Rightarrow \lambda = \frac{1}{2}\)
Therefore, \(\vec{\beta}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\) and \(\vec{\beta}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\)