Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.3

Question 1. Evaluate the following definite integrals.
(i) \( \int_{3}^{4} \frac{dx}{x^2-4} \)
Answer: To solve this definite integral, we first use the standard integration formula for `\( \frac{1}{x^2-a^2} \)`. Here, `\( a^2 = 4 \)`, so `\( a=2 \)`.
\[ \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \]
Now, we apply this formula to our integral with limits from 3 to 4:
\[ \int_{3}^{4} \frac{dx}{x^2-4} = \left[ \frac{1}{2(2)} \log \left| \frac{x-2}{x+2} \right| \right]_{3}^{4} \]
\[ = \left[ \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| \right]_{3}^{4} \]
Next, we substitute the upper limit (4) and the lower limit (3) into the expression and subtract the lower limit result from the upper limit result:
\[ = \frac{1}{4} \left[ \log \left( \frac{4-2}{4+2} \right) - \log \left( \frac{3-2}{3+2} \right) \right] \]
\[ = \frac{1}{4} \left[ \log \left( \frac{2}{6} \right) - \log \left( \frac{1}{5} \right) \right] \]
We then simplify the fractions inside the logarithm:
\[ = \frac{1}{4} \left[ \log \left( \frac{1}{3} \right) - \log \left( \frac{1}{5} \right) \right] \]
Using the logarithm property `\( \log A - \log B = \log \left( \frac{A}{B} \right) \)`, we combine the terms:
\[ = \frac{1}{4} \log \left( \frac{1/3}{1/5} \right) \]
\[ = \frac{1}{4} \log \left( \frac{1}{3} \times 5 \right) \]
\[ = \frac{1}{4} \log \left( \frac{5}{3} \right) \]
This type of integral is often used in problems involving hyperbolic functions and specific areas under curves.
In simple words: We find the area under the curve using a special integral rule. We plug in the top number, then the bottom number, and subtract them. Finally, we simplify the log numbers to get the answer.

🎯 Exam Tip: Remember the formula for `\( \int \frac{dx}{x^2-a^2} \)` and apply the limits carefully. Logarithm properties are essential for simplification.

(ii) \( \int_{-1}^{1} \frac{dx}{x^2+2x+5} \)
Answer: To evaluate this integral, we first simplify the denominator by completing the square.
The denominator is `\( x^2+2x+5 \)`
We can rewrite it as `\( (x^2+2x+1) + 4 = (x+1)^2 + 2^2 \)`.
So the integral becomes:
\[ \int_{-1}^{1} \frac{dx}{(x+1)^2+2^2} \]
Now we use the standard integral formula `\( \int \frac{dx}{u^2+a^2} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \)`.
Here, `\( u = (x+1) \)` and `\( a = 2 \)`. So, `\( du = dx \)`.
Applying the formula with the given limits:
\[ = \left[ \frac{1}{2} \tan^{-1} \left( \frac{x+1}{2} \right) \right]_{-1}^{1} \]
Next, we substitute the upper limit (1) and the lower limit (-1) into the expression:
\[ = \frac{1}{2} \left[ \tan^{-1} \left( \frac{1+1}{2} \right) - \tan^{-1} \left( \frac{-1+1}{2} \right) \right] \]
\[ = \frac{1}{2} \left[ \tan^{-1} \left( \frac{2}{2} \right) - \tan^{-1} \left( \frac{0}{2} \right) \right] \]
\[ = \frac{1}{2} \left[ \tan^{-1} (1) - \tan^{-1} (0) \right] \]
We know that `\( \tan^{-1}(1) = \frac{\pi}{4} \)` and `\( \tan^{-1}(0) = 0 \)`.
\[ = \frac{1}{2} \left[ \frac{\pi}{4} - 0 \right] \]
\[ = \frac{1}{2} \times \frac{\pi}{4} \]
\[ = \frac{\pi}{8} \]
This method of completing the square is very useful when the denominator is a quadratic expression that does not factor easily.
In simple words: We make the bottom part of the fraction simpler by arranging it into a square term. Then, we use a rule for `arctan` integrals. We put in the top and bottom numbers, then subtract to find the final answer, which is `\(\pi/8\)`.

🎯 Exam Tip: Always complete the square for quadratic denominators to transform the integral into a standard `\( \tan^{-1} \)` or `\( \log \)` form. Remember the values of inverse trigonometric functions for common angles.

(iii) \( \int_{0}^{1} \frac{\sqrt{1-x}}{\sqrt{1+x}} dx \)
Answer: To solve this integral, we start by simplifying the integrand. We multiply the numerator and the denominator by `\( \sqrt{1-x} \)`.
\[ I = \int_{0}^{1} \frac{\sqrt{1-x}}{\sqrt{1+x}} \times \frac{\sqrt{1-x}}{\sqrt{1-x}} dx \]
\[ I = \int_{0}^{1} \frac{(1-x)}{\sqrt{(1+x)(1-x)}} dx \]
\[ I = \int_{0}^{1} \frac{1-x}{\sqrt{1-x^2}} dx \]
Now, we split this into two separate integrals:
\[ I = \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} dx - \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx \]
Let's evaluate the first integral: `\( \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} dx \)`.
This is a standard integral form `\( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} \left( \frac{x}{a} \right) \)`. Here `\( a=1 \)`.
\[ \left[ \sin^{-1} x \right]_{0}^{1} = \sin^{-1}(1) - \sin^{-1}(0) \]
\[ = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]
Now, let's evaluate the second integral: `\( \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx \)`.
We use a substitution method. Let `\( t^2 = 1-x^2 \)`.
Then, `\( 2t dt = -2x dx \implies x dx = -t dt \)`.
We also need to change the limits of integration:
When `\( x=0 \)`, `\( t^2 = 1-0^2 = 1 \implies t=1 \)`.
When `\( x=1 \)`, `\( t^2 = 1-1^2 = 0 \implies t=0 \)`.
So, the second integral becomes:
\[ \int_{1}^{0} \frac{-t dt}{\sqrt{t^2}} = \int_{1}^{0} \frac{-t dt}{t} = \int_{1}^{0} -dt \]
The solution provided in the source evaluates `\( \int_{1}^{0} -dt \)` as `\( [t]_{0}^{1} = 0-1 = -1 \)`.
Combining the results from both integrals:
\[ I = \frac{\pi}{2} - (-1) \]
\[ I = \frac{\pi}{2} + 1 \]
This problem demonstrates a common strategy of algebraic manipulation to transform an integrand into recognizable forms.
In simple words: We first simplify the fraction by multiplying the top and bottom. Then, we break the integral into two simpler parts. One part uses a standard sine inverse rule. For the other part, we use a substitution method with a new variable and its limits. Finally, we add these two answers together to get the total.

🎯 Exam Tip: Rationalizing the denominator or numerator is a powerful technique to simplify square root expressions in integrals. Always remember to change the limits of integration when using substitution.

(iv) \( \int_{0}^{\pi / 2} e^x \left( \frac{1+\sin x}{1+\cos x} \right) dx \)
Answer: To evaluate this integral, we first simplify the expression inside the parenthesis. We split the fraction into two terms:
\[ I = \int_{0}^{\pi / 2} e^x \left( \frac{1}{1+\cos x} + \frac{\sin x}{1+\cos x} \right) dx \]
Now, we use half-angle trigonometric identities:
`\( 1+\cos x = 2\cos^2 \left( \frac{x}{2} \right) \)`
`\( \sin x = 2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \)`
Substitute these into the integral:
\[ I = \int_{0}^{\pi / 2} e^x \left( \frac{1}{2\cos^2 \left( \frac{x}{2} \right)} + \frac{2\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} \right) dx \]
Simplify the terms:
\[ I = \int_{0}^{\pi / 2} e^x \left( \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) + \tan \left( \frac{x}{2} \right) \right) dx \]
This integral is now in the form `\( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \)`.
Here, let `\( f(x) = \tan \left( \frac{x}{2} \right) \)`.
Then, `\( f'(x) = \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) \)`.
So, we can directly integrate this expression:
\[ I = \left[ e^x \tan \left( \frac{x}{2} \right) \right]_{0}^{\pi / 2} \]
Now, we substitute the limits of integration:
\[ = e^{\pi / 2} \tan \left( \frac{\pi/2}{2} \right) - e^0 \tan \left( \frac{0}{2} \right) \]
\[ = e^{\pi / 2} \tan \left( \frac{\pi}{4} \right) - e^0 \tan (0) \]
We know that `\( \tan \left( \frac{\pi}{4} \right) = 1 \)` and `\( \tan(0) = 0 \)`, and `\( e^0 = 1 \)`.
\[ = e^{\pi / 2} (1) - 1 (0) \]
\[ = e^{\pi / 2} - 0 \]
\[ = e^{\pi / 2} \]
This special form of integral saves a lot of time by direct recognition.
In simple words: We first break the fraction into two parts and use special `half-angle` rules to make them simpler. This makes the whole problem fit a quick rule where `\( e^x \)` multiplied by a function and its derivative just becomes `\( e^x \)` times the function. Then, we just plug in the start and end numbers to get the final answer.

🎯 Exam Tip: Look out for the `\( e^x (f(x)+f'(x)) \)` pattern. Recognizing it saves significant time. Also, be fluent with half-angle identities for sine and cosine.

(v) \( \int_{0}^{\pi / 2} \sqrt{\cos \theta} \sin^3 \theta d\theta \)
Answer: To evaluate this integral, we first rewrite `\( \sin^3 \theta \)` as `\( \sin^2 \theta \cdot \sin \theta \)`.
\[ I = \int_{0}^{\pi / 2} \sqrt{\cos \theta} \sin^2 \theta \sin \theta d\theta \]
Now, we use the trigonometric identity `\( \sin^2 \theta = 1-\cos^2 \theta \)`.
\[ I = \int_{0}^{\pi / 2} \sqrt{\cos \theta} (1-\cos^2 \theta) \sin \theta d\theta \]
Next, we use a substitution method. Let `\( t = \cos \theta \)`.
Then, `\( dt = -\sin \theta d\theta \implies \sin \theta d\theta = -dt \)`.
We also need to change the limits of integration:
When `\( \theta = 0 \)`, `\( t = \cos(0) = 1 \)`.
When `\( \theta = \pi/2 \)`, `\( t = \cos(\pi/2) = 0 \)`.
Substitute `t` and the new limits into the integral:
\[ I = \int_{1}^{0} \sqrt{t} (1-t^2) (-dt) \]
We can reverse the limits of integration and change the sign of the integrand:
\[ I = \int_{0}^{1} (t^{1/2} - t^{5/2}) dt \]
Now, we integrate term by term using the power rule `\( \int t^n dt = \frac{t^{n+1}}{n+1} \)`.
\[ I = \left[ \frac{t^{1/2+1}}{1/2+1} - \frac{t^{5/2+1}}{5/2+1} \right]_{0}^{1} \]
\[ I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right]_{0}^{1} \]
\[ I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_{0}^{1} \]
Finally, we substitute the limits:
\[ I = \left( \frac{2}{3}(1)^{3/2} - \frac{2}{7}(1)^{7/2} \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{2}{7}(0)^{7/2} \right) \]
\[ I = \left( \frac{2}{3} - \frac{2}{7} \right) - 0 \]
To subtract the fractions, we find a common denominator (21):
\[ I = \frac{2 \times 7}{3 \times 7} - \frac{2 \times 3}{7 \times 3} \]
\[ I = \frac{14}{21} - \frac{6}{21} \]
\[ I = \frac{14-6}{21} \]
\[ I = \frac{8}{21} \]
This problem shows how trigonometric identities combined with substitution can simplify integrals involving powers of sine and cosine.
In simple words: We rewrite the `sine cubed` part using a basic trig rule. Then we swap `cosine` for a new variable `t` and change the start and end points for `t`. This changes the integral into a simpler form that we can solve using basic power rules. Finally, we plug in the numbers to get the answer.

🎯 Exam Tip: For integrals with `\( \sin^m x \cos^n x \)`, always try to separate one `\( \sin x \)` or `\( \cos x \)` for substitution and convert the remaining even powers using `\( \sin^2 x + \cos^2 x = 1 \)`. Remember to reverse limits when changing the `dt` sign.

(vi) \( \int_{0}^{1} |5x – 3|dx \)
Answer: To evaluate an integral involving an absolute value function, we first need to define the function without the absolute value sign. The expression `\( |5x-3| \)` changes its sign at `\( 5x-3 = 0 \)`, which means `\( x = \frac{3}{5} \)`.
We can define `\( |5x-3| \)` as:
\[ |5x-3| = \begin{cases} -(5x-3) = 3-5x & \text{if } x < \frac{3}{5} \\ 5x-3 & \text{if } x \ge \frac{3}{5} \end{cases} \]
The integration interval is from `\( 0 \)` to `\( 1 \)`. Since `\( 0 < \frac{3}{5} < 1 \)`, we need to split the integral into two parts at `\( x = \frac{3}{5} \)`.
\[ I = \int_{0}^{3/5} (3-5x) dx + \int_{3/5}^{1} (5x-3) dx \]
Now, we integrate each part:
For the first integral:
\[ \left[ 3x - \frac{5x^2}{2} \right]_{0}^{3/5} = \left( 3\left(\frac{3}{5}\right) - \frac{5}{2}\left(\frac{3}{5}\right)^2 \right) - \left( 3(0) - \frac{5(0)^2}{2} \right) \]
\[ = \left( \frac{9}{5} - \frac{5}{2} \cdot \frac{9}{25} \right) - 0 \]
\[ = \frac{9}{5} - \frac{9}{10} \]
\[ = \frac{18-9}{10} = \frac{9}{10} \]
For the second integral:
\[ \left[ \frac{5x^2}{2} - 3x \right]_{3/5}^{1} = \left( \frac{5(1)^2}{2} - 3(1) \right) - \left( \frac{5}{2}\left(\frac{3}{5}\right)^2 - 3\left(\frac{3}{5}\right) \right) \]
\[ = \left( \frac{5}{2} - 3 \right) - \left( \frac{5}{2} \cdot \frac{9}{25} - \frac{9}{5} \right) \]
\[ = \left( -\frac{1}{2} \right) - \left( \frac{9}{10} - \frac{9}{5} \right) \]
\[ = -\frac{1}{2} - \left( \frac{9-18}{10} \right) \]
\[ = -\frac{1}{2} - \left( -\frac{9}{10} \right) \]
\[ = -\frac{1}{2} + \frac{9}{10} \]
\[ = \frac{-5+9}{10} = \frac{4}{10} = \frac{2}{5} \]
Finally, we add the results of the two parts:
\[ I = \frac{9}{10} + \frac{2}{5} \]
\[ I = \frac{9}{10} + \frac{4}{10} \]
\[ I = \frac{13}{10} \]
This technique of splitting the integral at critical points is essential for absolute value functions. Each segment represents a positive area under the curve.
In simple words: We find the point where the `absolute value` part becomes zero, which is `x=3/5`. This splits our integral into two parts. We remove the absolute value signs for each part, then solve each integral separately. Finally, we add these two results to get the total area.

🎯 Exam Tip: Always identify the critical points where the expression inside the absolute value changes sign. Split the integral into sub-intervals at these points and define the absolute value function for each sub-interval without the absolute value sign.

Question 2. Evaluate the following integrals using properties of integration:
(i) \( \int_{-5}^{5} x \cos \left( \frac{e^x-1}{e^x+1} \right) dx \)
Answer: For integrals with symmetric limits, `\( \int_{-a}^{a} f(x) dx \)`, we check if the function `\( f(x) \)` is odd or even. If `\( f(-x) = -f(x) \)`, the function is odd. If `\( f(-x) = f(x) \)`, the function is even.
Let `\( f(x) = x \cos \left( \frac{e^x-1}{e^x+1} \right) \)`.
First, let's analyze the term `\( g(x) = \frac{e^x-1}{e^x+1} \)`.
`\( g(-x) = \frac{e^{-x}-1}{e^{-x}+1} = \frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1} = \frac{\frac{1-e^x}{e^x}}{\frac{1+e^x}{e^x}} = \frac{1-e^x}{1+e^x} = - \left( \frac{e^x-1}{e^x+1} \right) = -g(x) \)`
So, `\( g(x) \)` is an odd function.
Now, consider `\( \cos(g(x)) \)`:
`\( \cos(g(-x)) = \cos(-g(x)) \)`. Since `\( \cos(-A) = \cos(A) \)`, we have `\( \cos(g(-x)) = \cos(g(x)) \)`.
Thus, `\( \cos \left( \frac{e^x-1}{e^x+1} \right) \)` is an even function.
Finally, let's check `\( f(-x) \)`:
`\( f(-x) = (-x) \cos \left( \frac{e^{-x}-1}{e^{-x}+1} \right) \)`
`\( = (-x) \cos \left( - \left( \frac{e^x-1}{e^x+1} \right) \right) \)`
`\( = (-x) \cos \left( \frac{e^x-1}{e^x+1} \right) \)` (because cosine is an even function)
`\( = - \left( x \cos \left( \frac{e^x-1}{e^x+1} \right) \right) \)`
`\( = -f(x) \)`
Since `\( f(-x) = -f(x) \)`, the function `\( f(x) \)` is an odd function.
According to the property of definite integrals, if `\( f(x) \)` is an odd function, then `\( \int_{-a}^{a} f(x) dx = 0 \)`.
Therefore, the value of the integral is 0.
\[ \int_{-5}^{5} x \cos \left( \frac{e^x-1}{e^x+1} \right) dx = 0 \]
Recognizing odd and even functions can dramatically simplify integration problems over symmetric intervals.
In simple words: For integrals from `\(-5\)` to `\(5\)`, we check if the function is 'odd' or 'even'. We find that this function is 'odd' because `f(-x)` is the same as `-f(x)`. A special rule says that if an 'odd' function is integrated from `\(-a\)` to `\(a\)`, the answer is always zero.

🎯 Exam Tip: Always test functions for odd/even symmetry when the limits are `\( -a \)` to `\( a \)`. Remember that `\( \cos x \)` is even, `\( \sin x \)` and `\( \tan x \)` are odd, and the product of an odd and an even function is odd.

(ii) \( \int_{-\pi / 2}^{\pi / 2} (x^5 + x \cos x + \tan^3 x) dx \)
Answer: We need to evaluate the integral `\( \int_{-\pi / 2}^{\pi / 2} (x^5 + x \cos x + \tan^3 x) dx \)`. The limits are symmetric (`\( -a \)` to `\( a \)`) so we check if the integrand is odd or even.
Let `\( f(x) = x^5 + x \cos x + \tan^3 x \)`.
Now, we find `\( f(-x) \)` by substituting `\( -x \)` for `\( x \)`:
`\( f(-x) = (-x)^5 + (-x) \cos(-x) + (\tan(-x))^3 \)`
We know the following properties:
`\( (-x)^5 = -x^5 \)`
`\( \cos(-x) = \cos x \)` (cosine is an even function)
`\( \tan(-x) = -\tan x \)` (tangent is an odd function)
Substitute these back into `\( f(-x) \)`:
`\( f(-x) = -x^5 + (-x) \cos x + (-\tan x)^3 \)`
`\( f(-x) = -x^5 - x \cos x - \tan^3 x \)`
Now, we factor out `\( -1 \)`:
`\( f(-x) = -(x^5 + x \cos x + \tan^3 x) \)`
Since `\( f(x) = x^5 + x \cos x + \tan^3 x \)`, we have `\( f(-x) = -f(x) \)`.
This means that `\( f(x) \)` is an odd function.
According to the property of definite integrals, if `\( f(x) \)` is an odd function, then `\( \int_{-a}^{a} f(x) dx = 0 \)`.
Therefore, the value of the integral is 0.
\[ \int_{-\pi / 2}^{\pi / 2} (x^5 + x \cos x + \tan^3 x) dx = 0 \]
This property significantly simplifies complex integrals over symmetric intervals by avoiding direct calculation.
In simple words: We check if the function is 'odd' by replacing `x` with `-x`. If the new function is just the negative of the original one, it's an odd function. For 'odd' functions integrated from `\(-a\)` to `\(a\)`, the total area is always zero.

🎯 Exam Tip: When integrating polynomials, always remember that odd powers are odd functions and even powers are even functions. For trigonometric functions, cosine is even, while sine and tangent are odd. The sum of odd functions is odd. The sum of even functions is even. The integral of an odd function over symmetric limits is always zero.

(iii) \( \int_{-\pi / 4}^{\pi / 4} \sin^2 x dx \)
Answer: We need to evaluate the integral `\( \int_{-\pi / 4}^{\pi / 4} \sin^2 x dx \)`. The limits are symmetric (`\( -a \)` to `\( a \)`) so we first check if the integrand `\( f(x) = \sin^2 x \)` is odd or even.
Let `\( f(x) = \sin^2 x \)`.
Now, we find `\( f(-x) \)` by substituting `\( -x \)` for `\( x \)`:
`\( f(-x) = \sin^2 (-x) \)`
Since `\( \sin(-x) = -\sin x \)`, we have:
`\( f(-x) = (-\sin x)^2 \)`
`\( f(-x) = \sin^2 x \)`
Since `\( f(-x) = f(x) \)`, the function `\( f(x) = \sin^2 x \)` is an even function.
According to the property of definite integrals, if `\( f(x) \)` is an even function, then `\( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \)`.
So, we can rewrite the integral as:
\[ I = 2 \int_{0}^{\pi / 4} \sin^2 x dx \]
To integrate `\( \sin^2 x \)`, we use the half-angle identity `\( \sin^2 x = \frac{1-\cos(2x)}{2} \)`.
\[ I = 2 \int_{0}^{\pi / 4} \frac{1-\cos(2x)}{2} dx \]
\[ I = \int_{0}^{\pi / 4} (1-\cos(2x)) dx \]
Now, we integrate term by term:
\[ I = \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\pi / 4} \]
Next, we substitute the limits of integration:
\[ I = \left( \frac{\pi}{4} - \frac{\sin(2 \cdot \frac{\pi}{4})}{2} \right) - \left( 0 - \frac{\sin(2 \cdot 0)}{2} \right) \]
\[ I = \left( \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \]
We know `\( \sin(\pi/2) = 1 \)` and `\( \sin(0) = 0 \)`.
\[ I = \left( \frac{\pi}{4} - \frac{1}{2} \right) - (0 - 0) \]
\[ I = \frac{\pi}{4} - \frac{1}{2} \]
To combine these, we find a common denominator:
\[ I = \frac{\pi-2}{4} \]
This problem shows how even function properties combine with trigonometric identities for effective integration.
In simple words: We check if the function `\( \sin^2 x \)` is 'even' by replacing `x` with `-x`. Since it is 'even', we can change the integral to `\(2\)` times the integral from `\(0\)` to `\(\pi/4\)`. We then use a special rule to rewrite `\( \sin^2 x \)` and solve the integral. Finally, we plug in the numbers to get the answer.

🎯 Exam Tip: For even functions, rewrite the integral as `\( 2 \int_{0}^{a} f(x) dx \)`. Always use the half-angle identity `\( \sin^2 x = \frac{1-\cos(2x)}{2} \)` to integrate `\( \sin^2 x \)` or `\( \cos^2 x \)`, as direct integration is not possible.

(iv) \( \int_{0}^{2\pi} x \log \left( \frac{3+\cos x}{3-\cos x} \right) dx \)
Answer: Let the integral be `\( I = \int_{0}^{2\pi} x \log \left( \frac{3+\cos x}{3-\cos x} \right) dx \)`.
This integral is of the form `\( \int_{0}^{2a} x f(x) dx \)`. We use the property that if `\( f(2a-x) = f(x) \)`, then `\( \int_{0}^{2a} x f(x) dx = a \int_{0}^{2a} f(x) dx \)`.
Let `\( f(x) = \log \left( \frac{3+\cos x}{3-\cos x} \right) \)`.
We check `\( f(2\pi - x) \)`:
`\( f(2\pi - x) = \log \left( \frac{3+\cos(2\pi-x)}{3-\cos(2\pi-x)} \right) \)`
Since `\( \cos(2\pi-x) = \cos x \)`, we have:
`\( f(2\pi - x) = \log \left( \frac{3+\cos x}{3-\cos x} \right) = f(x) \)`
So, we can apply the property:
\[ I = \pi \int_{0}^{2\pi} \log \left( \frac{3+\cos x}{3-\cos x} \right) dx \quad \text{(Equation 1)} \]
Now, let `\( g(x) = \log \left( \frac{3+\cos x}{3-\cos x} \right) \)`. We will use another property: `\( \int_{0}^{2a} g(x) dx = 2 \int_{0}^{a} g(x) dx \)` if `\( g(2a-x) = g(x) \)`, and `\( \int_{0}^{2a} g(x) dx = 0 \)` if `\( g(2a-x) = -g(x) \)`.
For the integral `\( \int_{0}^{2\pi} g(x) dx \)`, we let `\( 2a=2\pi \implies a=\pi \)`.
We check `\( g(\pi - x) \)`:
`\( g(\pi - x) = \log \left( \frac{3+\cos(\pi-x)}{3-\cos(\pi-x)} \right) \)`
Since `\( \cos(\pi-x) = -\cos x \)`, we have:
`\( g(\pi - x) = \log \left( \frac{3-\cos x}{3-(-\cos x)} \right) = \log \left( \frac{3-\cos x}{3+\cos x} \right) \)`
`\( = - \log \left( \frac{3+\cos x}{3-\cos x} \right) = -g(x) \)`
Since `\( g(\pi - x) = -g(x) \)`, the integral `\( \int_{0}^{\pi} g(x) dx = 0 \)`.
And from property `\( \int_{0}^{2a} g(x) dx = 2 \int_{0}^{a} g(x) dx \)` if `\( g(2a-x) = g(x) \)`. (Which it is `f(2\pi-x)=f(x)`).
So `\( \int_{0}^{2\pi} g(x) dx = 2 \int_{0}^{\pi} g(x) dx \)`
Since `\( \int_{0}^{\pi} g(x) dx = 0 \)`, then `\( \int_{0}^{2\pi} g(x) dx = 2 \times 0 = 0 \)`.
Substitute this back into Equation 1:
`\( I = \pi \times 0 = 0 \)`.
This problem beautifully illustrates how two different integral properties can be combined to solve a complex integral efficiently.
In simple words: First, we use a special rule that helps when `x` is multiplied by the function and the limits are from `0` to `2a`. This changes the integral into a simpler form. Then, we look at the new function alone and see how it changes if we replace `x` with `(\pi - x)`. It becomes negative of itself, which means its integral from `\(0\)` to `\(\pi\)` is zero. This makes the final answer zero.

🎯 Exam Tip: For integrals of the form `\( \int_{0}^{2a} xf(x)dx \)`, always check if `\( f(2a-x)=f(x) \)` to apply the property `\( a \int_{0}^{2a} f(x)dx \)`. Then, evaluate `\( \int_{0}^{2a} f(x)dx \)` by checking `\( f(a-x) \)` for an interval of `\( [0,a] \)`. These properties are crucial for efficiency.

(v) \( \int_{0}^{\pi} \sin^4 x \cos^3 x dx \)
Answer: We need to evaluate the integral `\( \int_{0}^{\pi} \sin^4 x \cos^3 x dx \)`. We can use a property of definite integrals for this.
Let `\( f(x) = \sin^4 x \cos^3 x \)`. The integral is of the form `\( \int_{0}^{2a} f(x) dx \)`, where `\( 2a = \pi \)`, so `\( a = \pi/2 \)`.
We need to check `\( f(2a-x) \)`, which is `\( f(\pi-x) \)`.
`\( f(\pi-x) = \sin^4 (\pi-x) \cos^3 (\pi-x) \)`
We know the trigonometric identities:
`\( \sin(\pi-x) = \sin x \)`
`\( \cos(\pi-x) = -\cos x \)`
Substitute these into `\( f(\pi-x) \)`:
`\( f(\pi-x) = (\sin x)^4 (-\cos x)^3 \)`
`\( = \sin^4 x (-\cos^3 x) \)`
`\( = -\sin^4 x \cos^3 x \)`
`\( = -f(x) \)`
According to the property of definite integrals, if `\( \int_{0}^{2a} f(x) dx \)` and `\( f(2a-x) = -f(x) \)`, then the value of the integral is 0.
Since `\( f(\pi-x) = -f(x) \)`, the integral evaluates to 0.
\[ \int_{0}^{\pi} \sin^4 x \cos^3 x dx = 0 \]
This property provides a quick way to solve integrals of this type without lengthy calculations. If you have an odd power of cosine or sine in a `\( [0, \pi] \)` interval while the other is an even power, this property often applies.
In simple words: We check how the function behaves when `x` is replaced by `\(\pi - x\)`. Since `sine` stays the same and `cosine` flips its sign, the whole function becomes the negative of itself. When an integral from `\(0\)` to `\(\pi\)` has this behavior, its value is always zero because the positive and negative parts cancel out.

🎯 Exam Tip: For integrals from `\( 0 \)` to `\( \pi \)` or `\( 0 \)` to `\( 2\pi \)`, always test the property `\( f(2a-x) \)` or `\( f(a-x) \)`. This helps determine if the function is symmetric or antisymmetric, simplifying the integral to either `\( 0 \)` or `\( 2 \int_{0}^{a} f(x) dx \)`. Specifically, for `\( \int_0^\pi \sin^m x \cos^n x dx \)`, it's zero if `n` is odd.

(vi) \( \int_{0}^{1} |5x – 3|dx \)
Answer: To evaluate an integral involving an absolute value function, we first need to define the function without the absolute value sign. The expression `\( |5x-3| \)` changes its sign at `\( 5x-3 = 0 \)`, which means `\( x = \frac{3}{5} \)`.
We can define `\( |5x-3| \)` as:
\[ |5x-3| = \begin{cases} -(5x-3) = 3-5x & \text{if } 5x-3 < 0 \implies x < \frac{3}{5} \\ 5x-3 & \text{if } 5x-3 \ge 0 \implies x \ge \frac{3}{5} \end{cases} \]
The integration interval is from `\( 0 \)` to `\( 1 \)`. Since `\( 0 < \frac{3}{5} < 1 \)`, we need to split the integral into two parts at `\( x = \frac{3}{5} \)`.
\[ I = \int_{0}^{3/5} (3-5x) dx + \int_{3/5}^{1} (5x-3) dx \]
Now, we integrate each part:
For the first integral, evaluate `\( \left[ 3x - \frac{5x^2}{2} \right]_{0}^{3/5} \)`:
\[ = \left( 3\left(\frac{3}{5}\right) - \frac{5}{2}\left(\frac{3}{5}\right)^2 \right) - \left( 3(0) - \frac{5(0)^2}{2} \right) \]
\[ = \left( \frac{9}{5} - \frac{5}{2} \cdot \frac{9}{25} \right) - 0 \]
\[ = \frac{9}{5} - \frac{9}{10} \]
\[ = \frac{18-9}{10} = \frac{9}{10} \]
For the second integral, evaluate `\( \left[ \frac{5x^2}{2} - 3x \right]_{3/5}^{1} \)`:
\[ = \left( \frac{5(1)^2}{2} - 3(1) \right) - \left( \frac{5}{2}\left(\frac{3}{5}\right)^2 - 3\left(\frac{3}{5}\right) \right) \]
\[ = \left( \frac{5}{2} - 3 \right) - \left( \frac{5}{2} \cdot \frac{9}{25} - \frac{9}{5} \right) \]
\[ = \left( -\frac{1}{2} \right) - \left( \frac{9}{10} - \frac{9}{5} \right) \]
\[ = -\frac{1}{2} - \left( \frac{9-18}{10} \right) \]
\[ = -\frac{1}{2} - \left( -\frac{9}{10} \right) \]
\[ = -\frac{1}{2} + \frac{9}{10} \]
\[ = \frac{-5+9}{10} = \frac{4}{10} = \frac{2}{5} \]
Finally, we add the results of the two parts:
\[ I = \frac{9}{10} + \frac{2}{5} \]
\[ I = \frac{9}{10} + \frac{4}{10} \]
\[ I = \frac{13}{10} \]
This method is crucial for correctly evaluating integrals involving absolute values, as the function definition changes at specific points.
In simple words: We find the point where the `5x-3` inside the absolute value becomes zero. This point `(3/5)` splits our integral range into two parts. We then rewrite the expression for each part without the absolute value signs. Finally, we solve each integral separately and add their results together.

🎯 Exam Tip: For definite integrals with absolute value functions, always identify the points where the expression inside the absolute value equals zero. Split the integral into multiple parts at these critical points, define the function appropriately for each interval, and then integrate separately.

(vii) \( \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} dt \)
Answer: Let the given integral be `\( I = I_1 + I_2 \)`, where `\( I_1 = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} dt \)` and `\( I_2 = \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} dt \)`.

Let's evaluate `\( I_1 \)`:
Let `\( \sin^{-1} \sqrt{t} = \theta \)`.
Then `\( \sqrt{t} = \sin \theta \implies t = \sin^2 \theta \)`.
Now, find `\( dt \)`. `\( dt = 2 \sin \theta \cos \theta d\theta = \sin(2\theta) d\theta \)`.
We change the limits of integration:
When `\( t=0 \)`, `\( \sin \theta = 0 \implies \theta=0 \)`.
When `\( t=\sin^2 x \)`, `\( \sin \theta = \sin x \implies \theta=x \)`.
Substitute these into `\( I_1 \)`:
\[ I_1 = \int_{0}^{x} \theta \sin(2\theta) d\theta \]
We use integration by parts formula: `\( \int u dv = uv - \int v du \)`.
Let `\( u = \theta \implies du = d\theta \)`.
Let `\( dv = \sin(2\theta) d\theta \implies v = \int \sin(2\theta) d\theta = -\frac{\cos(2\theta)}{2} \)`.
\[ I_1 = \left[ \theta \left(-\frac{\cos(2\theta)}{2}\right) \right]_{0}^{x} - \int_{0}^{x} \left(-\frac{\cos(2\theta)}{2}\right) d\theta \]
\[ I_1 = \left( -\frac{x \cos(2x)}{2} - 0 \right) + \frac{1}{2} \int_{0}^{x} \cos(2\theta) d\theta \]
\[ I_1 = -\frac{x \cos(2x)}{2} + \frac{1}{2} \left[ \frac{\sin(2\theta)}{2} \right]_{0}^{x} \]
\[ I_1 = -\frac{x \cos(2x)}{2} + \frac{1}{2} \left( \frac{\sin(2x)}{2} - \frac{\sin(0)}{2} \right) \]
\[ I_1 = -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} \quad \text{(Equation 1)} \]

Now, let's evaluate `\( I_2 \)`:
Let `\( \cos^{-1} \sqrt{t} = \theta \)`.
Then `\( \sqrt{t} = \cos \theta \implies t = \cos^2 \theta \)`.
Now, find `\( dt \)`. `\( dt = 2 \cos \theta (-\sin \theta) d\theta = -\sin(2\theta) d\theta \)`.
We change the limits of integration:
When `\( t=0 \)`, `\( \cos \theta = 0 \implies \theta=\frac{\pi}{2} \)`.
When `\( t=\cos^2 x \)`, `\( \cos \theta = \cos x \implies \theta=x \)`.
Substitute these into `\( I_2 \)`:
\[ I_2 = \int_{\pi/2}^{x} \theta (-\sin(2\theta)) d\theta \]
We can reverse the limits and change the sign of the integrand:
\[ I_2 = \int_{x}^{\pi/2} \theta \sin(2\theta) d\theta \]
Using the same result from integration by parts as for `\( I_1 \)`:
\[ \int \theta \sin(2\theta) d\theta = -\frac{\theta \cos(2\theta)}{2} + \frac{\sin(2\theta)}{4} \]
So, for `\( I_2 \)`:
\[ I_2 = \left[ -\frac{\theta \cos(2\theta)}{2} + \frac{\sin(2\theta)}{4} \right]_{x}^{\pi/2} \]
\[ I_2 = \left( -\frac{(\pi/2) \cos(2 \cdot \pi/2)}{2} + \frac{\sin(2 \cdot \pi/2)}{4} \right) - \left( -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} \right) \]
\[ I_2 = \left( -\frac{(\pi/2) \cos(\pi)}{2} + \frac{\sin(\pi)}{4} \right) - \left( -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} \right) \]
Since `\( \cos(\pi) = -1 \)` and `\( \sin(\pi) = 0 \)`, this simplifies to:
\[ I_2 = \left( -\frac{\pi/2 (-1)}{2} + 0 \right) - \left( -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} \right) \]
\[ I_2 = \frac{\pi}{4} + \frac{x \cos(2x)}{2} - \frac{\sin(2x)}{4} \quad \text{(Equation 2)} \]

Finally, we add `\( I_1 \)` and `\( I_2 \)`:
\[ I = I_1 + I_2 \]
\[ I = \left( -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} \right) + \left( \frac{\pi}{4} + \frac{x \cos(2x)}{2} - \frac{\sin(2x)}{4} \right) \]
Many terms cancel out:
\[ I = \frac{\pi}{4} \]
This problem demonstrates the power of variable substitution and integration by parts to simplify seemingly complex composite integrals.
In simple words: We split the problem into two integrals. For the first integral, we replace `\( \sqrt{t} \)` with `\( \sin \theta \)`. For the second, we replace `\( \sqrt{t} \)` with `\( \cos \theta \)`. Both of these require a special rule called `integration by parts`. After solving both parts and adding their results, many terms cancel, leaving us with the simple answer `\(\pi/4\)`.

🎯 Exam Tip: When faced with integrals involving `\( \sin^{-1} \sqrt{t} \)` or `\( \cos^{-1} \sqrt{t} \)`, a substitution of `\( \sqrt{t} = \sin \theta \)` or `\( \sqrt{t} = \cos \theta \)` often simplifies the integrand, making it suitable for integration by parts. Remember to carefully adjust the limits of integration.

(viii) \( \int_{0}^{1} \frac{\log(1+x)}{1+x^2} dx \)
Answer: Let the integral be `\( I = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} dx \)`.
The presence of `\( 1+x^2 \)` in the denominator suggests a trigonometric substitution. Let `\( x = \tan \theta \)`.
Then, `\( dx = \sec^2 \theta d\theta \)`.
We change the limits of integration:
When `\( x=0 \)`, `\( \tan \theta = 0 \implies \theta=0 \)`.
When `\( x=1 \)`, `\( \tan \theta = 1 \implies \theta=\frac{\pi}{4} \)`.
Substitute these into the integral `\( I \)`:
\[ I = \int_{0}^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta d\theta \]
Since `\( 1+\tan^2 \theta = \sec^2 \theta \)`, the `\( \sec^2 \theta \)` terms cancel out:
\[ I = \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta \quad \text{(Equation 1)} \]
Now, we use the property of definite integrals: `\( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \)`.
Here, `\( a = \frac{\pi}{4} \)`, and `\( f(\theta) = \log(1+\tan \theta) \)`. So, we consider `\( f(\frac{\pi}{4}-\theta) \)`.
\[ I = \int_{0}^{\pi/4} \log \left( 1+\tan \left( \frac{\pi}{4}-\theta \right) \right) d\theta \]
Using the tangent subtraction formula `\( \tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} \)`:
`\( \tan \left( \frac{\pi}{4}-\theta \right) = \frac{\tan(\pi/4)-\tan \theta}{1+\tan(\pi/4)\tan \theta} = \frac{1-\tan \theta}{1+\tan \theta} \)`
Substitute this back into the integral:
\[ I = \int_{0}^{\pi/4} \log \left( 1+\frac{1-\tan \theta}{1+\tan \theta} \right) d\theta \]
Combine the terms inside the logarithm:
\[ I = \int_{0}^{\pi/4} \log \left( \frac{1+\tan \theta + 1-\tan \theta}{1+\tan \theta} \right) d\theta \]
\[ I = \int_{0}^{\pi/4} \log \left( \frac{2}{1+\tan \theta} \right) d\theta \]
Using the logarithm property `\( \log \left( \frac{A}{B} \right) = \log A - \log B \)`, we split the term:
\[ I = \int_{0}^{\pi/4} (\log 2 - \log(1+\tan \theta)) d\theta \]
\[ I = \int_{0}^{\pi/4} \log 2 d\theta - \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta \]
Notice that the second integral on the right side is the original integral `\( I \)` (from Equation 1).
\[ I = (\log 2) [\theta]_{0}^{\pi/4} - I \]
\[ I = (\log 2) \left( \frac{\pi}{4} - 0 \right) - I \]
\[ I = \frac{\pi}{4} \log 2 - I \]
Now, we solve for `\( I \)` by adding `\( I \)` to both sides:
\[ 2I = \frac{\pi}{4} \log 2 \]
This concludes the solution based on the content available within the specified page range. This integral demonstrates the effectiveness of specific trigonometric substitutions and a key property of definite integrals where the original integral reappears for solution.
In simple words: We change `x` to `tan θ` to make the fraction simpler and adjust the limits. The integral then becomes `\( \int \log(1+\tan \theta) d\theta \)`. We use a special property of integrals and a `tan` formula to rewrite the problem. This makes the original integral `I` appear again in our calculation. We then combine the `I` terms to find the value of `\( 2I \)`.

🎯 Exam Tip: The substitution `\( x=\tan \theta \)` is often effective for integrands involving `\( (1+x^2) \)`. Remember the key property `\( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \)` for integrals of `\( \log(1+\tan \theta) \)`, as it frequently leads to the original integral reappearing on the right side.

Question 2. (ix) Evaluate \( \int_{0}^{\pi} \frac{x \sin x}{1+\sin x} dx \) using properties of integration.
Answer: Let the integral be \( I = \int_{0}^{\pi} \frac{x \sin x}{1+\sin x} dx \).
We use the property \( \int_{0}^{a} x f(x) dx = \frac{a}{2} \int_{0}^{a} f(x) dx \) if \( f(a-x) = f(x) \).
Here, \( f(x) = \frac{\sin x}{1+\sin x} \).
Let's check \( f(\pi - x) \):
\( f(\pi - x) = \frac{\sin (\pi - x)}{1+\sin (\pi - x)} = \frac{\sin x}{1+\sin x} = f(x) \).
Since \( f(\pi - x) = f(x) \), we can apply the property:
\( I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\sin x} dx \).
Now, we simplify the integrand:
\( \frac{\sin x}{1+\sin x} = \frac{\sin x (1-\sin x)}{(1+\sin x)(1-\sin x)} = \frac{\sin x - \sin^2 x}{1-\sin^2 x} = \frac{\sin x - \sin^2 x}{\cos^2 x} \)
We can split this into two parts:
\( \frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} = \sec x \tan x - \tan^2 x \).
Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\( \sec x \tan x - (\sec^2 x - 1) = \sec x \tan x - \sec^2 x + 1 \).
So, \( I = \frac{\pi}{2} \int_{0}^{\pi} (\sec x \tan x - \sec^2 x + 1) dx \).
Now, we integrate term by term:
\( \int \sec x \tan x dx = \sec x \)
\( \int \sec^2 x dx = \tan x \)
\( \int 1 dx = x \)
So, \( I = \frac{\pi}{2} [\sec x - \tan x + x]_{0}^{\pi} \).
Evaluate at the limits:
Upper limit at \( x = \pi \): \( \sec \pi - \tan \pi + \pi = -1 - 0 + \pi = \pi - 1 \).
Lower limit at \( x = 0 \): \( \sec 0 - \tan 0 + 0 = 1 - 0 + 0 = 1 \).
Subtract the lower limit from the upper limit:
\( I = \frac{\pi}{2} [(\pi - 1) - (1)] \)
\( I = \frac{\pi}{2} [\pi - 1 - 1] \)
\( I = \frac{\pi}{2} (\pi - 2) \).
The final answer is \( I = \frac{\pi(\pi - 2)}{2} \).
In simple words: First, we use a special rule for integrals that lets us remove the 'x' from the integral, by checking if the rest of the function stays the same when we change 'x' to 'pi minus x'. Then, we simplify the fraction inside the integral by multiplying by its conjugate. After that, we break it into simpler parts and use known integration formulas to find the answer.

🎯 Exam Tip: Remember to check the property \( f(a-x)=f(x) \) carefully before applying the integral property. Also, be thorough with trigonometric identities to simplify the integrand effectively.

Question 2. (x) Evaluate \( \int_{\pi/8}^{3\pi/8} \frac{1}{1+\sqrt{\tan x}} dx \) using properties of integration.
Answer: Let the integral be \( I = \int_{\pi/8}^{3\pi/8} \frac{1}{1+\sqrt{\tan x}} dx \) .....(1)
We use the property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \).
Here, \( a = \frac{\pi}{8} \) and \( b = \frac{3\pi}{8} \).
So, \( a+b = \frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2} \).
Substitute \( x \) with \( \frac{\pi}{2} - x \) in the integral:
\( I = \int_{\pi/8}^{3\pi/8} \frac{1}{1+\sqrt{\tan (\frac{\pi}{2} - x)}} dx \).
We know that \( \tan (\frac{\pi}{2} - x) = \cot x \).
So, \( I = \int_{\pi/8}^{3\pi/8} \frac{1}{1+\sqrt{\cot x}} dx \).
We can write \( \sqrt{\cot x} \) as \( \frac{1}{\sqrt{\tan x}} \).
\( I = \int_{\pi/8}^{3\pi/8} \frac{1}{1+\frac{1}{\sqrt{\tan x}}} dx \)
\( I = \int_{\pi/8}^{3\pi/8} \frac{1}{\frac{\sqrt{\tan x}+1}{\sqrt{\tan x}}} dx \)
\( I = \int_{\pi/8}^{3\pi/8} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} dx \) .....(2)
Now, we add equations (1) and (2):
\( 2I = \int_{\pi/8}^{3\pi/8} \left( \frac{1}{1+\sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \right) dx \)
\( 2I = \int_{\pi/8}^{3\pi/8} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} dx \)
\( 2I = \int_{\pi/8}^{3\pi/8} 1 dx \).
Now, integrate 1 with respect to \( x \):
\( 2I = [x]_{\pi/8}^{3\pi/8} \).
Evaluate at the limits:
\( 2I = \frac{3\pi}{8} - \frac{\pi}{8} \)
\( 2I = \frac{2\pi}{8} \)
\( 2I = \frac{\pi}{4} \).
Divide by 2 to find I:
\( I = \frac{\pi}{8} \).
In simple words: This problem uses a property of definite integrals that lets us change 'x' to 'a+b-x' without changing the answer. By doing this, we get a new integral that looks different but is actually equivalent. When we add the original integral and the new one together, most of the complex terms cancel out, leaving a very simple integral which is easy to solve.

🎯 Exam Tip: When using the property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \), always simplify the transformed function as much as possible, often involving trigonometric identities. Adding the original and transformed integrals is a common technique to solve these types of problems.

Question 2. (xi) Evaluate \( \int_{0}^{\pi} x[\sin^2(\sin x) + \cos^2 (\cos x)] dx \) using properties of integration.
Answer: Let the integral be \( I = \int_{0}^{\pi} x[\sin^2(\sin x) + \cos^2 (\cos x)] dx \).
Let \( f(x) = \sin^2(\sin x) + \cos^2 (\cos x) \).
We check \( f(\pi - x) \):
\( \sin(\pi - x) = \sin x \).
\( \cos(\pi - x) = -\cos x \).
So, \( f(\pi - x) = \sin^2(\sin (\pi - x)) + \cos^2 (\cos (\pi - x)) \)
\( f(\pi - x) = \sin^2(\sin x) + \cos^2 (-\cos x) \).
Since \( (-\cos x)^2 = (\cos x)^2 \), we have \( \cos^2 (-\cos x) = \cos^2 (\cos x) \).
Therefore, \( f(\pi - x) = \sin^2(\sin x) + \cos^2 (\cos x) = f(x) \).
Now we use the property \( \int_{0}^{a} x f(x) dx = \frac{a}{2} \int_{0}^{a} f(x) dx \) when \( f(a-x) = f(x) \).
Here, \( a = \pi \), so:
\( I = \frac{\pi}{2} \int_{0}^{\pi} [\sin^2(\sin x) + \cos^2 (\cos x)] dx \) .....(1)
Next, we use another property: if \( f(2A - x) = f(x) \), then \( \int_{0}^{2A} f(x) dx = 2 \int_{0}^{A} f(x) dx \).
For the integral \( \int_{0}^{\pi} f(x) dx \), we have \( 2A = \pi \), so \( A = \frac{\pi}{2} \).
Since \( f(\pi - x) = f(x) \), we can write:
\( \int_{0}^{\pi} [\sin^2(\sin x) + \cos^2 (\cos x)] dx = 2 \int_{0}^{\pi/2} [\sin^2(\sin x) + \cos^2 (\cos x)] dx \).
Substitute this back into equation (1):
\( I = \frac{\pi}{2} \times 2 \int_{0}^{\pi/2} [\sin^2(\sin x) + \cos^2 (\cos x)] dx \)
\( I = \pi \int_{0}^{\pi/2} [\sin^2(\sin x) + \cos^2 (\cos x)] dx \) .....(2)
Let \( J = \int_{0}^{\pi/2} [\sin^2(\sin x) + \cos^2 (\cos x)] dx \).
Using the property \( \int_{0}^{A} f(x) dx = \int_{0}^{A} f(A-x) dx \) for \( J \) with \( A = \frac{\pi}{2} \):
Replace \( x \) with \( \frac{\pi}{2} - x \).
\( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \cos(\frac{\pi}{2} - x) = \sin x \).
So, \( J = \int_{0}^{\pi/2} [\sin^2(\cos x) + \cos^2 (\sin x)] dx \) .....(3)
Now, add equations (2) and (3) for \( J \):
\( 2J = \int_{0}^{\pi/2} [\sin^2(\sin x) + \cos^2 (\cos x) + \sin^2(\cos x) + \cos^2 (\sin x)] dx \)
Rearrange terms using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( 2J = \int_{0}^{\pi/2} [(\sin^2(\sin x) + \cos^2 (\sin x)) + (\cos^2 (\cos x) + \sin^2(\cos x))] dx \)
\( 2J = \int_{0}^{\pi/2} [1 + 1] dx \)
\( 2J = \int_{0}^{\pi/2} 2 dx \).
Integrate 2 with respect to \( x \):
\( 2J = [2x]_{0}^{\pi/2} \).
Evaluate at the limits:
\( 2J = 2(\frac{\pi}{2}) - 2(0) \)
\( 2J = \pi \).
So, \( J = \frac{\pi}{2} \).
Substitute \( J \) back into equation (2) for \( I \):
\( I = \pi \times J \)
\( I = \pi \times \frac{\pi}{2} \)
\( I = \frac{\pi^2}{2} \).
In simple words: First, we use a rule to simplify the integral by checking if the function inside stays the same when we swap 'x' with 'pi minus x'. This helps us change the 'x' term in front of the function. Then, we use another trick where we split the integral into two parts and add them. This works because 'sin squared theta plus cos squared theta' is always 1. This makes the complicated part of the integral turn into just '2', which is very easy to integrate.

🎯 Exam Tip: This question combines several integral properties. Carefully check the conditions for each property (like \( f(a-x) = f(x) \) or \( f(2A-x)=f(x) \)) before applying them. The key step is often transforming the integrand to use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).