RD Sharma Class 12 Exercise 19.4(b) Definite Integrals Solutions Maths

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RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise

Definite Integrals Excercise: 19.4(B)

Definite Integrals Exercise 19.4 (a) Question 16

Answer : $\int_{a}^{b} x f(x) d x=\frac{a+b}{2} \int_{a}^{b} f(x) d x$
Given : $f(a+b-x)=f(x)$ and prove that
$\int_{a}^{b} x f(x)=\frac{a+b}{2} \int_{a}^{b} f(x) d x$
Hint : You must know the formula of $\int_{a}^{b} f(x) d x$
Solution : $I=\int_{a}^{b} x f(x) d x$
$\begin{aligned} &I=\int_{a}^{b}(a+b-x) f(a+b-x) d x\left(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right) \\ &I=\int_{a}^{b}(a+b-x) f(x) d x[f(a+b-x)=f(x)] \\ &I=\int_{a}^{b}(a+b) f(x) d x-\int_{a}^{b} x f(x) d x \end{aligned}$
$\begin{aligned} &I=\int_{a}^{b}(a+b) f(x) d x-I \\ &2 I=(a+b) \int_{a}^{b} f(x) d x \\ &I=\frac{a+b}{2} \int_{a}^{b} f(x) d x \end{aligned}$
$\int_{a}^{b} x f(x) d x=\frac{a+b}{2} \int_{a}^{b} f(x) d x \text { (proved) }$

Definite Integrals Exercise 19.4 (b) Question 3

Answer : $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given : $\int_{0}^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x$
Solution : $I=\int_{0}^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x$ .....(1)
$I=\int \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}$ .....(2)
Adding (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{\pi / 2} \frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \\ &2 I=[x]_{0}^{\pi / 2} \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \times \frac{\pi}{2} \\ &=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 4

Answer:- $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x$
Solution : $I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x} d x$ ....(1)
$\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x\\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\\ &I=\int_{0}^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \end{aligned}$
Adding (1) and (2)
$\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x+\int_{0}^{\pi / 2} \frac{\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x+\cos ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 . d x \end{aligned}$
$\begin{aligned} &2 I=[x]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 5

Answer:- $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given : $\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$
Solution : $I=\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$ ...(1)
$\therefore I=\int_{0}^{\pi / 2} \frac{\sin ^{n}\left(\frac{\pi}{2}-x\right)}{\sin ^{n}\left(\frac{\pi}{2}-x\right)+\cos ^{n}\left(\frac{\pi}{2}-x\right)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$I=\int_{0}^{\pi / 2} \frac{\cos ^{n} x}{\cos ^{n} x+\sin ^{n} x} d x$ ....(2)
Adding (1) and (2)
$\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x+\int_{0}^{\pi / 2} \frac{\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin ^{n} x+\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \end{aligned}$
$\begin{aligned} &2 I=[x]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 2

Answer : $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given : $\int_{0}^{\pi / 2} \frac{1}{1+\cot x} d x$
Solution : $\int_{0}^{\pi / 2} \frac{1}{1+\cot x} d x$
$\begin{aligned} &\int_{0}^{\pi / 2} \frac{d x}{1+\frac{\cos x}{\sin x}} \\ &\int_{0}^{\pi / 2} \frac{d x \cdot \sin x}{\sin x+\cos x} \end{aligned}$ .....(1)
We know $\left[\int_{0}^{a} f(x)=\int_{0}^{a} f(a-x) d x\right]$
$\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\\ &I=\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \end{aligned}$ ....(2)
Adding,
$\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x+\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \end{aligned}$
$2 I=[x]_{0}^{\pi / 2}$
$\begin{aligned} &I=\frac{1}{2} \times \frac{\pi}{2} \\ &=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 6

Answer:- $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x$
Solution :
$\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\frac{\sin x}{\cos x}}} d x \\ &I=\int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \end{aligned}$ ....(1)
$I=\int_{0}^{\pi / 2} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Adding (1) and (2)
$\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x+\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \end{aligned}$
$\begin{aligned} &2 I=[x]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 7

Answer:- $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} d x$
Solution : $\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} d x$
$x=a \sin \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[x=0, a \sin \theta=0, \theta=0]$
$d x=a \cos \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[x=a, 0 \sin \theta=a, \sin \theta=1, \theta=\pi / 2]$
$I=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+\sqrt{a^{2}-a^{2} \sin ^{2} \theta}} d \theta$
$\begin{aligned} &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+a \sqrt{1-\sin ^{2} \theta}} d \theta \\ &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+a \cos \theta} d \theta \\ &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a(\sin \theta+\cos \theta)} d \theta \end{aligned}$
$=\int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta$ .....(1)
$I=\int_{0}^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\sin \left(\frac{\pi}{2}-\theta\right)+\cos \left(\frac{\pi}{2}-\theta\right)} d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$I=\int_{0}^{\pi / 2} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta$ .....(2)
Adding both (1) and (2)
$\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta+\int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin \theta+\cos \theta}{\sin \theta+\cos \theta} d \theta \\ &2 I=\int_{0}^{\pi / 2} 1 . d \theta \end{aligned}$
$\begin{aligned} &2 I=[0]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 8

Answer:- 0
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x$
Solution : $\int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x$
Using substitute $x=\frac{1}{t}$
We get $-\int_{0}^{1} \frac{\log x}{1+x^{2}} d x=\int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x$
And $\therefore \int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x=0$

Definite Integrals Exercise 19.4 (b) Question 9

Answer : $\frac{\pi}{8} \log 2$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{1} \frac{\log x(1+x)}{1+x^{2}} d x$
Solution : $\int_{0}^{1} \frac{\log x(1+x)}{1+x^{2}} d x$
$\begin{aligned} &\text { Put } x=\tan \theta \\ &d x=\sec ^{2} \theta d \theta \end{aligned}$
Lower limit, $x=0 \text {, then } \theta=\tan ^{-1} 0=0$
Upper limit, $x=1, \text { then } \theta=\tan ^{-1}=\frac{\pi}{4}$
$\begin{aligned} &\therefore I=\int_{0}^{\frac{\pi}{4}} \frac{\log |1+\tan \theta|}{1+\tan ^{2} \theta} \sec ^{2} \theta d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \frac{\log |1+\tan \theta|}{\sec ^{2} \theta} \sec ^{2} \theta d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log |1+\tan \theta| d \theta \end{aligned}$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \log \left|1+\tan \left(\frac{\pi}{4}-\theta\right)\right| \cdot d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log \left|1+\frac{(1-\tan \theta)}{1+\tan \theta}\right| \cdot d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log \left|\frac{2}{1+\tan \theta}\right| \cdot d \theta \end{aligned}$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \log 2 d \theta-\int_{0}^{\frac{\pi}{4}} \log |1+\tan \theta| \cdot d \theta \\ &I=\log 2[\theta]_{0}^{\frac{\pi}{4}}-I \\ &2 I=\frac{\pi}{4} \log 2 \\ &I=\frac{\pi}{8} \log 2 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 10

Answer:- $\frac{\pi}{4}$
Hints:- You must know about the integration rules of functions .
Given:- $\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$
Solution : $\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$
We can write integrand using partial fractions
$\begin{aligned} &\frac{x}{\left(x^{2}+1\right)(x+1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x+1)} \\ &\frac{x}{\left(x^{2}+1\right)(x+1)}=\frac{(A x+B)(x+1)+C\left(x^{2}+1\right)}{\left(x^{2}+1\right)(x+1)} \end{aligned}$
Cancelling denominators
$x=(A x+B)(x+1)+C\left(x^{2}+1\right)$ ......(1)
Putting $x=1, \text { in (1) }$
$\begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &1=(A \times 1+B)(1+1)+C\left(1^{2}+1\right) \\ &1=2(A+B)+C(2) \\ &1=2 A+2 B+2 C \end{aligned}$ .....(2)
Putting $x=0 \text { in (1) }$
$\begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &0=(A \times 0+B)(0+1)+C(0+1) \end{aligned}$
$\begin{aligned} &0=B+C \\ &B=-C \end{aligned}$
Put in (2)
$\begin{aligned} &1=2 A-2 C+2 C \\ &A=\frac{1}{2} \end{aligned}$
Putting $x = -1$
$\begin{aligned} &-1=\left(A(-1)+B(-1+1)+C\left((-1)^{2}+1\right)\right. \\ &-1=(-A+B)(0)+C(2) \\ &-1=2 C \\ &C=\frac{-1}{2} \end{aligned}$
Put value of A and C in (2)
$\begin{aligned} &2 A+2 B+2 C=1 \\ &2 \times \frac{1}{2}+2 B+2 \times \frac{-1}{2}=1 \\ &1+2 B-1=1 \\ &2 B=1 \\ &B=\frac{1}{2} \end{aligned}$
$\therefore A=\frac{1}{2}, B=\frac{1}{2}, C=\frac{-1}{2}$
Hence we can write
$\begin{gathered} \frac{x}{(x+1)\left(x^{2}+1\right)}=\frac{\left(\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{-1}{2}}{(x+1)} \\ =\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2\left(x^{2}+1\right)}-\frac{1}{2(x+1)} \end{gathered}$
Solving
$\begin{aligned} &I_{3}=\frac{-1}{2} \int_{0}^{\infty} \frac{1}{x+1} d x \\ &=\frac{-1}{2}\left[\log |x+1|_{0}^{\infty}\right. \\ &=\frac{-1}{2}[\log (1)-\log (1) \mid] \\ &=0 \end{aligned}$
Hence,
$\begin{aligned} &I=I_{1}+I_{2}+I_{3} \\ &=0+\frac{\pi}{4}+0 \\ &=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 11

Answer:- $\frac{\pi^{2}}{4}$
Hints:- You must know the integration rules of trigonometric functions.
Given:- $\int_{0}^{\pi} \frac{x \tan x}{\sec x \cos e c x} d x$
Solution : $I=\int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x} \times \frac{1}{\sin x}} d x$
$I=\int_{0}^{\pi} x \sin ^{2} x \cdot d x$ ......(1)
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \sin ^{2}(\pi-x) \cdot d x\\ &I=\int_{0}^{\pi}(\pi-x) \sin ^{2} x d x \end{aligned}$ .......(2)
Adding (1) and (2)
$\begin{aligned} &I+I=\int_{0}^{\pi} \pi \sin ^{2} x \cdot d x \\ &2 I=\int_{0}^{\pi} \pi \sin ^{2} x \cdot d x \\ &2 I=\frac{\pi}{2} \int_{0}^{\pi}(1-\cos 2 x) \cdot d x \end{aligned}$
$\begin{aligned} &2 I=\frac{\pi}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi} \\ &2 I=\frac{\pi}{2}\left[\left(\pi-\frac{\sin 2 \pi}{2}\right)-\left(0-\frac{\sin 0}{2}\right)\right] \end{aligned}$
$\begin{aligned} &2 I=\frac{\pi}{2}[\pi] \\ &2 I=\frac{\pi^{2}}{2} \\ &I=\frac{\pi^{2}}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 12

Answer:- $\frac{\pi}{5}$
Hints:- You must know the integration rules of trigonometric functions and its limits
Given:- $\int_{0}^{\pi} \sin x \cos ^{4} x \cdot d x$
Solution : $\int_{0}^{\pi} \sin x \cos ^{4} x \cdot d x=I$ ......(1)
$I=\int_{0}^{\pi}(\pi-x) \sin (\pi-x) \cos ^{4}(\pi-x) \cdot d x\\$
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \sin x \cos ^{4} x \cdot d x\\ \end{aligned}$ $\text { .......(2) }[\because \sin (\pi-x)=\sin x]$
$\mathrm{Eq}(1)+\mathrm{Eg}(2) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\cos (\pi-x)=-\cos x]$
$\begin{aligned} &2 I=\int_{0}^{\pi} x \sin x \cdot \cos ^{4} x d x+\int_{0}^{\pi}(\pi-x) \sin x \cdot \cos ^{4} x \cdot d x \\ &2 I=\int_{0}^{\pi} \pi \sin x \cdot \cos ^{4} x \cdot d x \end{aligned}$
Let $\sin x = t$
$\begin{aligned} &x=0, x=\pi \\ &t=0, t=0 \\ &\cos x d x=d t \\ &2 I=\int_{0}^{0} \text { No benefit. } \end{aligned}$
Now, Let $\cos x = t$
$\begin{aligned} &x=0, t=1 \\ &x=\pi, t=-1 \end{aligned}$
$\begin{aligned} &2 I=\int_{1}^{-1} \pi t^{4} d t \\ &I=\frac{-\pi}{2} \int_{1}^{-1} t^{4} d t \\ &I=\frac{-\pi}{2}\left[\frac{t^{5}}{5}\right]_{1}^{-1} \end{aligned}$
$\begin{aligned} &=\frac{-\pi}{2}\left[\frac{-1}{5}-\frac{1}{5}\right] \\ &=\frac{-\pi}{2}\left[\frac{-2}{5}\right] \\ &=\frac{\pi}{5} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 13

Answer:- $\frac{2\pi}{3}$
Hints:- You must know the integration rules of trigonometric functions.
Given:- $\int_{0}^{\pi} x \sin ^{3} x \cdot d x$
Solution : $\int_{0}^{\pi} x \sin ^{3} x \cdot d x=I$
$I=\int_{0}^{\pi} x \sin ^{3} x . d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x)[\sin (\pi-x)]^{3} \cdot d x \\ &I=\int_{0}^{\pi}(\pi-x) \sin ^{3} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \sin ^{3} x \cdot d x-\int_{0}^{\pi} x \cdot \sin ^{3} x \cdot d x \end{aligned}$
$\begin{aligned} &I=\pi \int_{0}^{\pi} \sin ^{3} x \cdot d x-I \\ &2 I=\pi \int_{0}^{\pi} \sin ^{3} x \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{4}(3 \sin x-\sin 3 x) \cdot d x \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{8} \int_{0}^{\pi} 3 \sin x-\sin 3 x \cdot d x \\ &I=\frac{\pi}{8}\left[3(-\cos x)+\frac{\cos 3 x}{3}\right]_{0}^{\pi} \end{aligned}$
Putting limits
$\begin{aligned} &I=\frac{\pi}{8}\left[-3 \cos \pi+\frac{\cos 3 x}{3}\right]-\left[-3 \cos 0+\frac{\cos 3(0)}{3}\right] \\ &I=\frac{\pi}{8}\left[(-3)(-1)+\frac{(-1)}{3}\right]-\left[-3+\frac{1}{3}\right] \\ &I=\frac{\pi}{8} \times 2\left[3-\frac{1}{3}\right] \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{4}\left[\frac{9-1}{3}\right] \\ &I=\frac{\pi}{4} \times \frac{8}{3} \\ &=\frac{2 \pi}{3} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 14

Answer: $-\frac{\pi^{2}}{2} \log (2)$
Hints:- You must know the integration rules of trigonometric and logarithmic functions.
Given:- $\int_{0}^{\pi} x \log \sin x \cdot d x$
Solution : $\int_{0}^{\pi} x \log \sin x \cdot d x$ .....(1)
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \log \sin (\pi-x) \cdot d x\\ &I=\int_{0}^{\pi}(\pi-x) \log (\sin x) \cdot d x \end{aligned}$ .....(2)
Adding (1) and (2)
$2 I=\pi \int_{0}^{\pi} \log (\sin x) \cdot d x$
$\begin{aligned} &2 I=2 \pi \int_{0}^{\pi / 2} \log (\sin x) \cdot d x \\ &I=\pi\left(\frac{-\pi}{2} \log 2\right) \\ &I=\frac{-\pi^{2}}{2} \log (2) \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 15

Answer:- $\frac{(\pi-2) \pi}{2}$
Hints:- You must know the integration rules of trigonometric rule.
Given:- $\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$
Solution : $\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$
$I=\int_{0}^{\pi} \frac{(\pi-x) x \sin (\pi-x)}{1+\sin (\pi-x)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$\begin{aligned} &I=\int_{0}^{\pi} \frac{\pi \sin x}{1+\sin x} d x-I \\ &2 I=\int_{0}^{\pi} \frac{\pi \sin x \cdot(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ &2 I=\int_{0}^{\pi} \frac{\pi \sin x \cdot(1-\sin x)}{\left(1-\sin ^{2} x\right)} d x \end{aligned}$
$\begin{aligned} &\frac{2 I}{\pi}=\int_{0}^{\pi} \frac{\sin x-\sin ^{2} x}{\cos ^{2} x} d x \\ &\frac{2 I}{\pi}=\int_{0}^{\pi} \frac{\sin x}{\cos ^{2} x} d x-\int_{0}^{\pi} \frac{\sin ^{2} x}{\cos ^{2} x} d x \\ &\frac{2 I}{\pi}=\int_{0}^{\pi} \sec x \cdot \tan x \cdot d x-\int_{0}^{\pi} \tan ^{2} x \cdot d x \end{aligned}$
$\begin{aligned} &\frac{2 I}{\pi}=[\sec \pi-\sec 0]-\int_{0}^{\pi} \sec ^{2} x \cdot d x+\int_{0}^{\pi} 1 \cdot d x \\ &\frac{2 I}{\pi}=[-1-1]-[\tan ]_{0}^{\pi}-[x]_{0}^{\pi} \end{aligned}$
$\begin{aligned} &\frac{2 I}{\pi}=[-2]-[\tan \pi-\tan 0]+\pi \\ &\frac{2 I}{\pi}=[-2]-0+\pi \\ &\therefore I=\frac{(\pi-2) \pi}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 16

Answer:- $\frac{\alpha \pi}{\sin \alpha}$
Hints:- You must know the integration rules of trigonometric functions and its limits.
Given:- $\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x, \quad 0<\alpha<\pi$
Solution : $\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$ ....(1)
$\begin{aligned} &I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\cos \alpha \sin (\pi-x)} d x \\ &I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\cos \alpha \sin x} d x \end{aligned}$ .....(2)
Adding (1) and (2)
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{d x}{1+\cos \alpha \sin x} \\ &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} \frac{\pi}{2} d x}{\left(1+\tan ^{2} \frac{x}{2}\right)+2 \cos \alpha \tan \frac{x}{2}} \end{aligned}$
Put $\tan\frac{x}{2}=t$
$\begin{aligned} &\sec ^{2} \frac{\pi}{2} d x=2 d t \\ &2 I=\pi \int_{0}^{\pi} \frac{2 d t}{1+t^{2}+2 t \cos \alpha} \\ &2 I=2 \pi \int_{0}^{\pi} \frac{d t}{(t+\cos \alpha)^{2}+\sin ^{2} \alpha} \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}\left(\frac{t+\cos \alpha}{\sin \alpha}\right)\right]_{0}^{\infty} \\ &=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}(\infty)-\tan ^{-1}(\cot \alpha)\right] \\ &=\frac{\pi}{\sin \alpha}\left[\frac{\pi}{2}-\left(\frac{\pi}{2}-\alpha\right)\right] \\ &=\frac{\alpha \pi}{\sin \alpha} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 17

Answer:- $\frac{\pi^{2}}{4}$
Hints:- You must know about the integration rules of trigonometry functions.
Given:- $\int_{0}^{\pi} x \cos ^{2} x d x$
Solution : $\int_{0}^{\pi} x \cos ^{2} x d x=I$
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \cos ^{2}(\pi-x) d x \\ &I=\int_{0}^{\pi}(\pi-x) \cos ^{2} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \cos ^{2} x \cdot d x-\int_{0}^{\pi} x \cos ^{2} x \cdot d x \end{aligned}$
$\begin{aligned} &I=\pi \int_{0}^{\pi} \cos ^{2} x \cdot d x-I \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x+1}{2} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x}{2} \cdot d x+\int_{0}^{\pi} \frac{1}{2} \cdot d x \end{aligned}$
$\begin{gathered} =\pi\left[\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\pi}+\frac{1}{2}[x]_{0}^{\pi}\right] \\ 2 I=\frac{\pi}{4}[\sin 2 \pi-\sin 0]+\frac{\pi}{2}[\pi] \end{gathered}$
$\begin{aligned} &2 I=\frac{\pi^{2}}{4} \\ &I=\frac{\pi^{2}}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 18

Answer:- $\frac{\pi}{12}$
Hints:- You must know the integration rules of trigonometric functions and its limits.
Given:- $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x$
Solution : $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x=I$ ...(1)
$I=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot \left(\frac{\pi}{2}-x\right)^{3 / 2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(1+b-x) d x\right]$
$\begin{aligned} &I=\int_{\pi / 6}^{\pi / 3} \frac{1}{\tan ^{3 / 2}+1} d x \\ &I=\int_{\pi / 6}^{\pi / 3} \frac{\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \end{aligned}$ ...(2)
Adding (1) and (2)
$\begin{aligned} &2 I=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\cot ^{3 / 2} x} d x+\int_{\pi / 6}^{\pi / 3} \frac{\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \\ &2 I=\int_{\pi / 6}^{\pi / 3} \frac{1+\cot ^{1 / 2} x}{1+\cot ^{3 / 2} x} d x \\ &2 I=\int_{\pi / 6}^{\pi / 3} 1 \cdot d x \end{aligned}$
$\begin{aligned} &2 I=[x]_{\pi / 6}^{\pi / 3} \\ &2 I=[\pi / 3-\pi / 6] \\ &2 I=\frac{2 \pi-\pi}{6} \end{aligned}$
$I=\frac{\pi}{12}$

Definite Integrals Exercise 19.4 (b) Question 19

Answer:- $\frac{\pi}{4}$
Hints:- You must know the integration rules of trigonometric functions and its limits.
Given:- $\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x$
Solution : $I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x$ ....(1)
$I=\int_{0}^{\pi / 2} \frac{\tan ^{7}\left(\frac{\pi}{2}-x\right)}{\tan ^{7}\left(\frac{\pi}{2}-x\right)+\cot ^{7}\left(\frac{\pi}{2}-x\right)} d x$
$\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &...\text { (2) }\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \end{aligned}$
Adding both
$\begin{aligned} &2 I=\int_{0}^{\pi / 2} \frac{\tan ^{7} x+\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x \\ &2 I=\int_{0}^{\pi / 2} 1 \cdot d x \\ &2 I=[x]_{0}^{\pi / 2} \end{aligned}$
$\begin{aligned} &2 I=\frac{\pi}{2} \\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 20

Answer:- 3
Hints:- You must know about the rules of integration.
Given:- $\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x$
Solution : $\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x$ ....(1)
Then , $I=\int_{2}^{8} \frac{\sqrt{10-(2+8-x)}}{\sqrt{2+8-x}+\sqrt{10-(2+8-x)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x$ ...(2)
Adding (1) and (2)
$\begin{aligned} &2 I=\int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x \\ &2 I=\int_{2}^{8} 1 \cdot d x \\ &2 I=[x]_{2}^{8} \end{aligned}$
$\begin{aligned} &2 I=[8-2] \\ &I=\frac{6}{2} \\ &I=3 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 21

Answer:- $\frac{\pi}{3}$
Hints:- You must know the integration rules of trigonometric functions.
Given:- $\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x$
Solution : $\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x=I$ ....(1)
$\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \cdot \sin (\pi-x) \cdot \cos ^{2}(\pi-x) \cdot d x \\ &I=\int_{0}^{\pi}(\pi-x) \cdot \sin x \cdot \cos ^{2} x \cdot d x \end{aligned}$ .....(2)
Adding (1) and (2)
$\begin{aligned} &2 I=\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x+\int_{0}^{\pi}(\pi-x) \cdot \sin x \cdot \cos ^{2} x \cdot d x \\ &2 I=\int_{0}^{\pi} \pi \sin x \cdot \cos ^{2} x \cdot d x \end{aligned}$
Let $\cos x = t$
$\begin{aligned} &x=0, t=1 \\ &x=\pi, t=-1 \end{aligned}$
$\begin{aligned} &2 I=\int_{1}^{-1}-\pi t^{2} d t \\ &I=\frac{-\pi}{2}\left[\frac{t^{3}}{3}\right]_{1}^{-1} \\ &I=\frac{-\pi}{2}\left[\frac{-1}{3}-\frac{1}{3}\right] \end{aligned}$
$\begin{aligned} &I=\frac{-\pi}{2}\left[\frac{2}{3}\right] \\ &I=\frac{\pi}{3} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 22

Answer:- $\frac{\pi^{2}}{16}$
Hints:- You must know the integration rules of trigonometric functions and its limits.
Given:- $\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
Solution : $\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
$\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cdot \sin \left(\frac{\pi}{2}-x\right) \cdot \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x \\ &I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cdot \cos x \cdot \sin x}{\cos ^{4} x+\sin ^{4} x} d x \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\cos x \cdot \sin x}{\sin ^{4} x+\cos ^{4} x} d x-\int_{0}^{\pi / 2} \frac{x \cdot \cos x \cdot \sin x}{\sin ^{4} x+\cos ^{4} x} d x \\ &I=\frac{\pi}{2 \times 2} \int_{0}^{\pi / 2} \frac{2 \tan x \cdot \sec ^{2} x}{1+\left(\tan ^{2} x\right)^{2}} d x-I \end{aligned}$
Let $\tan^{2}x = z$
$\begin{aligned} &2 \tan x \sec ^{2} x \cdot d x=d z \\ &x=0, z=0 \\ &x=\frac{\pi}{2}, z=\infty \\ &2 I=\frac{\pi}{4} \int_{0}^{\infty} \frac{d z}{1+z^{2}} \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{8} \int_{0}^{\infty}\left[\tan ^{-1} z\right]_{0}^{\infty} \\ &I=\frac{\pi}{8} \int_{0}^{\infty}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &I=\frac{\pi}{8}\left[\frac{\pi}{2}-0\right] \\ &=\frac{\pi^{2}}{16} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 23

Answer:- 0
Hints:- You must know the integration rules of trigonometric functions .
Given:- $\int_{-\pi / 2}^{\pi / 2} x \cdot \cos ^{2} x \cdot d x$
Solution : $I=\int_{-\pi / 2}^{\pi / 2} x \cdot \cos ^{2} x \cdot d x$
$\begin{aligned} &f(x)=x \cdot \cos ^{2} x \\ &f(-x)=(-x) \cos ^{2}(-x) \\ &=-x \cos ^{2} x \\ &=-f(x) \end{aligned}$
Hence, $f(x)$ is an odd function.
Since, $\int_{-a}^{a} f(x) d x=0, \text { if } f(x)$ is an odd function
$\therefore I = 0$

Definite Integrals Exercise 19.4 (b) Question 24

Answer:- $\frac{3\pi}{8}$
Hints:- You must know the integration rules of trigonometric functions and its limits.
Given:- $\int_{-\pi / 2}^{\pi / 2} \sin ^{4} x \cdot d x$
Solution : $\int_{-\pi / 2}^{\pi / 2} \sin ^{4} x \cdot d x$
$\begin{aligned} &=\int_{-\pi / 2}^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\cos ^{2} 2 x-2 \cos ^{2} 2 x\right) \cdot d x \\ &=\frac{1}{4} \int_{-\pi / 2}^{\pi / 2}\left(1+\frac{1+\cos ^{4} x}{2}-2 \cos 2 x\right) \cdot d x \end{aligned}$
$\begin{aligned} &=\frac{3}{8} \int_{-\pi / 2}^{\pi / 2} d x+\frac{1}{8} \int \cos ^{4} x \cdot d x-\frac{1}{2} \int \cos 2 x \cdot d x \\ &\left.\left.\left.=\frac{3}{8} x .\right]_{-\pi / 2}^{\pi / 2}+\frac{1}{32} \sin ^{4} x\right]_{-\pi / 2}^{\pi / 2}-\frac{1}{2} \sin 2 x\right]_{-\pi / 2}^{\pi / 2} \\ &=\frac{3 \pi}{8} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 25 Subquestion (i)

Answer:- 0
Hints:- You must know the integration rules of logarithmic functions.
Given:- $\int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right) d x$
Solution :
$\begin{aligned} &f(x)=\log \left(\frac{2-x}{2+x}\right) \\ &f(-x)=\log \left(\frac{2-x}{2+x}\right) \end{aligned}$
$\begin{aligned} &=\log (2+x)-\log (2-x) \\ &=-[\log (2-x)-\log (2+x)] \\ &=-\log \left(\frac{2-x}{2+x}\right) \end{aligned}$
$\begin{aligned} &=-f(x) \\ f(-x) &=-f(x) \end{aligned}$
$f(x)$ is an odd function
So $=\int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right) \cdot d x=0$

Definite Integrals Exercise 19.4 (b) Question 25 Subquestion (ii)

Answer:- 0
Hints:- You must know the integral rules of logarithmic functions.
Given:- $\int_{-\pi}^{\pi}\left(1-x^{2}\right) \sin x \cdot \cos ^{2} x \cdot d x$
Solution : $\int_{-\pi}^{\pi}\left(1-x^{2}\right) \sin x \cdot \cos ^{2} x \cdot d x=f(x)$
$\begin{aligned} &f(-x)=-\left(1-x^{2}\right) \sin x \cdot \cos ^{2} x\\ &=-f(x) \text {, is an odd function }\\ &I=\int_{-\pi}^{\pi}\left(1-x^{2}\right) \sin x \cdot \cos ^{2} x . d x=0 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 26

Answer:- $\frac{\pi }{4}-\frac{1}{2}$
Hints:- You must know the integral rules of trignometric functions with its limits.
Given:- $\int_{-\pi / 4}^{\pi / 4} \sin ^{2} x \cdot d x$
Solution : $\int_{-\pi / 4}^{\pi / 4} \sin ^{2} x \cdot d x$
$\begin{aligned} &=\int_{-\pi / 4}^{\pi / 4} \frac{1-\cos 2 x}{2} \cdot d x \\ &=\frac{1}{2} \int_{-\pi / 4}^{\pi / 4}(1-\cos 2 x) \cdot d x \\ &=\frac{1}{2}[x-\sin 2 x]_{-\pi / 4}^{\pi / 4} \end{aligned}$
$\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{4}-\sin 2 \times \frac{\pi}{4}-\left(\frac{-\pi}{4}\right)+\sin \left(2 \times \frac{-\pi}{4}\right)\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}-1+0\right] \\ &=\frac{2 \pi}{2 \times 4}-\frac{1}{2} \\ &=\frac{\pi}{4}-\frac{1}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 27

Answer:- $-\pi \log 2$
Hints:- You must know the integral rules of logarithmic functions.
Given:- $\int_{0}^{\pi} \log (1-\cos x) d x$
Solution :
$\begin{aligned} &I=\int_{0}^{\pi} \log (1-\cos x) d x \\ &=\int_{0}^{\pi} \log \left(2 \sin ^{2} \frac{x}{2}\right) d x \\ &=\int_{0}^{\pi} \log 2 \cdot d x+2 \int_{0}^{\pi} \log \sin \frac{x}{2} \cdot d x \end{aligned}$
$\begin{array}{ll} \text { Let } t=\frac{x}{2} &\; \; \; \; \; \; x \rightarrow 0 ; t \rightarrow 0 \\ d t=\frac{1}{2} d x & \; \; \; \; \; \; x \rightarrow \pi ; t \rightarrow \frac{\pi}{2} \end{array}$
$\begin{aligned} &I=\log 2[x]_{0}^{\pi}+4 \int_{0}^{\frac{\pi}{2}} \log \sin t d t \\ &I=\pi \log 2+4 \times\left(\frac{-\pi}{2} \log 2\right) \\ &I=-\pi \log 2 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 28

Answer:- 0
Hints:- You must know the integration rules of trignometry functions.
Given:- $\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$
Solution : $f(x)=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$
$\begin{aligned} f(-x) &=\log \left(\frac{2+\sin x}{2-\sin x}\right) d x \\ f(-x) &=-\log \left(\frac{2-\sin x}{2+\sin x}\right) d x \\ &=f(-x) \\ &=-f(x) \end{aligned}$
$\therefore f(x)$ is an odd function
$\therefore \int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x=0$

Definite Integrals Exercise 19.4 (b) Question 29

Answer:- $\pi^{2}$
Hints:- You must know the integral rules of logarithmic functions.
Given:- $\int_{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x$
Solution : $\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x+\int_{-\pi}^{\pi} \frac{2 x \cdot \sin x}{1+\cos ^{2} x} d x$
$I_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; I_{2}$
$I_{1}=0$ [being an odd function]
$\begin{aligned} &I_{2}=2 \int_{0}^{\pi} \frac{2 x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &I_{2}=4 \int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}$
Where $I_{3}=4 I_{2}+0=4 I$
Let $I_{3}=\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x$
$\begin{aligned} &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin (\pi-x)}{1+\cos ^{2}(\pi-x)} \cdot d x \\ &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &=\int_{0}^{\pi} \frac{\pi \cdot \sin x}{1+\cos ^{2} x} \cdot d x-\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}$
$\begin{aligned} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x-I_{3} \\ 2 I_{3} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}$
Put $\cos x = t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x\rightarrow 0,t=1$
$-\sin x \cdot d x=d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t=-1$
$\begin{aligned} &2 I_{3}=-\pi \int_{1}^{-1} \frac{d t}{1+t^{2}} \\ &=\pi\left[\tan ^{-1} x\right]_{-1}^{1} \\ &2 I_{3}=\frac{\pi^{2}}{2} \end{aligned}$
$\begin{aligned} &I_{3}=\frac{\pi^{2}}{4} \\ &\therefore I=\pi^{2} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 30

Answer:- 0
Hints:- You must know the integral rules of trignometric functions and odd even functions.
Given:- $\int_{-a}^{a} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta, a>0$
Solution : $f(x)=\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right)$
$\begin{aligned} &f(-x)=\log \left(\frac{a+\sin \theta}{a-\sin \theta}\right) \\ &=-\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) \\ &=-f(x) \end{aligned}$
$f(x)$ is an odd function
$\therefore \int_{-a}^{a} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta=0$

Definite Integrals Exercise 19.4 (b) Question 31

Answer:- $2\log_{e}7$
Hints:- You must know the rules of integration .
Given:- $\int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x$
Solution : $I=\int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x$
$\begin{aligned} &=\int_{-2}^{2} \frac{3 x^{3}}{x^{2}+|x|+1}+\int_{-2}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x\\ &\text { Odd function } \quad \text { Even function } \end{aligned}$
$\begin{aligned} &=2 \int_{0}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x=2 \log _{e} 7 \\ &\frac{\pi}{2} \\ &\int_{0}^{1}\left(\tan ^{-1}(x)+\tan ^{-1}(1-x)-\tan ^{-1}(1-x)-\tan ^{-1}(x)\right) d x \end{aligned}$
$I=\frac{\pi}{2}$
$\begin{aligned} &I=\frac{16}{15} \sqrt{2} \\ &{\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\ &I=2 \sqrt{2}\left(\frac{4}{3}-\frac{4}{5}\right) \end{aligned}$
$\begin{aligned} &I=2 \int_{0}^{2} \frac{d t}{t} \\ &=2 \int_{0}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x \end{aligned}$
Put $x^{2}+|x|+1=t$
$\begin{gathered} 2 x+1 d x=d t \\ I=2 \int_{0}^{2} \frac{d t}{t} \\ =2\left[\log _{e}|t|\right]_{0}^{2} \end{gathered}$
$\begin{aligned} &=2\left[\log _{e}\left|x^{2}+\right| x|+1|\right]_{0}^{2} \\ &=2\left[\log _{e}(4+2+1)-\log (0+0+1)\right] \\ &=2\left[\log _{e} 7-\log _{e} 1\right] \\ &=2 \log _{e} 7 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 32 Subquestion (i)

Answer:- $\frac{\pi}{2}$
Hints:- You must know the integral rules of trignometric functions.
Given:- $\int_{-3 \pi / 2}^{-\pi / 2}\left\{\sin ^{2}(3 \pi+x)+(\pi+x)^{3}\right\} d x$
Solution : $\int_{-3 \pi / 2}^{-\pi / 2}\left\{\sin ^{2}(3 \pi+x)+(\pi+x)^{3}\right\} d x$
$I=\int_{-3 \pi / 2}^{-\pi / 2}\left\{(x+\pi)^{3}+\sin ^{2} x\right\} d x$ ....(1)
$\begin{aligned} &=\int_{-3 \pi / 2}^{-\pi / 2}\left(\left(\frac{-\pi}{2}-\frac{-3 \pi}{2}-x-\pi\right)^{3}+\sin ^{2}\left(\frac{-\pi}{2}-\frac{-3 \pi}{2}-x\right)\right) d x\\ &=\int_{-3 \pi / 2}^{-\pi / 2}\left(-(x+\pi)^{3}+\sin ^{2} x\right) d x \end{aligned}$ ......(2)
Add (1) and (2)
$\begin{aligned} &2 I=\int_{-3 \pi / 2}^{-\pi / 2} 2 \sin ^{2} x \cdot d x \\ &2 I=2 \int_{-3 \pi / 2}^{-\pi / 2}\left(\frac{1-\cos 2 x}{2}\right) \cdot d x \\ &2 I=2 \int_{-3 \pi / 2}^{-\pi / 2} \frac{1}{2} \cdot d x-2 \int_{-3 \pi / 2}^{-\pi / 2}\left(\frac{\cos 2 x}{2}\right) \cdot d x \end{aligned}$
$\begin{aligned} &\left.2 I=x]_{-3 \pi / 2}^{-\pi / 2}-\sin 2 x\right]_{-3 \pi / 2}^{-\pi / 2} \\ &2 I=\frac{-\pi}{2}-\left(\frac{-3 \pi}{2}\right)-\sin 2\left(\frac{-\pi}{2}\right)+\sin 2 x \cdot\left(\frac{-3 \pi}{2}\right) \\ &2 I=\frac{-\pi}{2}+\frac{3 \pi}{2}-\sin (-\pi)+\sin (-3 \pi) \end{aligned}$
$\begin{aligned} &2 I=\frac{2 \pi}{2}-0 \\ &I=\frac{\pi}{2} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 32 Subquestion (ii)

Answer:- 0
Hints:- You must know the integration rules of trignometric functions .
Given:- $\int_{0}^{1} \tan ^{-1}\left(\frac{1-2 x}{1+x-x^{2}}\right) d x$
Solution :
$\begin{aligned} &I=\int_{0}^{1} \tan ^{-1}\left(\frac{1-2 x}{1+x-x^{2}}\right) d x \\ &I=\int_{0}^{1} \tan ^{-1} \frac{-x+(1-x)}{1+x(1-x)} d x \\ &I=\int_{0}^{1} \tan ^{-1}(-x)+\tan ^{-1}(1-x) d x \end{aligned}$ ....(1)
$\begin{aligned} &I=\int_{0}^{1} \tan ^{-1}(1+x) d x+\tan ^{-1}(1-1+x) d x \\ &I=\int_{0}^{1} \tan ^{-1}(1-x)+\tan ^{-1}(x) d x \end{aligned}$ .....(2)
Adding both
$\begin{aligned} &2 I=\int_{0}^{1}\left(\tan ^{-1}(x)+\tan ^{-1}(1-x)-\tan ^{-1}(1-x)-\tan ^{-1}(x)\right) d x \\ &2 I=0 \\ &I=0 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 33

Answer:- $\frac{16}{15}\sqrt{2}$
Hints:- You must know the rules of integration.
Given:- $\int_{0}^{2} x \sqrt{2-x} \cdot d x$
Solution : $I=\int_{0}^{2} x \sqrt{2-x} \cdot d x$
$I=\int_{0}^{2}(2-x) \sqrt{x} \cdot d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$\begin{aligned} &I=\int_{0}^{2} 2 \sqrt{x} d x-\int_{0}^{2} x^{3 / 2} x \cdot d x \\ &\left.\left.=2 \times \frac{2}{3} x^{3 / 2}\right]_{0}^{2}-\frac{2}{5} x^{5 / 2}\right]_{0}^{2} \\ &I=2 \sqrt{2}\left(\frac{4}{3}-\frac{4}{5}\right) \\ &I=\frac{16}{15} \sqrt{2} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 34

Answer:- 0
Hints:- You must know about the integration rules of logarithm functions.
Given:- $I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$
Solution :
$\begin{aligned} & I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x \\ &I=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x \end{aligned}$ ....(1)
$\begin{aligned} &I=\int_{0}^{1} \log \frac{(1-(1-x))}{1-x} d x \\ &I=\int_{0}^{1} \log \frac{x}{1-x} d x \end{aligned}$ .....(2)
Adding both
$\begin{aligned} &2 I=\int_{0}^{1} \log \frac{1-x}{x} d x+\int_{0}^{1} \log \frac{x}{1-x} d x \\ &2 I=\int_{0}^{1} \log \left(\frac{1-x}{x} \cdot \frac{x}{1-x}\right) d x \\ &2 I=\int_{0}^{1} \log 1 \cdot d x \end{aligned}$
$\begin{aligned} &2 I=0 \\ &I=0 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 35

Answer:- $\frac{2}{\pi}$
Hints:- You must know the rules of integration.
Given:- $\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x$
Solution : $f(x)=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x$
$|f(x)|=f(x) \text { when } x \geq 0 \text { and }-f(x) \text { when } x<0$
$\text { When } f(x)=x \cos (\pi x) \text {, we have }$
$\begin{aligned} &|x \cos (\pi x)|=\left\{\begin{array}{l} x \cos (\pi x) ;-1 \leq x \leq 1 / 2 \\ -x \cos (\pi x) ;-1 / 2 \leq x<0 \\ x \cos \pi x ; 0 \leq x<1 / 2 \\ -x \cos (\pi x) ; 1 / 2 \leq x<1 \end{array}\right. \\ &I=\int_{-1}^{1}|x \cdot \cos \pi x| \cdot d x \end{aligned}$
$\\I=\int_{-1}^{-1 / 2} x \cdot \cos (\pi x)-\int_{-1 / 2}^{0} x \cdot \cos (\pi x) \cdot d x+\int_{0}^{1 / 2} x \cdot \cos (\pi x) \cdot d x-\int_{-1 / 2}^{1} x \cdot \cos (\pi x) \cdot d x$
Now we can easily compile indefinite integral
$\begin{aligned} I &=\int x \cdot \cos (\pi x) \cdot d x \\ &=\frac{x \cdot \cos (\pi x)}{\pi}+\frac{\cos (\pi x)}{\pi} \\ \end{aligned}$
$\therefore I^{\prime}=\left(I_{1 / 2}-I_{-1}\right)-\left(I_{0}-I_{-1 / 2}\right)+\left(I_{1 / 2}-I_{0}\right)-\left(I_{1}-I_{1 / 2}\right)$
Where
$\begin{gathered} I_{-1}=\frac{-1}{\pi^{2}} \\ I_{-1 / 2}=\frac{1}{2 \pi} \\ I_{0}=\frac{1}{\pi^{2}} \end{gathered}$
$\begin{aligned} &I_{1 / 2}=\frac{1}{2 \pi} \\ &I_{1}=\frac{-1}{\pi^{2}} \end{aligned}$
Putting values
We get $I=\frac{2}{\pi}$

Definite Integrals Exercise 19.4 (b) Question 36

Answer:- $\frac{\pi^{2}}{2 \sqrt{2}}$
Hints:- You must know the integration rules of trigonometric functions.
Given:- $\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x$
Solution : $\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x$ ....(1)
Then
$\begin{gathered} I=\int_{0}^{\pi} \frac{\pi-x}{1-\sin ^{2}(\pi-x)}+\cos ^{7}(\pi-x) \cdot d x \\ \quad I=\int_{0}^{\pi} \frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x \cdot d x \end{gathered}$ ....(2)
Adding (1) +(2)
$\begin{aligned} &\left.2 I=\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x+\frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x\right) \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{1+\sin ^{2} x} d x \end{aligned}$
Dividing numerator and denominator by $\cos^{2} x$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{\sec ^{2} x+\tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \end{aligned}$ $\left[\because \int_{0}^{2 a} f(x) d x=\int_{0}^{2} \int_{0}^{a} f(x)\right]$
Put $\tan x = z$
Then $\sec^{2}x.dx=dz\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \text{when}x\rightarrow 0,z\rightarrow 0$
$x=\frac{\pi}{2},z\rightarrow \infty$
$\begin{aligned} &2 I=2 \pi \int_{0}^{\infty} \frac{d z}{1+(\sqrt{2} z)^{2}} \\ &\left.2 I=2 \pi \times \frac{\tan ^{-1} \sqrt{2} z}{\sqrt{2}}\right]_{0}^{\infty} \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{\sqrt{2}} \times\left(\tan ^{-1} \infty-\tan ^{-1} 0\right) \\ &I=\frac{\pi}{\sqrt{2}} \times\left(\frac{\pi}{2}-0\right) \\ &I=\frac{\pi^{2}}{2 \sqrt{2}} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 37

Answer:

Answer:- $\frac{\pi(\pi / 2-\alpha)}{\cos \alpha}$
Hints:- You must know the integral rules of trigonometric functions.
Given:- $\int_{0}^{\pi} \frac{x}{1+\sin \alpha \sin x} d x$
Solution : $I=\int_{0}^{\pi} \frac{x}{1+\sin \alpha \sin x} d x$
$\begin{aligned} &I=\int_{0}^{\pi} \frac{\pi-x}{1+\sin \alpha \sin (\pi-x)} d x \\ &I=\int_{0}^{\pi} \frac{\pi}{1+\sin \alpha \sin x} d x-\int_{0}^{\pi} \frac{x}{1+\sin \alpha \sin x} d x \\ &I=\int_{0}^{\pi} \frac{\pi}{1+\sin \alpha \sin x}-I \end{aligned}$
$\begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{1+\sin \alpha \sin x} \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{\sin \alpha \sin x+1} d x \end{aligned}$
Substituting $\sin x=\frac{2 \tan x / 2}{1+\tan ^{2} x / 2}$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{1+\tan ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan ^{2} x / 2} d x \\ &I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan x / 2} d x \end{aligned}$
Let $\tan \frac{x}{2}=t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { Also when } x \rightarrow 0, t \rightarrow 0$
$\sec ^{2} \frac{x}{2}=2 d t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t \rightarrow \infty$
$\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2}+2 t \sin \alpha+1} \\ &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(t+\sin \alpha)^{2}+\cos ^{2} \alpha} d t \\ &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1}\left(\frac{1+\sin \alpha}{\cos \alpha}\right)\right]_{0}^{\infty} \end{aligned}$
$\begin{aligned} &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1} \infty-\tan ^{-1}(\tan \alpha)\right] \\ &I=\frac{\pi}{\cos \alpha}\left[\frac{\pi}{2}-\alpha\right] \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 38

Answer:- 0
Hints:- You must know the integral rules of trignometric functions and its limits.
Given:- $\int_{0}^{2 \pi} \sin ^{100} x \cos ^{|10|} x \cdot d x$
Solution : $\int_{0}^{2 \pi} \sin ^{100} x \cos ^{|10|} x \cdot d x$
Suppose : $f(x)=\sin ^{100} x \cos ^{\left |1 0 \right |} x$
Now $f(2 \pi-x)=\sin ^{100}(2 \pi-x) \cos ^{\left |1 0 \right |}(2 \pi-x)$
$\begin{aligned} &=(-\sin x)^{100}(\cos x)^{|10|} \\ &=\sin ^{100} x \cdot \cos ^{|10|} x \\ &=f(x) \end{aligned}$
$\begin{aligned} \therefore I &=\int_{0}^{2 \pi} \sin ^{100} x \cos ^{10} x \cdot d x \\ &=2 \int_{0}^{\pi} \sin ^{100} x \cos ^ {10} x \cdot d x \end{aligned}$
Again
$\begin{aligned} f(\pi-x) &=\sin ^{100}(\pi-x) \cdot \cos ^{|10|}(\pi-x) \\ &=\sin ^{100} x \cdot(-\cos x)^{|10|}x \\ &=-\sin ^{100} x \cdot \cos ^{10}x \\ &=-f(x) \end{aligned}$
Its and odd function
$\begin{aligned} \therefore I &=2 \times 0 \\ &=0 \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 39

Answer:- $\frac{\pi}{4}(a+b)$
Hints:- You must know the integral rules of trignometric functions.
Given:- $\int_{0}^{\pi / 2} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x$
Solution : $I=\int_{0}^{\pi / 2} \frac{a \cdot \sin x+b \cos x}{\sin x+\cos x}$ ....(1)
$I=\int_{0}^{\pi / 2} \frac{a \cdot \sin (\pi / 2-x)+b \cos \left(\pi / 2^{-x)}\right.}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x$
$I=\int_{0}^{\pi / 2} \frac{a \cdot \cos x+b \sin x}{\sin x+\cos x} d x$ .....(2)
Adding both
$\begin{aligned} &2 I=\int_{0}^{\pi / 2}(a+b) \frac{\cos x+\sin x}{\sin x+\cos x} d x \\ &2 I=\int_{0}^{\pi / 2}(a+b) \cdot d x \\ &2 I=(a+b)[x]_{0}^{\pi / 2} \end{aligned}$
$\begin{aligned} &2 I=(a+b)\left(\frac{\pi}{2}\right) \\ &I=\frac{(a+b) \pi}{4} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 40

Answer:- $\frac{1}{\pi}\left(\frac{5}{2}-\frac{1}{\pi}\right)$
Hints:- You must know the integral rules of trignometric functions.
Given:- $\int_{0}^{3 / 2}|x \cos \pi x| d x$
Solution : $\int_{0}^{3 / 2}|x \cos \pi x| d x$
$\begin{aligned} &0<x<\frac{1}{2} \\ &0<\pi x<\frac{\pi}{2} \\ &\cos \pi x>0 \rightarrow x \cos (\pi x)>0 \end{aligned}$
$\begin{aligned} &|x \cos \pi x|=x \cos \pi x \\ &\frac{1}{2}<x<\frac{3}{2} \end{aligned}$
$\begin{aligned} &\frac{\pi}{2}<\pi x<\frac{3 \pi}{2} \\ &\cos \pi x<0 \rightarrow x \cos \pi x<0 \\ &|x \cos \pi x|=-x \cos \pi x \end{aligned}$
$\begin{aligned} &I=\int_{0}^{3 / 2}|x \cos \pi x| \cdot d x \\ &I=\int_{0}^{3 / 2} x \cos \pi x+\int_{1 / 2}^{3 / 2}-(x \cos \pi x) \\ &I=\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2} x \cos \pi x \cdot d x \end{aligned}$
$\begin{aligned} &\int x \cos \pi x=x \frac{\sin \pi x}{\pi}-\int \frac{\sin \pi x}{\pi} \\ &\int x \cos \pi x=\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}} \end{aligned}$
$\begin{aligned} &I=\left[\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}}\right]_{0}^{\frac{1}{2}}-\left[\frac{x}{\pi} \sin \pi x+\frac{\cos \pi x}{\pi^{2}}\right]_{\frac{1}{2}}^{3 / 2} \\ &I=\left[\frac{1}{\pi}\left(\frac{1}{2}-0\right)+\frac{1}{\pi^{2}}(0-1)\right]-\left[\frac{1}{\pi}\left(\frac{3}{2}(-1)\right)+\frac{1}{\pi^{2}}(0-0)\right] \end{aligned}$
$\begin{aligned} &=\left[\frac{1}{2 \pi}-\frac{1}{\pi^{2}}\right]-\left[\frac{-2}{\pi}\right] \\ &=\left[\frac{5}{2 \pi}-\frac{1}{\pi^{2}}\right] \\ &=\frac{1}{\pi}\left[\frac{5}{2}-\frac{1}{\pi}\right] \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 41

Answer:- $\frac{1}{\pi}$
Hints:- You must know the integral rules of trignometric functions.
Given:- $\int_{0}^{1}|x \sin \pi x| d x$
Solution : $\int_{0}^{1}|x \sin \pi x| d x$
$\begin{aligned} &=\left[-x \frac{\cos x \pi}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{0}^{1} \\ &=\left[-1 \frac{\cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}-0+0\right] \end{aligned}$
$\begin{aligned} &=\left[(-1) \frac{-1}{\pi}+0\right] \\ &=\frac{1}{\pi} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 42

Answer:- $\frac{2}{\pi}+\frac{1}{\pi^{2}}$
Hints:- You must know the integration rules of trignometric functions.
Given:- $\int_{0}^{3 / 2}|x \sin \pi x| \cdot d x$
Solution : $\int_{0}^{3 / 2}|x \sin \pi x| \cdot d x$
$\begin{aligned} &=\int_{0}^{1} x \sin \pi x \cdot d x-\int_{1}^{3 / 2} x \sin \pi x \cdot d x \\ &=\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{0}^{1}-\left[\frac{-x \cos x \pi}{\pi}+\frac{\sin \pi x}{\pi}\right]_{1}^{3 / 2} \end{aligned}$
$\begin{aligned} &=\left[\frac{-1 \cos \pi}{\pi}+\frac{\sin \pi}{\pi^{2}}-0+0\right]-\left[\frac{-3}{2} \frac{\cos ^{3 \pi / 2}}{\pi}+\frac{\cos \pi}{\pi}+\frac{\sin ^{3 \pi / 2}}{\pi^{2}}-\frac{\sin \pi}{\pi}\right]_{1}^{3 / 2} \\ &=\frac{1}{\pi}+\frac{3}{2} \cdot 0+\frac{1}{\pi}+\frac{1}{\pi^{2}}+0 \\ &=\frac{2}{\pi}+\frac{1}{\pi^{2}} \end{aligned}$

Definite Integrals Exercise 19.4 (b) Question 43

Answer:- Proved
Hints:- You must know the rules of continuous integral functions.
Given:- $f(2 a-x)=f(x)$
Prove : $\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$
Solution : $\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x$
So L.H.S,
$\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x$ ....(1)
Let $x=2 a-t \; \; \; \; \; \; \; \; \; \quad \text { when, } x=2 a, t=0$
$dx=-dt \; \; \; \; \; \; \; \; \; \; \; \; \; \; x = a, t=a$
Subscription
$\int_{0}^{2 a} f(x) d x=-\int_{a}^{0} f(2 a-t) d t$
Using the property of integration $\left[\int_{0}^{b} f(x) d x=-\int_{b}^{a} f(x) d t\right]$
Substituting in eq (1) $\left[\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(2 a-t) d t\right]$
$\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{2 a} f(2 a-x) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int_{0}^{2 a} f(x) d x=\int_{0}^{2 a} f(2 a-x) d x\right]$
Now using the property
$\begin{aligned} &\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x=\int_{a}^{b} f(x)+g(x) \cdot d x \\ &\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x)+f(2 a-x) \cdot d x \end{aligned}$
Since $f(2a-x)=f(x)$
$\begin{aligned} &\therefore \int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x)+f(x) \cdot dx \\ &\therefore \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x \end{aligned}$
Hence proved

Definite Integrals Exercise 19.4 (b) Question 44

Answer:

Answer:- Proved
Hints:- You must know the integration rules of trignometric functions.
Given:- $f(2a-x)=-f(x)$
Prove : $\int_{0}^{2 a} f(x) d x=0$
Solution : Let $I=\int_{0}^{2 a} f(x) d x$
Using additive property
$I=\int_{0}^{a} f(x) d x+\int_{a}^{2 a} f(x) d x$
Consider the integral $\int_{a}^{2 a} f(x) d x$
$\begin{aligned} &\text { Let } x=2 a-t, \text { then } d x=-d t \\ &x=a, t=a \text { and } x=2 a, t=0 \end{aligned}$
$\begin{aligned} \therefore \int_{a}^{2 a} f(x) d x &=-\int_{a}^{0} f(2 a-t) d t \\ &=\int_{0}^{a} f(2 a-t) d t \\ &=\int_{0}^{a} f(2 a-x) d x \end{aligned}$
We have $f(2a-x)=-f(x)$
$\begin{aligned} &I=\int_{0}^{a} f(x) d x-\int_{0}^{a} f(x) d x=0 \\ &I=0 \end{aligned}$
Hence, $\int_{0}^{2 a} f(x) d x=0$

Definite Integrals Exercise 19.4 (b) Question 45 Subquestion (i)

Answer:- Proved
Hints:- You must know the rules of integration.
Given:- If $f$ is an integral function, prove $\int_{-a}^{a} f\left(x^{2}\right) d x=2 \int_{0}^{a} f\left(x^{2}\right) d x$
Solution:- $f$ is an integrable function
$\begin{aligned} &I=\int_{-a}^{a} f\left(x^{2}\right) d x \\ &f(x)=f\left(x^{2}\right) \\ &f(-x)=f\left(-x^{2}\right)=f\left(x^{2}\right) \end{aligned}$
Hence it is an even function
Using integration property, if function is an even function
$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$
Therefore,
$\int_{-a}^{a} f\left(x^{2}\right) d x=2 \int_{0}^{a} f\left(x^{2}\right) d x$
Hence proved

Definite Integrals Exercise 19.4 (b) Question 45 Subquestion (ii)

Answer:- Proved
Hints:- You must know the rules of integration.
Given:- $f$ is an integrable function
Solution:- We have, $f$ is an integrable function
$I=\int_{-a}^{a} x f\left(x^{2}\right) d x$
Let
$\begin{aligned} &f(x)=x f\left(x^{2}\right) \\ &f(-x)=-x f\left(x^{2}\right) \\ &=-f(x) \end{aligned}$
Hence it is an odd function
Using integration property,
If function is odd, $\int_{-a}^{a} f(x) d x=0$
$\therefore \int_{-a}^{a} x \cdot f\left(x^{2}\right) d x=0$
Hence proved

Definite Integrals Exercise 19.4 (b) Question 46

Answer:- Proved
Hints:- You must know the continuity of integral functions.
Given:- $f(x)$ is a continuous on $\left [ 0,2a \right ]$
Solution : $I=\int_{0}^{2 a} f(x) d x$
By additive property
$I=\int_{0}^{a} f(x) d x+\int_{0}^{2 a} f(x) d x$
Consider the integral $\int_{0}^{2 a} f(x) d x$
$\begin{aligned} &\text { Let } x=2 a-t, \text { then } d x=-d t \\ &\text { When } x=a, t=a \quad x=2 a, t=0 \end{aligned}$
Hence
$\begin{aligned} &\int_{a}^{2 a} f(x) d x=-\int_{a}^{0} f(2 a-t) d t \\ &\int_{0}^{a} f(2 a-t) d t=\int_{0}^{a} f(2 a-x) d x \end{aligned}$
Therefore,
$\begin{aligned} &I=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x \\ &=\int_{0}^{a}\{f(x)+f(2 a-x)\} \cdot d x \end{aligned}$
Hence proved.

Definite Integrals Exercise 19.4 (b) Question 47

Answer:- Proved
Hints:- You must know the rules of integration with its limits.
Given:- $f(a+b-x)=f(x)$
Prove $\int_{a}^{b} x f(x) d x=\frac{a+b}{2} \int_{a}^{b} f(x) d x$
Solution : $I=\int_{a}^{b} x f(x) d x$ ......(1)
Using property of definite integral
$I=\int_{a}^{b}(a+b-x) \cdot f(a+b-x) d x$
It is given that, $f(a+b-x)=f(x)$
$\begin{aligned} &I=\int_{a}^{b}(a+b-x) \cdot f(x) d x \\ &I=(a+b) \int_{a}^{b} f(x) \cdot d x-\int_{a}^{b} x f(x) d x \end{aligned}$ .....(2)
Adding (1) and (2)
$\begin{aligned} &2 I=(a+b) \int_{a}^{b} f(x) \cdot d x \\ &I=\frac{(a+b)}{2} \int_{a}^{b} f(x) \cdot d x \end{aligned}$
Hence proved.

Definite Integrals Exercise 19.4 (b) Question 48

Answer:- Proved
Hints:- You must know the continuity rules of integration.
Given:- $f(x)$ is a continuous function on $[-a,a]$
Solution:- $\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x)+f(-x) d x$
$\begin{aligned} &{\left[\begin{array}{ll} f(-x)=f(x) & f(x) \rightarrow \text { even } \\ f(-x)=-f(x) & f(x) \rightarrow \text { odd } \end{array}\right]} \\ &\int_{-a}^{a} f(x) d x=\left\{\begin{array}{lr} \int_{0}^{a} f(x)+f(-x) d x & f(x) \rightarrow \text { even } \\ \int_{0}^{a} f(x)+f(-x) d x & f(x) \rightarrow \text { odd } \end{array}\right] \end{aligned}$
$\int_{-a}^{a} f(x) d x=\left\{\begin{array}{ll} \int_{0}^{a} 2 f(x) d x & f(x) \rightarrow \text { even } \\ 0 & f(x) \rightarrow \text { odd } \end{array}\right]$
Hence $\left.\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x)+f(-x)\right] d x$ is true because this function is not zero

Definite Integrals Exercise 19.4 (b) Question 49

Answer:- Proved
Hints:- You must know the rules of integration for trignometric functions.
Given:- Prove $\int_{0}^{\pi} x f(\sin ) d x=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \cdot d x$
Solution : Let $I=\int_{0}^{\pi} x f(\sin ) d x$ ....(1)
$\begin{aligned} I &=\int_{0}^{\pi}(\pi-x) f(\sin (\pi-x)) d x \\ &=\int_{0}^{\pi}(\pi-x) f(\sin x) d x \end{aligned}$ ......(2)
Adding (1) and (2)
$2 I=\int_{0}^{\pi}(x+\pi-x) f(\sin x) d x$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} f(\sin x) d x \\ &I=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) d x \\ &\therefore \int_{0}^{\pi} f(\sin x) d x=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \cdot d x \end{aligned}$
Hence proved.

Definite Integrals Exercise 19.4 (b) Question 50

Answer:- $\frac{2}{(n+1)(n+2)(n+3)}$
Hints:- You must know the rules of integration with its limits.
Given:- Prove $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
Hence evaluate $\int_{0}^{1} x^{2}(1-x)^{n} d x$
Solution : Let $I=\int_{0}^{a} f(x) d x$ ....(1)
$\begin{aligned} &\text { Put } x=a-t \\ &\therefore d x=-d t \end{aligned}$
When, $x=0,t=a-0=a$
$x=a,t=a-a=0$
$\begin{aligned} I=\int_{0}^{a} f(x) d x &=\int_{a}^{0} f(a-t)(-d t) \\ &=-\int_{a}^{0} f(a-t) d t \\ &=\int_{0}^{a} f(a-x) d x \end{aligned}$ $\left[\because \int_{a}^{b} f(x) \cdot d x=-\int_{b}^{a} f(x) \cdot d x\right]$
$\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
Now, $\int_{0}^{1} x^{2}(1-x)^{n} d x$
$\begin{aligned} I &=\int_{0}^{1}\left(1-x^{2}\right)[1-(1-x)]^{n} d x \\ &=\int_{0}^{1}\left(1-2 x+x^{2}\right) x^{n} d x \\ &=\int_{0}^{1}\left(x^{n}-2 x^{n+1}+x^{n+2}\right) d x \end{aligned}$
$\begin{aligned} &=\left[\frac{x^{n+1}}{n+1}-2 \cdot \frac{x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3}\right]_{0}^{1} \\ &=\left[\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}\right] \end{aligned}$
$\begin{aligned} &=\frac{(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)}{(n+1)(n+2)(n+3)} \\ &=\frac{n^{2}+5 n+6-2 n^{2}-8 n-6+n^{2}+3 n+2}{(n+1)(n+2)(n+3)} \\ &=\frac{2}{(n+1)(n+2)(n+3)} \end{aligned}$

The mathematics syllabus of the class 12 students consists of a challenging chapter, Definite Integrals. It is the 19th chapter and has five exercises. The fourth exercises consist of ex 19.4(a) and ex.19.4(b). The latter portion has 53 questions given in the textbook. The concept present in this exercise is to evaluate and prove the definite integrals of the given sums. The students can use the RD Sharma Class 12 Chapter 19 Exercise 19.4(b) reference material whenever a doubt strikes them.

The solutions are provided for every sum given in the textbook. The concepts are explained clearly, and the sums are solved in every possible method to arrive at the solution. The RD Sharma solution follow the NCERT pattern, and this gives a valid reason why the CBSE students must use it. The RD Sharma Class 12th Exercise 19.4(b) consists of numerous practice sums for the students to check their understanding capability about the concept of the Definite Integrals.

If you find any difficulties in solving integration sums, good practice must be undergone. A teacher or a tutor is not required 24 x 7 when you have the Class 12 RD Sharma Chapter 19 Exercise 19.4(b) Solution material. All the solutions provided in this book are given by wise experts who have a strong foundation in their respective subjects. Once integration becomes easier to solve, soon you can witness yourself crossing your benchmark score and achieving good marks. And to attain this level, you need the RD Sharma Class 12th Exercise 19.4(b) book.

The RD Sharma Class 12 Solutions Definite Integrals Ex 19.4(b) books are more convenient for the students to refer to as they are available as a freebie. The students can visit the Career 360 website to find books free of charge.

Teachers use the RD Sharma Class 12 Solutions Chapter 19 Ex 19.4(b) to prepare questions for the exams, and the students use it to solve homework and assignment sums. Therefore, this recourse material would be of great help to you in your exam preparation. Hence, visit the Career 360 website now and get a copy for yourself.

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Frequently Asked Question (FAQs)

1. What makes RD Sharma solution books so unique?

Experts provide the solution given in the RD Sharma books.
Numerous practice questions are available.
It is accessed free of cost.

2. Which is the solution book that the class 12 students who have doubts about chapter 19 can refer to?

The RD Sharma Class 12th Exercise 19.4(b) solution book is the best choice for the students who have doubts in the Definite Integrals chapter.

3. How many questions are answered from ex 19.4(b) in the RD Sharma books?

The RD Sharma Class 12th Exercise 19.4(b) material consists of answers for all the 53 questions present in this exercise.

4. What makes it easier for the CBSE board students to use the RD Sharma books for doubt clarifications?

The RD Sharma books are based on the NCERT curriculum, this makes the CBSE students attune to these books for their reference.

5. How can I download the RD Sharma reference book from the Career 360 website?

Browse for the career 360 website and search the name of the solution book. Once you get it, you can access its content and click the Download button to save the PDF file on your device