RD Sharma Solutions Class 12 Mathematics Chapter 18 RE

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:47 PM IST

The RD Sharma books are the most prescribed solution materials by the CBSE board schools to their students. It helps them in solving sums that are complicated, and they can recheck their answers effortlessly. Moreover, in class 12, chapter 18 in mathematics has Revision Exercise (RE) for the students to practice better in this chapter. The questions given in this are challenging to solve; hence, they can use the RD Sharma Class 12th RE book as a reference.

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RD Sharma Class 12 Solutions Chapter 18 RE Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:RE
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter 18 RE Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:RE

Indefinite Integrals Exercise Revision Exercise Question 1

Answer:
$\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{8}{2}}\right]+c$
Given:
$\int \frac{1}{\sqrt{x}+(\sqrt{x+1})} d x$
Hint:
Do Rationalization
Solution:
we have,
$\int \frac{1}{\sqrt{x}+(\sqrt{x+1})} d x$
$=\int \frac{1}{\sqrt{x}+\sqrt{x+1}} \times \frac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}} d x \quad(\because \text { rationalizing })$
$\begin{aligned} &=\int \frac{\sqrt{x}-\sqrt{x+1}}{x-(x+1)} d x \ldots\left[(a-b)(a+b)=\left(a^{2}-b^{2}\right)\right] \\ &=\int-\frac{(\sqrt{x+1}-\sqrt{x)}}{-1} d x \end{aligned}$
$\begin{aligned} &=\int \sqrt{x+1}-\sqrt{x} d x \\ &=\frac{(x+1)^{\frac{8}{2}}}{\frac{3}{2}}-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c \end{aligned}$
$=\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{8}{2}}\right]+c$

Indefinite Integrals Exercise Revision Exercise Question 2

Answer:
$x+\frac{x^{2}}{2}+\frac{x^{8}}{3}+\frac{x^{4}}{4}+c$
Hint:
Use the formula $a^{2}-b^{2}=(a-b)(a+b)$
Solution: we have
$\begin{aligned} &=\int \frac{1-x^{4}}{1-x} d x \\ &=\int \frac{\left(1-x^{2}\right)\left(1+x^{2}\right)}{1-x} d x \\ &=\int \frac{(1-x)(1+x)\left(1+x^{2}\right)}{1-x} d x \\ &=\int(1+x)\left(1+x^{2}\right) d x \\ &=\int 1+x^{2}+x+x^{3} d x \\ &=x+\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x^{4}}{4}+c \\ &=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+c \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 3

Answer:
$\int-\frac{1}{1+x}-\frac{1}{2(x+1)^{2}}+C$
Given:
$\int \frac{x+2}{(x+1)^{3}} d x$
Hint:
Use power rule for integration.
Solution: we have
$\int \frac{x+2}{(x+1)^{3}} d x$
$\begin{aligned} &=\int \frac{x+1+1}{(x+1)^{3}} d x \\ &=\int \frac{x+1}{(x+1)^{3}}+\frac{1}{(x+1)^{3}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{1}{(x+1)^{2}}+\frac{1}{(x+1)^{3}} d x \\ &=-\frac{1}{x+1}-\frac{1}{2(x+1)^{2}}+c \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 4

Answer:
$\frac{1}{3}(4 x+7)^{\frac{2}{3}}-\frac{1}{2} \sqrt{4 x+7}+c$
Given:
$\int\left(\frac{(8 x+13)}{\sqrt{4 x+7}} d x\right.$
Hint:
Separate the terms and integrate
Solution:
$\int \frac{(8 x+13)}{\sqrt{4 x+7}} d x$
$=\int \frac{8 x+14-1}{\sqrt{4 x+7}} d x$
$=\int \frac{2(4 x+7)-1}{\sqrt{4 x+7}} d x$
$=\int \frac{2(4 x+7)}{\sqrt{4 x+7}}-\frac{1}{\sqrt{4 x+7}} d x$
$\begin{aligned} &=2 \int \sqrt{4 x+7} d x-\int(4 x+7)^{\frac{1}{2}} d x \\ &=\frac{2\left((4 x+7)^{\frac{3}{2}}\right)}{4 \times \frac{3}{2}}-2 \frac{(4 x+7)^{\frac{1}{2}}}{4}+c \\ &=\frac{\left((4 x+7)^{\frac{3}{2}}\right)}{3}-\frac{(4 x+7)^{\frac{1}{2}}}{2}+c \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 5

Answer:
$-\frac{1}{x}+\log |x+1|+c$
Given:
$\int \frac{1+x+x^{2}}{x^{2}(1+x)} d x$
Hint:
Use partial function method.
Solution:
$I=\int \frac{1+x+x^{2}}{x^{2}(1+x)} d x$
using partial function $\begin{aligned} &\frac{1+x+x^{2}}{x^{2}(1+x)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{c}{1+x} \\ &1+x+x^{2}=A\left(x+x^{2}\right)+B(1+x)+C x^{2} \\ &1+x+x^{2}=A x^{2}+A x+B+B x+C x^{2} \\ &1+x+x^{2}=B+(A+B) x+(A+C) x^{2} \\ &B=1, A+1=1 \therefore A=0, A+C=1 \therefore C=1 \end{aligned}$

now,

$\frac{1+x+x^{2}}{x^{2}(1+x)}=\int \frac{1}{x^{2}}+\frac{1}{1+x} d x$

$=-\frac{1}{x}+\log |x+1|+c$

Indefinite Integrals Exercise Revision Exercise Question 7

Answer:
$x-\tan x+\sec x+c$
Given :
$\int \frac{\sin x}{1+\sin x} d x$
Hint:
Do rationalization and then use trigonometry identities.
Solution:
$\int \frac{\sin x}{1+\sin x} d x$
On rationalising,
$\int \frac{\operatorname{Sin} x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x$
$\int \frac{\operatorname{Sin} x-\sin ^{2}}{1-\sin ^{2} x} d x$
$=\int \frac{\sin x}{\cos ^{2} x}-\tan ^{2} \mathrm{x} d x$
$=\int \frac{\sin x}{\cos x} \times \frac{1}{\cos x}-\left(\sec ^{2} x-1\right) d x$
$=\int\left(\sec x \tan x-\sec ^{2} x+1\right) d x$
$=\sec x-\tan x+x+c$

Indefinite Integrals Exercise Revision Exercise Question 8

Answer:
$\frac{x^{3}}{3}-\tan ^{-1} x+c$
Given:
$\int \frac{x^{4}+x^{2}-1}{x^{2}+1} d x$
$=\int \frac{x^{2}\left(x^{2}+1\right)-1}{x^{2}+1} d x$
$=\int \frac{x^{2}\left(x^{2}+1\right)}{x^{2}+1}-\frac{1}{x^{2}+1} d x$
$=\int x^{2} d x-\int \frac{1}{x^{2}+1} d x$
$=\frac{x^{3}}{3}-\tan ^{-1} x+c$

Indefinite Integrals Exercise Revision Exercise Question 9

Answer:
$\sin 2 x+\tan x-2 x+c$
Given:
$\int \sec ^{2} x \cos ^{2} 2 x d x$
Hint:
Use trigonometry identity.
Solution:
$\int \sec ^{2} x \cos ^{2}(2 x) d x$
$=\int \sec ^{2} x\left(2 \cos ^{2} x-1\right)^{2} d x \quad\left[\because \cos 2 x=2 \cos ^{2} x-1\right]$
$=\int \sec ^{2} x\left(4 \cos ^{4} x-4 \cos ^{2} x+1\right) d x$
$=\int 4 \cos ^{2} x-4+\sec ^{2} x d x \quad\left[\because \frac{1}{\sec ^{2} x}=\cos ^{2} x\right]$
$=4 \int \cos ^{2} x-1 d x+\int \sec ^{2} x d x$
$=2 \int 2 \cos ^{2} x d x-4 \int d x+\int \sec ^{2} x d x$
$=2 \int(1+\cos 2 \mathrm{x}) \mathrm{dx}-4 \int \mathrm{dx}+\int \sec ^{2} \mathrm{x} \mathrm{dx}$
$=2 x+2 \frac{\sin 2 x}{2}-4 x+\tan x+c$
$=\sin 2 x-2 x+\tan x+c$

Indefinite Integrals Exercise Revision Exercise Question 10

Answer:
$-\cot x-\sin 2 x-2 x+c$
Given:
$\int \operatorname{cosec}^{2} x \cos ^{2} 2 x d x$
Hint:
Use trigonometric identities and integrate it by parts.
Solution:
$\int \operatorname{cosec}^{2} x \cdot \cos ^{2} 2 x d x$
$=\int \operatorname{cosec}^{2} x\left(1-2 \sin ^{2} x\right)^{2} d x \quad\left[\because \cos 2 \mathrm{x}=1-2 \sin ^{2} \mathrm{x}\right]$
$=\int \operatorname{cosec}^{2} x\left(1+4 \sin ^{4} x-4 \sin ^{2} x\right) d x \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$=\int\left(\operatorname{cosec}^{2} x+4 \sin ^{2} x-4\right) d x \quad\left[\because \sin x=\frac{1}{\operatorname{cosec} x}\right]$
$=\int \operatorname{cosec}^{2} x d x+2 \int 1-\cos 2 x d x-4 \int d x$
$=-\cot x+2 x-\sin 2 x-4 x+c$
$=-\cot x-\sin 2 x-2 x+c$

Indefinite Integrals Exercise Revision Exercise Question 11

Answer:
$\frac{3 x}{8}+\frac{\sin 8 x}{64}-\frac{\sin 4 x}{8}+c$
Given:
$\int \sin ^{4} 2 x d x$
Hint:
You must know about trigonometric identities.
Solution:
$\int\left[\sin ^{2}(2 x)\right]^{2} d x$
$=\int \frac{(1-\cos 4 x)^{2}}{4} d x \quad \because\left[\cos 2 x=1-2 \sin ^{2} x\right]$
$=\int \frac{1-2 \cos 4 x+\cos ^{2} 4 x}{4} d x$
$=\int \frac{1-2 \cos 4 x}{4} d x+\int \frac{1+\cos 8 x}{8} d x:\left[1+\cos 2 x=2 \cos ^{2} x\right]$
$=\frac{1}{4} \int d x-\frac{1}{2} \int \cos 4 x d x+\frac{1}{8} \int d x+\frac{1}{8} \int \cos 8 x d x$
$=\frac{3}{8} x-\frac{\sin 4 x}{8}+\frac{\sin 8 x}{64}+c$

Indefinite Integrals Exercise Revision Exercise Question 12

Answer:
$\frac{\sin 3 x}{3}-\frac{\sin ^{8} 3 x}{9}+c$
Given:
$\int \cos ^{3} 3 x d x$
Hint:
Let the term and derivate it and then integrate it.
Solution:
$\int \cos ^{3} 3 x d x$
$=\int \cos ^{2} 3 x \cos 3 x d x$
$=\int\left(1-\sin ^{2} 3 x\right) \cos 3 x d x \quad\left[\because \cos ^{2} x+\sin ^{2} x=1\right]$
now,let $\mathrm{u}=\sin 3 \mathrm{x}$
$Differentiate w.r.t x$
$\frac{\mathrm{du}}{\mathrm{dx}}=3 \cos 3 \mathrm{x}$
$d u=3 \cos 3 x d x$
Now,
$\frac{1}{3} \int 3\left(1-\sin ^{2} 3 x\right) \cos 3 x d x$
$=\frac{1}{3} \int 1-u^{2} d u \quad(p u t u \& d u)$
$=\frac{1}{3} \int d u-\frac{1}{3} \int u^{2} d u$
$=\frac{1}{3} u-\frac{1}{3} \times \int u^{2} d u$
$=\frac{1}{3} \sin 3 x-\frac{\sin ^{8} 3 x}{9}+c$

Indefinite Integrals Exercise Revision Exercise Question 13

Answer:
$\frac{1}{b^{2}} \log \left(a^{2}+b^{2} \sin ^{2} x\right)+c$
Given:
$\int \frac{\sin 2 x}{a^{2}+b^{2} \sin ^{2} x} d x$
Hint:
Let the denominator and then integrate the equation.
Solution:
$\int \frac{\sin 2 x}{a^{2}+b^{2} \sin ^{2} x} d x$
$let u=a^{2}+b^{2} \sin ^{2} x$
$differentiate it with respect to \mathrm{x}$
$\frac{d u}{d x}=2 b^{2} \sin x \cos x$
$\frac{d u}{d x}=b^{2} \sin 2 x$
$\frac{1}{b^{2}} d u=\sin 2 x d x$
$now$
$I=\frac{1}{b^{2}} \int \frac{1}{u} d u \quad($ put $u \& d u)$
$=\frac{1}{b^{2}} \log |u|+c$
$=\frac{1}{b^{2}} \log \left|a^{2}+b^{2} \sin ^{2} x\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 15

Answer:
$\frac{\left(\sin ^{-1} x\right)^{4}}{4}+c$
Given:
$\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x$
Hint:
We must know about substitution formula to integrate.
Solution:
$\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x$
Let $\mathrm{u}=\sin ^{-1} \mathrm{x}$
Differentiate it with respect to x
$\frac{d u}{d x}=\frac{1}{\sqrt{1-x^{2}}}$
$d u=\frac{1}{\sqrt{1-x^{2}}} d x$
$now, \int u^{3} d u$ $(p u t u \& d u)$
$=\frac{u^{4}}{4}+c$
$\frac{\left(\sin ^{-1} x\right)^{4}}{4}+c$

Indefinite Integrals Exercise Revision Exercise Question 16

Answer:
$x-\log \left|e^{x}+1\right|+c$
Given:
$\int \frac{1}{e^{x}+1} d x$
Hint:
Do integration by separation.
Solution:
$\int \frac{1}{e^{x}+1} d x$
$=\int \frac{\left(e^{x}+1\right)\left(-e^{x}\right)}{e^{x}+1} d x$
$=\int d x-\int \frac{e^{x}}{e^{x}+1} d x$
$now, let e^{x}+1=u$
Differentiate it with respect to x
$\frac{d u}{d x}=e^{x}$
$d u=e^{x} d x$
now,
$I=\int d x-\int \frac{1}{u} d u$
$=x-\log |u|+c$
$=x-\log \left|e^{x}+1\right|+c \quad\left(\because u=e^{x}+1\right)$

Indefinite Integrals Exercise Revision Exercise Question 17

Answer:
$2 \log \left|e^{x}+1\right|-x+c$
Given:
$\int \frac{e^{x}-1}{e^{x}+1} d x$
Hint:
Use partial fraction method.
Solution:
$\int \frac{e^{x}-1}{e^{x}+1} d x$
$\operatorname{let} e^{\mathrm{x}}=u$
$e^{x} d x=d u$ $(differentiate with respect to x)$
$d x=\frac{d u}{u}$
$now,$
$I=\int \frac{u-1}{u+1} \times \frac{d u}{u} \quad(p u t u \& d u)$
use partial fraction
$u-1=A(1+u)+B u$
$u-1=A+A u+B u$
$A=-1 \& B+A=1 \quad \therefore B=2 \quad (on comparing)$
So, $\int \frac{e^{x}-1}{e^{x}+1} d x$
$=-\int \frac{1}{u}+\frac{2}{u+1} d u$
$=-\log |\mathrm{u}|+2 \log |\mathrm{u}+1|+\mathrm{c}$
$=-\log \left|\mathrm{e}^{\mathrm{x}}\right|+2 \log \left|\mathrm{e}^{\mathrm{x}}+1\right|+\mathrm{c}$
$=-2 \log \left|\mathrm{e}^{x}+1\right|-x+c \ldots(\operatorname{loge}=1)$

Indefinite Integrals Exercise Revision Exercise Question 18

Answer:
$\tan ^{-1}\left(e^{x}\right)+c$
Given:
$\int \frac{1}{e^{x}+e^{-x}} d x$
Hint:
Use substitution method.
Solution:
$\int \frac{1}{e^{x}+e^{-x}} d x$
$=\int \frac{1}{e^{x}+\left(\frac{1}{e^{x}}\right)} d x$
$=\frac{e^{x}}{\left(e^{x}\right)^{2}+1} d x$
$let u=e^{x}$
$d u=e^{x} d x$ $(on differentiating)$
$now,$
$I=\int \frac{1}{u^{2}+1} d u$
$=\tan ^{-1} u+c$
$=\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$

Indefinite Integrals Exercise Revision Exercise Question 19

Answer:
$\log |\sin x|-\frac{\sin x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+c$
Given:
$\int \frac{\cos ^{7} x}{\sin x} d x$
Hint:
Use trigonometric identities and substitution method of integration.
Solution:
$\int \frac{\cos ^{7} x}{\sin x} d x$
Let $t=\sin x$
$d t=\cos x d x$ $(on differentiating)$
$now,$
$I=\int \frac{\cos ^{7} x}{\sin x} d x$
$=\int \frac{\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right)}{t} d t$
$=\int \frac{\left(1-t^{2}\right)^{2}\left(1-t^{2}\right)}{t} d t$
$=\int \frac{\left(1+t^{4}-2 t^{2}\right)\left(1-t^{2}\right)}{t} d t$
$=\int \frac{1-3 t^{2}+3 t^{4}-t^{6}}{t} d t$
$=\int \frac{1}{t}-3 t+3 t^{3}-t^{5} d t$
$=\operatorname{logt}-\frac{3 t^{2}}{2}+\frac{3 t^{4}}{4}-\frac{t^{6}}{6}+c$
$=\log |\sin x|-\frac{\sin x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+c$

Indefinite Integrals Exercise Revision Exercise Question 20

Answer:
$-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}+c$
Given:
$\int \sin x \sin 2 x \sin 3 x d x$
Hint:
Use the formula $2 \sin A \sin B=-\cos (A+B)+\cos (A-B)\rfloor$
Solution:
$\int \sin x \sin 2 x \sin 3 x d x$
$=\int \frac{1}{2}[\cos (x-2 x)-\cos (x+2 x)] \sin 3 x d x$
$=\int \frac{1}{2}(\cos (-\mathrm{x})-\cos 3 \mathrm{x}) \sin 3 \mathrm{x} \mathrm{dx}$
$=\frac{1}{2} \int \cos x \sin 3 x-\cos 3 x \sin 3 x d x$
$=\frac{1}{2} \int \frac{1}{2}(\sin 4 x+\sin 2 x) d x-\frac{1}{2} \int \frac{1}{2}(\sin 6 x) d x$
$=\frac{1}{4}\left[\int(\sin 4 x+\sin 2 x-\sin 6 \mathrm{x}) \mathrm{dx}\right]$
$=\frac{1}{4}\left[-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}+\frac{\cos 6 x}{6}\right]+c$
$-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 \mathrm{x}}{24}+\mathrm{c}$

Indefinite Integrals Exercise Revision Exercise Question 21

Answer:
$I=\frac{1}{4}\left(x+\frac{\sin 6 x}{6}+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c$
Given:
$\int \cos x \cos 2 x \cos 3 x d x$
Hint:
To solve this equation we will use trigonometry method.
Solution:
$\int(\cos x \cos 2 x) \cos 3 x d x$
$I=\int \frac{\cos (x+2 x)+\cos (x-2 x)}{2} \cos 3 x d x \quad\left[\because \cos C \cdot \cos D=\frac{1}{2} \cos (C+D)+\cos (C-D)\right)$
$I=\frac{1}{2} \int(\cos 3 x+\cos x) \cos 3 x d x$
$I=\frac{1}{2}\left(\int \cos ^{2} 3 x d x+\int \cos x \cos 3 x\right) d x$
$I=\frac{1}{2} \int \frac{1+\cos 6 x}{2} d x+\frac{1}{2} \int \cos (x+3 x)+\cos (x-2 x) d x$
$I=\frac{1}{2} \int 1+\frac{\cos 6 x}{2} d x+\frac{1}{2} \int(\cos 4 x+\cos (2 x)) d x$
$I=\frac{1}{2} \int \frac{1}{2} d x+\int \frac{\cos 6 x}{2} d x+\frac{1}{2} \int \cos 4 x d x+\frac{1}{2} \int \cos 2 x d x$
$I=\frac{1}{2}\left(\frac{1}{2} x+\frac{1}{2} \frac{\sin 6 x}{6}+\frac{1}{2} \frac{\sin 4 x}{4}+\frac{1}{2} \frac{\sin 2 x}{2}\right)+c$
$I=\frac{1}{2}\left(\frac{x}{2}+\frac{\sin 6 x}{12}+\frac{\sin 4 x}{8}+\frac{\sin 2 x}{4}\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 22

Answer:
$\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]$
Given:
$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
Hint:
To solve this equation we use partial function method.
Solution:
$\int \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}} d x \quad \because\left[\sin x=\frac{\tan x}{\sec x}, \cos x=\frac{1}{\sec x}\right]$
$\int \frac{\frac{\tan x}{\sec x}+\frac{1}{\sec x}}{\sqrt{\frac{\operatorname{stan} x}{\sec x} \times \frac{1}{\sec x}}} d x$
$\int \frac{\tan x+1}{\sqrt{2(\tan x)}} d x$
$\frac{1}{\sqrt{2}} \int \frac{\tan x+1}{\sqrt{\tan x}} \frac{\tan ^{2} x+1}{\tan ^{2} x+1} d x$
$\frac{1}{\sqrt{2}} \int \frac{\sec ^{2} x(\tan x+1)}{\sqrt{\tan x}\left(\tan ^{2} x+1\right)} d x \quad\left[\because \tan x=u, \sec ^{2} x d x=d u\right]$
$=\frac{1}{\sqrt{2}} \int \frac{u+1}{\sqrt{u}\left(u^{2}+1\right)} d u$
$\left.=\frac{1}{\sqrt{2}} \int \frac{v^{2}+1}{\sqrt{u}\left(v^{4}+1\right)} 2 \sqrt{u} d v_{\cdots . .[\text { put } u}=v^{2}, d u=2 v d v\right]$
$=\sqrt{2} \int \frac{v^{2}+1}{v^{4}+1} d v$
$=\sqrt{2} \int \frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)} d v$
$=\frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)}=\frac{A}{v^{2}-\sqrt{2} v+1}+\frac{B}{v^{2}+\sqrt{2} v+1}$
$=\sqrt{2}\left(\frac{1}{2}\right)\left(\int \frac{1}{v^{2}-\sqrt{2} v+1} d v+\int \frac{1}{v^{2}+\sqrt{2} v+1} d v\right)$
$=\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} \int \frac{1}{v-\left(\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}} d v+\int \frac{1}{v+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)} d v$
$\left[\frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]$
$=\frac{1}{\sqrt{2}} \sqrt{2} \frac{\tan ^{-1}\left(v-\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}+\sqrt{2} \frac{\tan ^{-1}\left(v+\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}$
$=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right]$
$=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right] \quad[\because v=\sqrt{u}, u=\tan x]$
$=\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]$

Indefinite Integrals Exercise Revision Exercise Question 23

Answer:
$-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+c$
Given:
$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$
Hint:
To solve this equation we use sin x formula and substitute method.
Solution:
$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$ $\left[(\cos x+\sin x)^{2}=\cos ^{2} x+\sin ^{2} x+2 \cos x \sin x\right]=1+\sin 2 x$
$\left[\sin 2 x=(\cos x+\sin x)^{2}-1, \quad t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x\right.$
$=\int \frac{\sin x-\cos x}{\left(\sqrt{ \left.(\cos x+\sin x)^{2}-1\right)}\right)} d x$
Let $t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x$
$=-\frac{d t}{\sqrt{t^{2}-1}}$ $\left[\because \frac{1}{\sqrt{t^{2}-a^{2}}} d t=\log \left|t+\sqrt{t^{2}-a^{2}}\right|+c\right.$
$=-\log \left|t+\sqrt{t^{2}-1}\right|+c$
$=-\log \left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^{2}-1}\right|+c$
$=-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+c$

Indefinite Integrals Exercise Revision Exercise Question 24

Answer:
$\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \log \left|\frac{\sin (\mathrm{x}-\mathrm{a})}{\sin (\mathrm{x}-\mathrm{b})}\right|+\mathrm{c}$
Given:
$\int \frac{1}{(\sin x-a) \sin (x-b)} d x$
Hint:
To solve the statement we have to use formula such as sin(A-B).
Solution:
$\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\sin (x-a)(\sin (x-b))} d x$
$=\frac{1}{\sin (a-b)} \int \frac{\sin [(x-b)-(x-a)]}{\sin (x-a)(\sin (x-b))} d x$
$=\frac{1}{\sin (a-b)} \int \frac{\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)}{\sin (x-a)(\sin (x-b))} d x$
$=\frac{1}{\sin (a-b)} \int \frac{\cos (x-a) d x}{\sin (x-a)}-\frac{\cos (x-b)}{\sin (x-b)} \mathrm{d} \mathrm{x}$
$=\frac{1}{\sin (a-b)} \log |\sin (x-a)|-\log |\sin (x-b)|+c$
$=\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \log \left|\frac{\sin (x-\mathrm{a})}{\sin (x-\mathrm{b})}\right|+\mathrm{c}$

Indefinite Integrals Exercise Revision Exercise Question 25

Answer:
$\frac{1}{\sin (a-b)} \log |\sec (x-b)|-\log |\sec (x-a)|+c$
Given:
$\int \frac{1}{\cos (x-a) \cos (x-b)} d x$
Hint:
To solve this statement we have to convert cos into tan.
Solution:
$\frac{1}{\cos (x-a) \cos (x-b)}=\frac{1}{\sin (a-b)} \frac{\sin (x-b)-(x-a)}{\cos (x-a) \cos (x-b)}$
$=\frac{1}{\sin (a-b)} \frac{\sin (x-b) \cdot \cos (x-a)-\sin (x-a) \cos (x-b)}{\cos (x-a) \cos (x-b)}$
$=\frac{1}{\sin (a-b)} \frac{\sin (x-b)}{\cos (x-b)}-\frac{\sin (x-a)}{\cos (x-a)}$
$=\frac{1}{\sin (a-b)} \cdot \tan (x-b)-\tan (x-a)$
$I=\int \frac{1}{\cos (x-a) \cos (x-b)} d x$
$=\frac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x$
$=\frac{1}{\sin (a-b)} \log |\sec (x-b)|-\log |\sec (x-a)|+c$

Indefinite Integrals Exercise Revision Exercise Question 26

Answer:
$-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$
Given:
$\int \frac{\sin x}{\sqrt{1+\sin x}} d x$
Hint:
To solve the statement we have to use formula like sin x and 1+cos x
Solution:
$I=\int \sqrt{\sin x+1}-\frac{1}{\sqrt{\sin x+1}} d x$
$=\int \sqrt{\sin x+1} d x-\int \frac{d x}{\sqrt{\sin x+1}}$
$I_{1}=\int \sqrt{\sin x+1} d x$
$=\int \sqrt{1+\sin x} \cdot \frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}} d x$
$=\frac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{1-\sin x}} d x$
$=\frac{\sqrt{\cos ^{2} x}}{\sqrt{1-\sin x}} d x$
$=\frac{\cos x}{\sqrt{1-\sin x}} d x$
$=\sqrt{1-\sin x}=u$
$\frac{1}{2 \sqrt{1-\sin x}} \cos x d x=d u$
$\frac{\cos x}{\sqrt{1-\sin x}}=-2 d u$
$I_{1}=\int-2 d u=-2 u=-2 \sqrt{1-\sin x}$
$I_{2}=\int \frac{d x}{\sqrt{1+\sin x}}=\int \frac{d x}{\sqrt{1+\cos \left(\frac{\pi}{2}-x\right)}}$ $\left[\because \sin x=\cos \left(\frac{\pi}{2}-x\right)\right],\left[1+\cos x=2 \cos ^{2} x \frac{x}{2}\right]$
$=\int \frac{d x}{\sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}}$
$=\int \frac{d x}{\sqrt{2}\left(\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\right.}$ $\left[\because \int \sec x d x=\ln |\sec x+\tan x|+c\right]$
$=\int \frac{1}{\sqrt{2}} \sec \left(\frac{\pi}{4}-\frac{x}{2}\right) d x$
$=\frac{1}{\sqrt{2}} \cdot \frac{1}{-\frac{1}{2}} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$

$=-\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$
$\mathrm{I}_{1}-\mathrm{I}_{2}=\mathrm{l}=-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 27

Answer:
$\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+c$
Given:
$\int \frac{\sin x}{\cos 2 x} d x$
Hint:
To solve the equation we will use substitution method.
Solution:
$I=\int \frac{\sin x}{\cos 2 x} d x$
$=\int \frac{\sin x}{2 \cos ^{2} x-1} d x$
$I=-\int \frac{d t}{2 t^{2}-1} \ldots[\mathrm{put} \cos \mathrm{x}=\mathrm{t},-\sin x \mathrm{dx}=\mathrm{d} \mathrm{t}]$
$I=\frac{1}{2} \int \frac{d t}{\frac{1}{2}-t^{2}}$
$I=-\frac{1}{2} \int \frac{d t}{t^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}$
$I=-\frac{1}{2} \frac{\sqrt{2}}{2} \log \left|\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right|+c$
$I=\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+\mathrm{C}$

Indefinite Integrals Exercise Revision Exercise Question 28

Answer:
$\frac{1}{2} \tan ^{2} x-\log |\sec x|+c$
Given:
$\int \tan ^{3} x d x$
Hint:
To solve this statement we have to change tan into sec form.
Solution:
$\int \tan ^{2} x \cdot \tan x d x$
$\int\left(\sec ^{2} x-1\right) \tan x d x \quad\left(\tan ^{2} x=\sec ^{2} x-1\right)$
$\int \sec ^{2} x \tan x d x-\int \tan x d x$

$I_{1}=\int \sec ^{2} x \tan x d x$ $\left[\because \tan x=t, \sec ^{2} x d x=d t, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$
$I_{1}=\int t d t$
$=\frac{t^{2}}{2}=\frac{1}{2} \tan ^{2} x$
$I=I_{1}+I_{2}$
$=\frac{1}{2} \tan ^{2} x-\log |\sec x|+c$

Indefinite Integrals Exercise Revision Exercise Question 29

Answer:
$\frac{\tan ^{8} x}{3}-\tan x+x+c$
Given:
$\int \tan ^{4} x d x$
Hint:
To solve this statement we have to change tan into sec and then their formula.
Solution:
$\int \tan ^{4} x d x$
$I=\int \tan ^{2} x \cdot \tan ^{2} x \cdot d x \quad\left[\because \tan ^{2} x=\sec ^{2} x-1\right]$
$\mathrm{I}=\int \tan ^{2} x\left(\sec ^{2} x-1\right) d x$
$=\int \tan ^{2} x \sec ^{2} x d x-\int \tan ^{2} x d x$
$=\int \tan ^{2} x \sec ^{2} x d x-\int\left(\sec ^{2} x-1\right) d x$
$=\int \tan ^{2} x \sec ^{2} x d x-\int \sec ^{2} x d x+\int d x$
$\tan x=t, \sec ^{2} d x=d t$
$\int t^{2} d t-\int \sec ^{2} x d x+\int d x$
$=\frac{t^{8}}{3}-\tan x+x+c$
$\frac{\tan ^{8} x}{3}-\tan \mathrm{x}+\mathrm{x}+\mathrm{c}$

Indefinite Integrals Exercise Revision Exercise Question 30

Answer:
$\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$
Given:
$\int \tan ^{5} x d x$
Hint:
To solve this statement we have to change tan into sec and then use formula such tan²x.
Solution:
$I=\int \tan ^{5} x d x$
$I=\int \tan ^{3} x \tan ^{2} x d x$
$=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x \quad\left[\because \tan ^{2} x=\sec ^{2} x-1\right]$
$=\int \tan ^{3} x \sec ^{2} x d x-\int \tan ^{3} x d x$
$=\int \tan ^{3} x \sec ^{2} x d x-\int \tan x \tan ^{2} x d x$
$I_{1}$ $I_{2}$
$\operatorname{let} \tan x=t, \sec ^{2} x d x=d t$
$=\int t^{3} d t-\int \tan x \tan ^{2} x d x$
$=\frac{t^{4}}{4}-\int \tan x\left(\sec ^{2} x-1\right) d x$
$=\frac{t^{4}}{4}-\int \tan x \sec ^{2} x d x+\int \tan x d x$
$=\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$
$=\frac{\tan ^{4} x}{4}-\frac{(\tan x)^{2}}{2}+\log |\sec x|+c$

Indefinite Integrals Exercise Revision Exercise Question 31

Answer:
$-\frac{1}{3} \cot ^{3} x+\cot x+x+c$
Given:
$\int \cot ^{4} x d x$
Hint:
To solve the given statement split the given term $\left(\cot ^{4} x\right)$ into $\left(\cot ^{2} x \cot ^{2} x\right)$
Solution:
$\int \cot ^{2} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$I=\int \cot ^{2} x \operatorname{cosec}^{2} x d x-\int \cot ^{2} d x$
$I=\int \cot ^{2} x \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) d x$ $\left[\because \cot x=p, \operatorname{cosec}^{2} x d x=-d p\right.$
$I=-\int p^{2} d p+\int d p+\int d x$
$I=-\frac{p^{8}}{3}+p+x+c$
$I=-\frac{1}{3} \cot ^{3} x+\cot x+x+c$

Indefinite Integrals Exercise Revision Exercise Question 32

Answer:
$-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+c$
Given:
$\int \cot ^{5} x d x$
Hint:
To solve the given statement we will split into $\cot ^{3} x \cot ^{2} x$
Solution:
$\int \cot ^{3} x \cot ^{2} x d x$
$I=\int \cot ^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$
$=\int \cot ^{3} x \operatorname{cosec}^{2} x d x-\int \cot x\left(\operatorname{cosec}^{2} x-1\right) d x$
$=\int \cot ^{3} x \operatorname{cosec}^{2} x d x-\int \cot x \operatorname{cosec}^{2} x d x+\int \cot x d x$
$=-\int p^{3} d p+\int p d p+\int \cot x d x \quad[\because \cot x=p, \operatorname{cosec} x d x=d p]$
$=-\frac{p^{4}}{4}+\frac{p^{2}}{2}+\log \mid(\sin x)+c$
$=-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+c$

Indefinite Integrals Exercise Revision Exercise Question 33

Answer:
$\log |x-1|-\frac{2}{(x-1)}-\frac{1}{2(x-1)^{2}}+c$
Given:
$\int \frac{x^{2}}{(x-1)^{3}} d x$
Hint:
To solve the statement we will use partial fractions
Solution:
$Let \frac{x^{2}}{(x-1)^{3}} d x=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}$
$x^{2}=A(x-1)^{2}+B(x-1)+c$
$x^{2}=x^{2} A+x(2 A+B)+A+B+C$
On equating the coeffiecients of $x^{2}: A=1$
On equating the coeffiecients of $x: B=-2$
On equating the constants : $C=1$
Thus, $\int \frac{x^{2}}{(x-1)^{3}} d x=\int \frac{1}{(x-1)} d x+\int \frac{-2}{(x-1)^{2}} d x+\int \frac{1}{(x-1)^{3}} d x$

$=\log |x-1|+\frac{2(x-1)^{-2+1}}{-2+1}-\frac{1}{2(x-1)^{2}}+c$
$=\log |x-1|-\frac{2}{(x-1)}-\frac{1}{2(x-1)^{2}}+c$

Indefinite Integrals Exercise Revision Exercise Question 34

Answer:
$I=\frac{(2 x+3)^{2}}{8}-\frac{3(2 x+3)^{1}}{4}+c$
Given:
$\int x \sqrt{2 x+3} d x$
Hint
To solve the statement we have to put root statement to $\mathrm{t}: \sqrt{2 x+3}=t$
Solution:
$\sqrt{2 x+3}=t$
$put 2 x+3=t^{2}=>x=\frac{t^{2}-3}{2}$
$2 d x=2 t d t$
$d x=t d t$
$I=\int \frac{t^{2}-3}{2} t d t$
$I=\frac{1}{2} \int\left(t^{2}-3\right) t d t$
$I=\frac{1}{2} \int\left(t^{3}-3 t\right) d t$
$I=\frac{1}{2}\left(\frac{t^{4}}{4}-\frac{3 t^{2}}{2}\right)$
$I=\frac{t^{4}}{8}-\frac{3 t^{2}}{4}+c$
$I=\frac{(2 x+3)^{2}}{8}-\frac{3(2 x+3)^{4}}{4}+c$

Indefinite Integrals Exercise Revision Exercise Question 35

Answer:
$\frac{1}{2}\left|\log \left(1+x^{2}\right)\right|+\frac{1}{2\left(1+x^{2}\right)}+c$
Given:
$\int \frac{x^{3}}{\left(1+x^{2}\right)^{2}} d x$
Hint:
To solve the statement we will suppose x in term of t.
Solution:
$\int \frac{x^{2} x}{\left(1+x^{2}\right)^{2}} d x$
$Let 1+x^{2}=t$
$2 x=\frac{d t}{d x}$
$x d x=\frac{d t}{2}$
$I=\frac{1}{2} \int \frac{t-1}{t^{2}} d t$
$I=\frac{1}{2} \int \frac{t}{t^{2}} d t-\int \frac{1}{t^{2}} d t$
$I=\frac{1}{2} \int \frac{1}{t} d t-\int t^{-2} d t$
$I=\frac{1}{2} \log t-\frac{1}{2} \frac{\left(t^{-2+1}\right)}{-2+1}+c$
$I=\frac{1}{2}\left|\log \left(1+x^{2}\right)\right|+\frac{1}{2\left(1+x^{2}\right)}+c$

Indefinite Integrals Exercise Revision Exercise Question 36

Answer:
$\frac{\sin ^{6} x^{2}}{12}+c$
Given:
$\int x \sin ^{5} x^{2} \cos x^{2} d x$
Hint:
To solve this statement we have to suppose or assume sin x and cos x as v and dv.
Solution:
$I=\int x \sin ^{5} x^{2} \cos x^{2} d x$
$I=\frac{1}{2} \int \sin ^{5} t costdt \quad\left[\because x^{2}=t, 2 x d x=d t, x d x=\frac{d t}{2}\right]$
$I=\frac{1}{2} \int v^{5} d v$ $[\because \sin t=v, \cos t d t=d v]$
$I=\frac{1}{2} \frac{v^{6}}{6}+c$
$I=\frac{1}{2} \frac{\left(\sin ^{6} t\right)}{6}+c$
$I=\frac{\sin ^{6} x^{2}}{12}+c$

Indefinite Integrals Exercise Revision Exercise Question 37

Answer:
$\frac{\cos ^{5} x}{5}-\frac{\cos ^{7} x}{7}+C$
Given:
$\int \sin ^{3} x \cos ^{4} x d x$
Hint:
To solve the statement we will convert cos into sin and use some formula of 2sin(A+B) and 2sin(A-B)
Solution:
$\int \sin ^{3} x\cos ^{4}x d x$
put
$\int \sin x \cdot \sin ^{2} x \cdot \cos ^{4} x d x$
$p u t \: \cos x=t, \sin x d x=d t$
$\int \sin x\left(1-\cos ^{2} x\right) \cdot \cos ^{4} x d x\left[\sin ^{2} x=1-\cos ^{2} x\right]$
$\int\left(1-t^{2}\right) t^{4} d t$
$\int t^{4}-t^{6} d t$
$\int t^{4} d t-\int t^{6} d t$
$\int \frac{t^{5}}{5}-\frac{t^{7}}{7}$
$t=\cos x$
$\frac{\cos ^{5} x}{5}-\frac{\cos ^{7} x}{7}+C$

Indefinite Integrals Exercise Revision Exercise Question 38

Answer:
$\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x-\cos x+c$
Given:
$\int \sin ^{5} x d x$
Hint:
To solve the statement we have to convert sin x into cos x.
Solution:
$\int \sin x(\sin x)^{4} d x$
$\int \sin x\left(\sin ^{2} x\right)^{2} d x$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1, \sin ^{2} \theta=1-\cos ^{2} \theta\right]$
$\int \sin x\left(1-\cos ^{2} x\right)^{2} d x$
Let cosx= t
$-\sin x d x=d t$
$\sin x d x=-d t$
$I=-\int\left(1-t^{2}\right)^{2} d t$
$I=-\int\left(1+t^{4}-2 t^{2}\right) d t$
$I=\int\left(2 t^{2}-t^{4}-1\right) d t \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$
$I=\frac{2 t^{3}}{3}-\frac{t^{5}}{5}-t+c$
$I=\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x-\cos x+c$

Indefinite Integrals Exercise Revision Exercise Question 39

Answer:
$\sin x+\frac{\sin ^{5} x}{5}-\frac{2 \sin ^{8} x}{3}+c$
Given:
$\int \cos ^{5} x d x$
Hint:
To solve the statement we have to convert cos term in sin
Solution:
$\int \cos ^{4} x \cos x d x$
$I=\left(\cos ^{2} x\right)^{2} \cos x d x \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right]$
$I=\int\left(1-\sin ^{2} x\right)^{2} \cos x d x$
$\sin x=t, \cos x d x=d t$
$I=\int\left(1-t^{2}\right)^{2} d t$
$I=\int\left(1+t^{4}-2 t^{2}\right) d t$
$I=t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}+c$
$I=\sin x+\frac{\sin ^{5} x}{5}-\frac{2 \sin ^{3} x}{3}+c$

Indefinite Integral Exercise Rivision Exercise Question 66

Answer :
$I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c$
Hint: To solve the given solution we multiply and divide the given statement with sin x.
Given : $\int \frac{1}{\sin x(2+3 \cos x)} d x$
Solution :
$\begin{aligned} &I=\int \frac{1}{\sin x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x}{\sin ^{2} x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x d x}{\left(1-\cos ^{2} x\right)(2+3 \cos x)} \end{aligned}$
$\begin{aligned} &t=\cos x, d t=-\sin x d x \\ &I=\int \frac{-d t}{\left(1-t^{2}\right)(2+3 t)} \\ &I=\int \frac{-d t}{(1+t)(1-t)(2+3 t)} \end{aligned}$
$\\\frac{1}{(1+t)(1-t)(2-3 t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{c}{2+3 t}=A(1-t)(2+3 t)+B(1+t)(2+3 t)+C(1+t)(1-t)$
$\begin{array}{ll} A=\frac{1}{(1-(-1))(2-3)}=-\frac{1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1+t=0, t=-1]} \\ B=\frac{1}{(1+1)(2+3)}=\frac{1}{10} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1-t=0, t=1]} \\ C=\frac{1}{1-\left(\frac{4}{9}\right)}=\frac{9}{5} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because 2+3 t=0, t=-\frac{2}{3}\right]} \end{array}$
$\begin{aligned} &I=-\int d t\left(\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{2+3 t}\right) \\ &I=-\left[A \int \frac{1}{1+t} d t+B \int \frac{1}{1-t} d t+C \int \frac{1}{2+3 t} d t\right. \end{aligned}$
$\begin{aligned} &I=-\left[-\frac{1}{2} \ln |1+t|+\frac{1}{10} \ln |1-t|+\frac{9}{5} \times \frac{1}{3} \ln |2+3 t|+c\right. \\ &I=+\frac{1}{2} \ln |1+t|-\frac{1}{10} \ln |1-t|-\frac{3}{5} \ln |3 t+2|+c \end{aligned}$
$I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c$

Indefinite Integrals Exercise Revision Exercise Question 40

Answer:
$\frac{2(\sin x)^{\frac{3}{2}}}{3}-\frac{2}{7}(\sin x)^{\frac{7}{2}}+c$
Given:
$\int \sqrt{\sin x} \cos ^{3} x d x$
Hint:
To solve this equation we have to suppose sin x in term of t and cos x into 2t.
Solution:
$I=\int \sqrt{\sin x} \cos ^{3} x d x$ $\left[\because \sqrt{\sin x}=t, \sin x=t^{2}, \cos x d x=2 t d t\right.$
$I=\int \sqrt{\sin x}\left(1-\sin ^{2} x\right) \cos x d x$
$I=\int t\left(1-t^{4}\right) 2 t d t$
$I=\int 2 t^{2} d t-\int 2 t^{6} d t$
$I=\frac{2 t^{3}}{3}-\frac{2 t^{7}}{7}+c$
$I=\frac{2(\sin x)^{\frac{3}{2}}}{3}-\frac{2}{7}(\sin x)^{\frac{7}{2}}+c$

Indefinite Integrals Exercise Revision Exercise Question 41

Answer:
$\tan ^{-1}\left(\tan ^{2} x\right)+c$
Given:
$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$
Hint:
In this statement we have to convert sin in term of tan.
Solution:
$I=\frac{2 \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
$I=\frac{2 \sin x \cos x}{\cos ^{4} x\left(\tan ^{4} x+1\right)} d x$
$I=2 \int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x$
$I=2 \int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x$ $\left[\because \operatorname{let} \tan x=t, \sec ^{2} x d x=d t\right]$
$I=2 \int \frac{t d t}{t^{4}+1}$
$I=\int \frac{d y}{y^{2}+1}\left[\because \mathrm{t}^{2}=\mathrm{y}, 2 \mathrm{tdt}=\mathrm{dy}\right]$
$=\tan ^{-1} y+c$
$=\tan ^{-1}\left(t^{2}\right)+c$
$=\tan ^{-1}\left(\tan ^{2} x\right)+c$

Indefinite Integral Exercise Rivision Exercise Question 67

Answer :
$=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c$
Hint : To solve the given statement we will use partial fraction rule.
Given : $\int \frac{1}{\sin x+\sin 2 x} d x$
Solution :
$\begin{aligned} &I=\int \frac{1}{\sin x+2 \sin x \cos x} d x \\ &I=\int \frac{\sin x}{\sin ^{2} x(1+2 \cos x)} d x \\ &I=\int \frac{\sin x}{\left(1-\cos ^{2} x\right)(1+2 \cos x)} d x \end{aligned}$
$\\\frac{1}{(1-t)(1+t)(1+2 t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{c}{1+2 t}=A(1+t)(1+2 t)+B(1-t)(1+2 t)+C(1-t)(1+t)$
$\begin{aligned} &1=A+2 A t+A t+2 A t^{2}+B+2 B t-B t-2 B t^{2}+C-C t^{2} \\ &0=2 A-2 B-C \ldots \ldots \ldots . . (1)\end{aligned}$
$\begin{aligned} &0=3 A+B \ldots \ldots \ldots \ldots \text { (2) } \quad B=-3 A \\ &1=A+B+C \ldots \ldots \ldots \text { (3) } \end{aligned}$
On solving, $A=\frac{1}{6}, B=-\frac{1}{2}, C=\frac{4}{3}$
$\int \frac{1}{(1-t)(1+t)(1+2 t)} d t=-\int \frac{1}{6(1-t)}+\int \frac{1}{2(1+t)}-\int \frac{4}{3} \frac{1}{(1+2 t)} d t$
$\begin{aligned} &=-\frac{1}{6}-\ln (1-t)+\frac{1}{2} \ln (1+t)-\frac{4}{3} \times \frac{1}{2} \ln (1+2 t) \\ &=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 42

Answer:
$\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C^{\prime}$
$\text { where } C^{\prime}=C-\ln a$

Given:
$\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x$
Hint:
In this statement we have to assume x as asecθ.
Solution:
$I=\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x$
$\text { putting } x=a \sec \theta$
$\Rightarrow d x=a \sec \theta \tan \theta d \theta$
$I=\int \frac{a \sec \theta \tan \theta d \theta}{\sqrt{a^{2} \sec ^{2} \theta-a^{2}}}$
$=\int \frac{a \sec \theta \tan \theta d \theta}{a \cdot \tan \theta}$
$=\int \sec \theta \tan \theta d \theta$
$=\ln |\sec \theta+\tan \theta|+C$
$=\ln \left|\sec \theta+\sqrt{\sec ^{2} \theta-1}\right|+C$
$=\ln \left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^{2}}-1\right|+C$
$=\ln \left|\frac{x+\sqrt{x^{2}-a^{2}}}{a}\right|+C$
$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C$
$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|-\ln a+C$
$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C^{\prime}$
$\text { where } C^{\prime}=C-\ln a$

Indefinite Integrals Exercise Revision Exercise Question 43

Answer:
$\log \left(\frac{x}{a}+\sqrt{1+\left(\frac{x^{2}}{a^{2}}\right)}\right)+c$
Given:
$\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x$
Hint:
In this statement we assume x as atanθ.
Solution:
$I=\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x$ $\left[\because x=\operatorname{atan} \theta, d x=\operatorname{asec}^{2} \theta d \theta\right]$
$I=\int \frac{1}{\sqrt{a^{2} \tan ^{2} \theta+a^{2}}} \operatorname{asec}^{2} \theta d \theta$
$I=\int \sec \theta d \theta \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\left[1+\tan ^{2} \theta=\sec ^{2} \theta\right]$
$I=\log |\sec \theta+\tan \theta|+c$
$I=\log \left(\frac{x}{a}+\sqrt{1+\left(\frac{x^{2}}{a^{2}}\right)}\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 44

Answer:
$I=\frac{1}{4} \tan ^{-1}\left(x+\frac{1}{2}\right)+c$
Given:
$\int \frac{1}{4 x^{2}+4 x+5} d x$
Hint:
To solve this equation we will use split term method.
Solution:
$I=\int \frac{d x}{4 x^{2}+4 x+5}$
$I=\int \frac{d x}{4\left(x^{2}+x\right)+5}$
$I=\int \frac{d x}{4\left(x+\frac{1}{2}\right)^{2}+4}$ $\left[\because\left(x^{2}+\frac{1}{2}\right)^{2}=x^{2}+\frac{1}{4}+\frac{2 x}{2}=x^{2}+x+\frac{1}{4}-\left(\frac{1}{4}\right)\right]$
$I=\frac{1}{4} \int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+(1)^{2}}$
$I=\frac{1}{4} \tan ^{-1} \frac{\left(x+\frac{1}{2}\right)}{1}+\mathrm{c}$

Indefinite Integrals Exercise Revision Exercise Question 46

Answer:
$I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c$
Given:
$\int \frac{1}{1-x-4 x^{2}} d x$
Hint:
To solve this statement we have to do whole square method.
Solution:
$\int \frac{1}{1-4\left(\frac{x}{4}+x^{2}\right)} d x$
$I=\int \frac{1}{1-4\left(x+\frac{1}{8}\right)^{2}+\frac{1}{16}} d x$
$I=\int \frac{1}{\frac{17}{6}-4\left(x+\frac{1}{8}\right)^{2}} d x$
$I=\frac{1}{4} \int \frac{1}{\left(\frac{\sqrt{17}}{8}\right)^{2}-\left(x+\frac{1}{8}\right)^{2}} d x$ $\left[\because \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right]$
$I=\frac{1}{4} \frac{4}{\sqrt{17}} \log \left|\frac{x+\frac{1}{8}+\frac{\sqrt{17}}{8}}{\frac{\sqrt{17}}{8}-x-\frac{1}{8}}\right|+c$
$I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 47

Answer:
$\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c$
Given:
$\int \frac{1}{3 x^{2}+13 x-10} d x$
Hint:
In this equation we have to make whole square statement from quadratic equation.
Solution:
$I=\frac{1}{3} \int \frac{1}{x^{2}+\frac{18 x}{8}-\frac{10}{8}} d x$
$I=\frac{1}{3} \int \frac{1}{\left(x+\frac{18}{6}\right)^{2}-\frac{169}{36}-\frac{10}{3}} d x$
$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{169+120}{36}}$
$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{289}{36}}$
$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$
$I=\frac{1}{3} \frac{1}{2 \times \frac{17}{6}} \log \left|\frac{x+\frac{18}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+c$
$I=\frac{1}{17}+\log \left|\frac{x-\frac{4}{6}}{x+\frac{30}{6}}\right|+c$
$I=\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 48

Answer:
$I=-\ln \left|(\cos x-1)+\sqrt{\cos ^{2} x-2 \cos x+3}\right|+c$
Given:
$\int \frac{\sin x}{\sqrt{\cos ^{2} x-2 \cos x-3}} d x$
Hint:
To solve this statement we have to use standard formula.
Solution:
$I=\int \frac{\sin x}{\sqrt{\cos ^{2} x-2 \cos x-3}} d x$ $[\because \cos x=t, d t=-\sin x d x=>\sin x d x=-d t]$
$I=-\int \frac{d t}{\sqrt{t^{2}-2 t-3}}$
$I=-\int \frac{d t}{\sqrt{(t-1)^{2}-1-3}}$

$I=-\int \frac{d t}{\sqrt{(t-1)^{2}-4}}$
$I=-\int \frac{d t}{\sqrt{(t-1)^{2}-(2)^{2}}}$ $\left[\because \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right.$
$I=-\ln \mid(t-1)+\sqrt{(t-1)^{2}-(2)^{2} \mid+} c$
$I=-\ln \left|(t-1)+\sqrt{t^{2}-2 t-3}\right|+c$
$I=-\ln \left|(\cos x-1)+\sqrt{\cos ^{2} x-2 \cos x+3}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 49

Answer:
$I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c$
Given:
$\int \sqrt{\operatorname{cosec} x-1} d x$
Hint:
To solve this statement we have to convert cosec into sin and then we apply standard formula.
Solution:
$\int \sqrt{\operatorname{cosec} x-1} d x$
$I=\int \sqrt{\frac{1}{\sin x}-1} d x$
$I=\int \sqrt{\frac{1-\sin x}{\sin x}}-d x$
$I=\int \frac{\sqrt{1-\sin x} \sqrt{1+\sin x}}{\sqrt{\sin x} \sqrt{1+\sin x}} d x$
$I=\int \frac{\sqrt{1-\sin ^{2} x}}{\sqrt{\sin x(1+\sin x)}} d x$
$I=\int \frac{\sqrt{\cos ^{2} x}}{\sqrt{\sin x(1+\sin x)}} d x$
$I=\int \frac{\cos x}{\sqrt{\sin x(1+\sin x)}} d x$
sin x = t ,
dt = cos x dx
$\therefore I=\int \frac{d t}{\sqrt{t}(t+1)}=\int \frac{d t}{\sqrt{t^{2}+t}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}$
$I=\ln \left|t+\frac{1}{2}+\sqrt{t^{2}+t}\right|+c \quad\left[\because \int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\ln \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$ $I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 50

Answer:
$I=\sin ^{-1}\left(\frac{x+1}{2}\right)+c$
Given:
$\int \frac{1}{\sqrt{3-2 x-x^{2}}} d x$
Hint

Solution:
$\int \frac{1}{\sqrt{3-2 x-x^{2}}} d x$
$I=\int \frac{1}{\sqrt{3+1-(x+1)^{2}}} d x$
$I=\int \frac{1}{\sqrt{4-(x+1)^{2}}} d x$
$I=\int \frac{d x}{\sqrt{(2)^{2}-(x+1)^{2}}}$
$t=x+1$
$d t=d x$
$=\int \frac{d t}{\sqrt{2^{2}-t^{2}}} \quad\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+c\right.$
$=\sin ^{-1}\left(\frac{t}{2}\right)+c$
$I=\sin ^{-1}\left(\frac{x+1}{2}\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 52

Answer:
$I=5 \sqrt{x^{2}-9 x+20}+\frac{59}{2} \log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|+c$
Given:
$\int \frac{5 x+7}{\sqrt{(x-5)(x-4)}} d x$
Hint:
To solve this equation we have to do differentiation method.
Solution:
$I=\int \frac{5 x+7}{\sqrt{(x-5)(x-4)}} d x$
$I=\int \frac{5 x+7}{\sqrt{x^{2}-9 x+20}} d x$
Take 5/2 common from numerator and add and subtract 9,
$I=\frac{5}{2} \int \frac{2 x+\frac{14}{5}-9+9}{\sqrt{x^{2}-9 x+20}} d x$
$I=\frac{5}{2} \int \frac{(2 x-9)+\frac{14}{5}+9}{\sqrt{x^{2}-9 x+20}} d x$
$I=\frac{5}{2} \int \frac{(2 x-9)}{\sqrt{x^{2}-9 x+20}} d x+\frac{5}{2} \int \frac{\frac{59}{5}}{\sqrt{x^{2}-9 x+20}} d x$
$\text { put } \sqrt{x^{2}-9 x+20}=t$
$\frac{d t}{d x}=\frac{1}{2 \sqrt{x^{2}-9 x+20}} \times(2 x-9)$
$d x=\frac{d t \times 2}{2 x-9} \sqrt{x^{2}-9 x+20}$
$I_{1}=\frac{5}{2} \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} \times d t \frac{\left(2 \sqrt{x^{2}-9 x+20}\right)}{2 x-9}$
$I_{1}=5 \int d t$
$I_{1}=5 t$
$I_{1}=5 \sqrt{x^{2}-9 x+20}$
$I_{2}=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$
$=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+\frac{81}{4}-\frac{81}{4}+20}} d x$
$=\frac{59}{2} \int \frac{1}{\sqrt{x^{2}-9 x+\frac{81}{4}+\left(20-\frac{81}{4}\right)}} d x$
$=\frac{59}{2} \int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x$
$=\left[\because \int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right.$
$\text { let } x-\frac{9}{2}=u$
$d x=d u$
$I=\frac{59}{2} \int \frac{d u}{\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}}$
$=\frac{59}{2} \log \left|u+\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}\right|+c$
$=\frac{59}{2} \log \left|u+\sqrt{u^{2}-\left(\frac{1}{2}\right)^{2}}\right|+c$
$\text { Adding } I_{1} \text { and } I_{2}$
$I=5 \sqrt{x^{2}-9 x+20}+\frac{59}{2} \log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 53

Answer:
$I=\left[\sqrt{x^{2}+x}+\ln (\sqrt{x}+\sqrt{x+1})\right]+c$

ùTo solve this equation we have to take x = u².
Solution:
$I=\int \sqrt{\frac{(x+1)}{x}} \mathrm{dx}$
$\text { Let } x=u^{2}$
$d x=2 u d u$
$I=\int \sqrt{\frac{\left(u^{2}+1\right)}{\mathrm{u}^{2}}} 2 \mathrm{udu}$

$I=\int 2 \sqrt{u^{2}+1} d u$ $\left[\because \int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln x+\sqrt{x^{2}+a^{2}}+c\right]$
$I=2\left[\frac{u}{2} \sqrt{u^{2}+1}+\frac{1}{2} \ln \left(u+\sqrt{u^{2}+1}\right)\right]+c$
$I=2\left[\frac{\sqrt{x}}{2} \sqrt{x+1}+\frac{1}{2} \ln (\sqrt{x}+\sqrt{x+1})\right]+c$
$I=\left[\sqrt{x^{2}+x}+\ln (\sqrt{x}+\sqrt{x+1})\right]+c$

Indefinite Integrals Exercise Revision Exercise Question 54

Answer:
$I=\sin ^{-1} \sqrt{x}+\frac{1}{2} \sin 2 \sin ^{-1} \sqrt{x}+c$
Given:
$\int \sqrt{\frac{1-x}{x}} d x$
Hint:
To solve this question we have to assume that x as sin²θ
Solution:
$I=\int \sqrt{\frac{1-x}{x}} d x$
$x=\sin ^{2} \theta \quad d x=2 \sin \theta \cos \theta d \theta$
$I=\int \frac{\sqrt{1-\sin ^{2} \theta}}{\sqrt{\sin ^{2} \theta}} d \theta(2 \sin \theta \cos \theta)$
$I=\int \frac{\sqrt{\cos ^{2} \theta}}{\sqrt{\sin ^{2} \theta}}(2 \sin \theta \cos \theta) d \theta$
$I=\int \cos \theta 2 \cos \theta d \theta$
$I=\int 2 \cos ^{2} \theta d \theta$
$I=\int(1+\cos 2 \theta) d \theta$
$I=\int 1 d \theta+\int \cos 2 \theta d \theta$
$I=\theta+\frac{1}{2} \sin 2 \theta+c$
$I=\sin ^{-1} \sqrt{x}+\frac{1}{2} \sin 2 \sin ^{-1} \sqrt{x}+c$

Indefinite Integrals Exercise Revision Exercise Question 55

Answer:
$I=\frac{2}{a^{\frac{3}{2}}}\left[\frac{(a-2)(1-\sqrt{a x})}{2}+\frac{(1-\sqrt{a x})^{2}}{2}+(1-a) \ln |(1-\sqrt{a x})|\right]+c$
Given:
$\int \frac{\sqrt{a}-\sqrt{x}}{1-\sqrt{a x}} d x$
Hint:
To solve this equation we have to suppose denominator as t.
Solution:
$I=\int \frac{\sqrt{a}-\sqrt{x}}{1-\sqrt{a x}} d x$
$1-\sqrt{a x}=t$
$-\sqrt{a x}=t-1$
$-\frac{a}{2 \sqrt{a x}} d x=d t$
$d x=\frac{-2 \sqrt{a x}}{a} d t$
$d x=\frac{2(t-1)}{a} d t$
$I=\frac{1}{\sqrt{a}} \int \frac{a-\sqrt{a x}}{1-\sqrt{a x}} d x$
$I=\frac{1}{\sqrt{a}} \int \frac{a+(t-1)}{t}\left(\frac{2(t-1)}{a}\right) d t$
$I=\frac{2}{a^{\frac{3}{2}}} \int \frac{(t-1)(a+t-1)}{t} d t$
$I=\frac{2}{a^{\frac{3}{2}}} \int \frac{a t+t^{2}-t-a-t+1}{t} d t$
$I=\frac{2}{a^{\frac{3}{2}}} \frac{t(a-2)+t^{2}+(1-a)}{t} d t$
$I=\frac{2}{a^{\frac{3}{2}}} \int\left(a-2+t+\frac{1-a}{t}\right) d t$
$I=\frac{2}{a^{\frac{3}{2}}}\left[(a-2) t+\frac{t^{2}}{2}+(1-a) \ln |t|\right]+c$
$I=\frac{2}{a^{\frac{3}{2}}}\left[\frac{(a-2)(1-\sqrt{a x})}{2}+\frac{(1-\sqrt{a x})^{2}}{2}+(1-a) \ln |(1-\sqrt{a x})|\right]+c$

Indefinite Integrals Exercise Revision Exercise Question 56

Answer:
$\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$
Given:
$\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$
Hint:
To solve this statement we have to evaluate the team by partial fraction.
Solution:
$\int \frac{d x}{(\sin x-2 \cos x)(2 \sin x+\cos x)}$
Dividing cos²x with numerator and denominator.
$I=\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x-2 \cos x}{\cos x} \times \frac{2 \sin x+\cos x}{\cos x}}$
$I=\int \frac{\sec ^{2} x}{(\tan x-2)(2 \tan x+1)} d x$
$I=\int \frac{d t}{(t-2)(2 t+1)}$ $\left[\because t=\tan x, d t=\sec ^{2} x d x\right]$
$\frac{1}{(t-1)(2 t+1)}=\frac{a}{t-2}+\frac{b}{2 t+1}$
$=\frac{(2 a+b) t+(a-2 b)}{(t-2)(2 t+1)}$
$2 a+b=0 \ldots \ldots \ldots \ldots \text { (1) }$
$a-2 b=1 \ldots \ldots \ldots \ldots(2)$
$\text { --------------------}$
$1 \times 2+(2)=$
$4 a+a=1, \quad a=\frac{1}{5}$
$b=-\frac{2}{5}$
$I=\int\left[\frac{1}{5} \times \frac{1}{t-2}-\frac{2}{5} \times \frac{1}{2 t+1}\right] d t$
$I=\frac{1}{5} \log |t-2|-\frac{2}{5} \times \frac{1}{2} \log (2 t+1)+c$
$I=\frac{1}{5} \log \left|\frac{t-2}{2 t+1}\right|+c$
$I=\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 57

Answer:
$I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c$
Given:
$\int \frac{d x}{4 \sin ^{2} x+4 \sin x \cos x+5 \cos ^{2} x}$
Hint:
To solve this statement we have to divide numerator and denominator by cos²θ.
Solution:
$I=\int \frac{1}{4 \sin ^{2} x+4 \sin x \cos x+5 \cos ^{2} x} d x$
Dividing numerator and denominator by cos²x.
$I=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+4 \sin x \cos x \frac{1}{\cos ^{2} x}+\frac{5 \cos ^{2} x}{\cos ^{2} x}} d x$
$I=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+4 \tan x+5} d x$
$I=\int \frac{\sec ^{2} x}{(2 \tan x+1)^{2}+4} d x$
$\left[\because 2 \tan x+1=t, d t=2 \sec ^{2} x d x, \sec ^{2} x d x=\frac{d t}{2}\right.$
$I=\int \frac{d t\left(\frac{1}{2}\right)}{t^{2}+2^{2}}=\frac{1}{4} \int \frac{d t}{t^{2}+2^{2}}$
$I=\frac{1}{2} \times \frac{1}{2} \tan ^{-1} \frac{t}{2}+c$
$I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c$

Indefinite Integrals Exercise Revision Exercise Question 58

Answer:
$\left|=\frac{b}{a^{2}+b^{2}} \ln \right| \cos x+b \sin x \mid+\frac{a}{a^{2}+b^{2}} x+c$
Given:
$\int \frac{1}{a+\operatorname{btan} x} d x$
Hint:
To solve this equation, we have to use standard method form.
Solution:
$I=\int \frac{1}{a+b \tan x} d x$
$I=\int \frac{1}{a+b \frac{\sin x}{\cos x}} d x=\int \frac{\cos x}{a \cos x+\operatorname{ssin} x} d x$
$\text { Let } \cos x=A \frac{d}{d x}(a \cos x+b \sin x)+B(a \cos x+b \sin x)$
$=A(-\operatorname{asin} x+b \cos x)+B(\operatorname{acos} x+b \sin x)$
$=(A b+B a) \cos x+(B b-A a) \sin x$
$=A b+B a=1 \ldots \ldots . .(1) \quad B b-A a=0 . \ldots(2)$
On solving (1) and (2),
$B=\frac{a}{a^{2}+b^{2}}$
$A=\frac{b}{a^{2}+b^{2}}$
$I=\int \frac{A(-a \sin x+b \cos x)+B(\operatorname{acos} x+b \sin x)}{a \cos x+b \sin x} d x$
$I=\int A \frac{(-a \sin x+b \cos x)}{a \cos x+\operatorname{bsin} x}+B d x$
$I=\int A \frac{(-a \sin x+b \cos x)}{\operatorname{acos} x+\operatorname{bsin} x} d x+\int B d x$
$I=A \int \frac{(-a \sin x+b \cos x)}{a \cos x+b \sin x} d x+B \int d x \quad[\because d t=-a \sin x+b \cos x]$
$I=A \int \frac{d t}{t}+B x$
$I=\frac{b}{a^{2}+b^{2}} \ln |t|+\frac{a}{a^{2}+b^{2}} x c$
$I=\frac{b}{a^{2}+b^{2}} \ln |a \cos x+b \sin x|+\frac{a}{a^{2}+b^{2}} x+c$

Indefinite Integrals Exercise Revision Exercise Question 59

Answer:
$I=-\frac{1}{2} \log |1+2 \cot x|+c$
Given:
$\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
Hint:
To solve this equation we have to take sin²x common and also use sin2x formula.
Solution:
$I=\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
$I=\int \frac{d x}{\sin ^{2} x+2 \sin x \cos x}$
$I=\int \frac{d x}{\sin ^{2} x\left(1+\frac{2 \cos x}{\sin x}\right)}$
$I=\int \frac{\operatorname{cosec}^{2} x d x}{1+2 \cot x}$
$\text { Let } \cot x=t=>-\operatorname{cosec}^{2} x d x=d t$
$I=-\int \frac{d t}{1+2 t}$
$I=-\frac{1}{2} \log |1+2 t|+c \quad\left[\int \frac{d x}{x}=\log |x|+c\right]$ $\left[\int \frac{d x}{a x+b}=\log \frac{(\mathrm{ax}+\mathrm{b})}{a}+c\right]$
$I=-\frac{1}{2} \log |1+2 \cot x|+c$

Indefinite Integrals Exercise Revision Exercise Question 60

Answer:
$I=\frac{4}{5} x+\frac{3}{5} \log (2 \sin x+\cos x)+c$
Given:
$\int \frac{\sin x+2 \cos x}{2 \sin x+\cos x} d x$
Hint:
To solve this statement we have to use formula of sin θ+ 2cos θ and also substitute method.
Solution:
$\sin x+2 \cos x=A \frac{d}{d x}(\mathrm{D})+\mathrm{B}(\mathrm{D})$
$\sin x+2 \cos x=A(2 \cos x-\sin x)+B(2 \sin x+\cos x)$
$\sin x=>1=2 B-A \ldots . .(1) \quad \cos x=>2=2 A+B \ldots \ldots .(2)$
$\text { On solving (1) \& (2), }$ $A=\frac{3}{5} \quad \& B=\frac{4}{5}$
$I=\frac{3}{5} \int \frac{2 \cos x-\sin x}{2 \sin x+\cos x}+\frac{4}{5} \int \frac{2 \sin x+\cos x}{2 \sin x+\cos x} d x$
$I=\frac{3}{5} \int \frac{d t}{t}+\frac{4}{5} \int d x$
$I=\frac{3}{5} \log t+\frac{4}{5} x+c$
$I=\frac{4}{5} x+\frac{3}{5} \log (2 \sin x+\cos x)+c$

Indefinite Integrals Exercise Revision Exercise Question 61

Answer:
$I=\frac{1}{4} \ln \left|x^{4}+\sqrt{x^{8}+4}\right|+c$
Hint :
To solve this given statement split the x? term to (x?)² then put x? equal to t.
Given:
$\int \frac{x^{8}}{\sqrt{x^{8}+4}} d x$
Solution:
$I=\int \frac{x^{8}}{\sqrt{x^{8}+4}} d x$
$I=\int \frac{x^{3}}{\sqrt{\left(x^{4}\right)^{2}+4}} d x$
$I=\int \frac{\frac{1}{4} d t}{\sqrt{t^{2}+4}}$ $\left[\because x^{4}=t, d t=4 x^{3} d x, x^{3} d x=\frac{d t}{4}\right]$
$I=\frac{1}{4} \int \frac{d t}{\sqrt{t^{2}+2^{2}}}$
$I=\frac{1}{4} \ln \left|t+\sqrt{t^{2}+4}\right|+c$
$I=\frac{1}{4} \ln \left|x^{4}+\sqrt{x^{8}+4}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 62

Answer:
$\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}$
Hint:
To solve the given statement we will take $\cos 2 x \operatorname{as}\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)$
Given:
$\int \frac{1}{2-3 \cos 2 x} d x$
Solution:
$I=\int \frac{1}{2-3\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)} d x$
$I=\int \frac{1+\tan ^{2} x}{2\left(1+\tan ^{2} x\right)-3\left(1-\tan ^{2} x\right)} d x$
$I=\int \frac{\sec ^{2} x}{-1+\operatorname{stan}^{2} x} d x$
$I=\int \frac{\sec ^{2} x}{5\left(\tan ^{2} x-\frac{1}{5}\right)} d x$
$I=\int \frac{1}{5} \frac{d t}{t^{2}-\frac{1}{5}}$ $\left[\because \tan x=t, d t=\sec ^{2} x d x\right]$
$I=\frac{1}{5} \int \frac{d t}{t^{2}-\left(\frac{1}{\sqrt{5}}\right)^{2}}$
$I=\frac{1}{5} \frac{1}{2\left(\frac{1}{\sqrt{5}}\right)} \ln \left|\frac{t-\left(\frac{1}{\sqrt{5}}\right)}{t+\left(\frac{1}{\sqrt{5}}\right)}\right|+c$
$I=\frac{1}{2 \sqrt{5}} \ln \left|\frac{\tan x-\left(\frac{1}{\sqrt{5}}\right)}{\tan x+\left(\frac{1}{\sqrt{5}}\right)}\right|+\mathrm{C}$

Indefinite Integrals Exercise Revision Exercise Question 63

Answer:
$\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{3}}{2}\right)}{\sin x+\left(\frac{\sqrt{3}}{2}\right)}\right|+c$
Hint:
To solve this given statement we will write cos²x in terms of sin²x
Given:
$\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x$
Solution:
$\int \frac{\cos x}{\frac{1}{4}-\cos ^{2} x} d x$
$I=\int \frac{\cos x}{\frac{1}{4}-\left(1-\sin ^{2} x\right)} d x$
$I=\int \frac{\cos x}{-\frac{3}{4}+\sin ^{2} x} d x$

$\sin x=t \quad d t=\cos x d x$
$\therefore I=\int \frac{d t}{t^{2}-\frac{3}{4}}$
$I=\int \frac{d t}{t^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$I=\frac{1}{2\left(\frac{\sqrt{3}}{2}\right)} \ln \left|\frac{t-\left(\frac{\sqrt{3}}{2}\right)}{t+\left(\frac{\sqrt{3}}{2}\right)}\right|+c$
$I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sin x-\left(\frac{\sqrt{8}}{2}\right)}{\sin x+\left(\frac{\sqrt{8}}{2}\right)}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 64

Answer:
$I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \left(\frac{x}{2}\right)}{\sqrt{3}-\tan \left(\frac{x}{2}\right)}\right|+c$
Hint
We will write $\cos x \operatorname{as}\left(\frac{1-\tan ^{2} \frac{x \lambda}{2}}{1+\tan ^{2} \frac{x}{2}}\right)$
Given:
$\int \frac{1}{1+2 \cos x} d x$
Solution:
$I=\int \frac{1}{1+2 \cos x} d x$
$I=\int \frac{1}{1+2\left(\frac{1-\tan ^{2 \frac{x}{2}}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
$I=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}+2-2 \tan ^{2} \frac{x}{2}} d x$
$I=\int \frac{\sec ^{2} \frac{x}{2}}{3-\tan ^{2} \frac{x}{2}} d x$
$I=\int \frac{2 d t}{3-t^{2}}$ $\left[\because \tan \frac{\mathrm{x}}{2}=t, d t=\sec ^{2}\left(\frac{x}{2}\right)\left(\frac{1}{2}\right) d x\right]$
$I=2 \int \frac{d t}{(\sqrt{3})^{2}-t^{2}}$ $\left[\because \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t\right]$
$I=2\left(\frac{1}{2 \sqrt{3}}\right) \ln \left|\frac{\sqrt{3}-t}{\sqrt{3}+t}\right|+c$
$I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+c$
$I=\frac{1}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \left(\frac{x}{2}\right)}{\sqrt{3}-\tan \left(\frac{x}{2}\right)}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 65

Answer:
$I=\frac{1}{\sqrt{3}} \log \left|\frac{\tan \left(\frac{x}{2}\right)-2-\sqrt{3}}{\tan \left(\frac{x}{2}\right)-2+\sqrt{3}}\right|+c$

Hint:

To solve the above equation we will write sin x as $\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}$
Given:
$\int \frac{1}{1-2 \sin x} d x$
Solution:
$I=\int \frac{1}{1-2 \sin x} d x$
$I=\int \frac{1}{1-\frac{4 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}} d x$
$I=\int \frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$ $\left[\because \sin x=\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\right.$
$I=\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$
$I=\int \frac{2}{2} \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$
$I=2 \int \frac{d t}{1+t^{2}-4 t}$ $\left[\because t=\tan \left(\frac{x}{2}\right), \frac{d t}{d x}=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right), d t=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x\right]$
$I=2 \int \frac{d t}{\left(t^{2}-2 \cdot 2 \cdot t+2^{2}\right)-2^{2}+1}$
$I=2 \int \frac{d t}{(t-2)^{2}-3}$
$I=2 \int \frac{d u}{u^{2}-(\sqrt{3})^{2}}$ $[\because t-2=u, d t=d u]$
$\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right.$
$I=2 \times \frac{1}{2 \sqrt{3}} \log \left|\frac{u-\sqrt{3}}{u+\sqrt{3}}\right|+c$
$I=\frac{1}{\sqrt{3}} \log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|+c$
$I=\frac{1}{\sqrt{3}} \log \left|\frac{\tan \left(\frac{x}{2}\right)-2-\sqrt{3}}{\tan \left(\frac{x}{2}\right)-2+\sqrt{3}}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 69

Answer: $\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c$

Hint:

To solve the given statement write $\sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right) \text { and } \sin \operatorname{as} 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$
Given:
$\int \frac{1}{5-4 \sin x} d x$
Solution:
$I=\int \frac{1}{5-4 \sin x} d x$
$=\int \frac{1}{5\left(\sin ^{2}\left(\frac{x}{2}\right)+\cos ^{2}\left(\frac{x}{2}\right)-\frac{4}{5}\left(2 \sin \frac{x}{2} \cos \left(\frac{x}{2}\right)\right)\right.} d x$
$=\int \frac{d x}{5 \sin ^{2}\left(\frac{x}{2}\right)+5 \cos ^{2}\left(\frac{x}{2}\right)-8 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}$
$\text { Divide numerator and denominator by } \cos ^{2}\left(\frac{x}{2}\right)$
$=\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{\operatorname{stan}^{2}\left(\frac{x}{2}\right)+5-8 \tan \left(\frac{x}{2}\right)} d x$
$\text { let } \tan \left(\frac{x}{2}\right)=t$
$\frac{\sec ^{2}\left(\frac{x}{2}\right)}{2} d x=d t$
$=2 \int \frac{1}{5 t^{2}+5-8 t} d t$
$=2 \times \frac{1}{5} \int \frac{1}{t^{2} \frac{(-8 t)}{5}+1} d t$
$=\frac{2}{5} \int \frac{1}{t^{2}-\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t$
$=\frac{2}{5} \int \frac{d t}{\left(t-\left(\frac{4}{5}\right)^{2}\right)+\left(\frac{3}{5}\right)^{2}}$
$=\frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t-\left(\frac{4}{5}\right)}{\frac{3}{5}}\right)+c$
$=\frac{2}{3} \tan ^{-1} \frac{(5 t-4)}{3}+c$
$=\frac{2}{3} \tan ^{-1} \frac{\operatorname{5tan}\left(\frac{x}{2}\right)-4}{3}+c$

Indefinite Integrals Exercise Revision Exercise Question 70

Answer:
$\tan x+\frac{(\tan x)^{3}}{3}+c$
Hint:
To solve the given statement split the sec? x into sec² x sec²x then apply the formula.
Given:
$\int \sec ^{4} x d x$
Solution:
$I=\int \sec ^{4} x d x$
$1+\tan ^{2} x=\sec ^{2} x$
$=\int\left(1+\tan ^{2} x\right) \sec ^{2} x d x$
$=\int\left(1+t^{2}\right) d t$ $\left[\because \tan x=t, \sec ^{2} x d x=d t, \int x^{n} d x=\frac{x^{n}+1}{n+1}+c\right]$
$=t+\frac{t^{3}}{3}+c$
$=\tan x+\frac{(\tan x)^{3}}{3}+c$

Indefinite Integrals Exercise Revision Exercise Question 71

Answer:
$-\frac{\cot 2 x}{2}-\frac{\cot ^{8} 2 x}{6}+c$
Hint:
To solve the given question we will split the cosec? 2x into cosec² 2x cosec² 2x.
Given:
$\int \operatorname{cosec}^{4} 2 x d x$
Solution:
$=\int \operatorname{cosec}^{4} 2 x d x$
$=\int \operatorname{cosec}^{2} 2 x \operatorname{cosec}^{2} 2 x d x$
$=\int\left(1+\cot ^{2} 2 x\right) \operatorname{cosec}^{2} 2 x d x$
$p u t \cot 2 x=t=>-\operatorname{cosec}^{2} 2 x \cdot 2 d x=d t$
$=\frac{-1}{2} \int\left(1+t^{2}\right) \cdot d t$
$=-\frac{1}{2} t-\frac{t^{3}}{6}+c$
$=-\frac{\cot 2 x}{2}-\frac{\cot ^{3} 2 x}{6}+c$

Indefinite Integrals Exercise Revision Exercise Question 72

Answer:
$\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C$
Hint:
To solve the given statement multiply and divide the equation by sin x.
Given:
$\int \frac{1+\sin x}{\sin x(1+\cos x)} d x$
Solution:
$I=\int \frac{(1+\sin x)}{\sin x(1+\cos x)} d x$
$\text { putting }$
$\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$I=\int \frac{\left(1+\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}{\frac{\left(2 \tan \frac{x}{2}\right)}{\left(1+\tan ^{2} \frac{x}{2}\right)}\left(1+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
$I=\int \frac{\left(1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right)\left(1+\tan ^{2} \frac{x}{2}\right)}{\left(2 \tan \frac{x}{2}\right)\left(1+\tan ^{2} \frac{x}{2}+1-\tan ^{2} \frac{x}{2}\right)} d x$
$=\frac{1}{4} \int \frac{\left(1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right) \sec ^{2} \frac{x}{2}}{\tan \frac{x}{2}} d x$
$\text { putting } \tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x=d t$
$\Rightarrow \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t$
$I=\frac{1}{4} \int \frac{\left(1+t^{2}+2 t\right) \cdot(2 d t)}{t}$
$=\frac{1}{2} \int\left(\frac{1}{t}+t+2\right) d t$
$=\frac{1}{2}\left[\ln |t|+\frac{t^{2}}{2}+2 t\right]+C$
$=\frac{1}{2}\left[\ln \left|\tan \frac{x}{2}\right|+\frac{\tan ^{2}\left(\frac{x}{2}\right)}{2}+2 \tan \left(\frac{x}{2}\right)\right]+C\left[\because t=\tan \frac{x}{2}\right]$
$=\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C$

Indefinite Integrals Exercise Revision Exercise Question 73

Answer:
$2\left(\frac{1}{\sqrt{3}}\right) \tan ^{-1} \frac{\left(\tan \left(\frac{x}{2}\right)\right)}{\sqrt{3}}$
Hint:
To solve the given statement we will write $\cos x \operatorname{as} \frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}$
Given:
$\int \frac{1}{2+\cos x} d x$
Solution:
$\cos x=\frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}$
$I=\int \frac{1}{2+\frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}} d x$
$I=\frac{\int \sec ^{2}\left(\frac{x}{2}\right)}{\tan ^{2}\left(\frac{x}{2}\right)+3} d x$ $\left[\because \tan \left(\frac{x}{2}\right)=t, \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right)^{1} d x=d t\right]$
$I=2 \int \frac{d t}{t^{2}+(\sqrt{3})^{2}}$
$=2\left(\frac{1}{\sqrt{3}}\right) \tan ^{-1} \frac{\left(\tan \left(\frac{\mathrm{x}}{2}\right)\right)}{\sqrt{3}}$

Indefinite Integrals Exercise Revision Exercise Question 74

Answer:
$\sqrt{x^{2}+a x}+\frac{a}{2} \log \left|\left(x+\frac{a}{2}\right)+\sqrt{x^{2}+a x}\right|+c$
Hint:
To solve the given statement multiply and divide the equation by a +x.
Given:
$\int \sqrt{\frac{a+x}{x}} d x$
Solution:
$=\int \frac{a+x}{\sqrt{x(a+x)}} d x$
$=\frac{1}{2} \int \frac{2(a+x)}{\sqrt{x^{2}+a x}} d x$
$=\frac{1}{2}\left[\int \frac{2 x+a}{\sqrt{x^{2}+a x}} d x+\int \frac{a}{\sqrt{x^{2}+a x}} d x\right]$
$\left[d\left(\sqrt{x^{2}+a x}\right)=\frac{-1}{2 \sqrt{x^{2}+a x}}(2 x+a)\right]$
$\left[d\left(\sqrt{x^{2}+a x}\right)=\frac{a}{\sqrt{x^{2}+a x+\left(\frac{a^{2}}{4}\right)-\left(\frac{a^{2}}{4}\right)}}\right]$
$\left[d\left(\sqrt{x^{2}+a x}\right)=\frac{a}{\sqrt{\left(x+\frac{a}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}}}\right]$
$=\frac{1}{2} \cdot 2 \sqrt{x^{2}+a x}+\int \frac{a}{\sqrt{\left(x+\frac{a}{2}\right)^{2}-\left(\frac{a}{2}\right)}} d x$
$=\sqrt{x^{2}+a x}+a \log \left|\left(x+\frac{a}{2}\right)+\sqrt{x^{2}+a x}\right|$
$=\sqrt{x^{2}+a x}+\frac{a}{2} \log \left|\left(x+\frac{a}{2}\right)+\sqrt{x^{2}+a x}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 75

Answer:
$-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c$
Hint:
To solve the given solution we will use the partial fraction.
Given:
$\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}$
Solution:
$\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}$
$=\int \frac{\frac{-8}{2}(-4 x+1)+\left(\frac{13}{2}\right)}{\sqrt{6+x-2 x^{2}}} d x$
$=-\frac{3}{2}\left[\int \frac{1-4 x}{\sqrt{6+x-2 x^{2}}} d x+\frac{13}{2} \int \frac{1}{\sqrt{6+x-2 x^{2}}} d x\right.$
$=-\frac{3}{2}\left[\int \frac{d t}{\sqrt{t}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{x}{2}\right)-x^{2}}} d x\right.$
$=-\frac{3}{2} \times 2 \sqrt{t}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3-\left(x^{2}-\left(\frac{x}{2}\right)+\frac{1}{16}-\frac{1}{16}\right)}}$
$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{1}{16}\right)-\left(x-\left(\frac{1}{4}\right)\right)^{2}}}$
$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{7}{4}\right)^{2}-\left(x-\left(\frac{1}{4}\right)\right)^{2}}} d x$
$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{x-\left(\frac{1}{4}\right)}{\frac{7}{4}}\right)+c$
$=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 76

Answer:
$-\frac{1}{3 \cos ^{8} x}-\cos x-\frac{2}{\cos x}+c$
Hint:
To solve the given statement we will split sin 5x into sin? x sin x.
Given:
$\int \frac{\sin ^{5} x}{\cos ^{4} x} d x$
Solution:
$=\int \frac{\sin ^{4} x \sin x}{\cos ^{4} x} d x$
$=\int \frac{\left(1-\cos ^{2} x\right)^{2} \sin x}{\cos ^{4} x} d x \quad[\text { put } \cos x=t,-\sin x d x=d t]$
$=\int \frac{\left(1-t^{2}\right)^{2}(-d t)}{t^{4}}$
$=\int \frac{\left(1+t^{4}-2 t^{2}\right)(-d t)}{t^{4}}$
$=\int\left(\frac{1}{t^{4}}+1-\left(\frac{2}{t^{2}}\right)\right)(-d t)$
$=-\int \frac{1}{t^{4}} d t-\int d t+2 \int \frac{1}{t^{2}} d t$
$=-\frac{t^{-8}}{-3}-t+2\left(-\frac{t}{t}\right)+c$
$=-\frac{1}{3 \cos ^{8} x}-\cos x-\frac{2}{\cos x}+c$

Indefinite Integrals Exercise Revision Exercise Question 77

Answer:
$I=\log (\sin x)-\sin ^{2} x+\frac{\sin ^{4} x}{4}+c$
Hint:
to solve the given statement we will split cos? x into cos? x cos x then put cos? x = (1-sin²x)².
Given:
$\int \frac{\cos ^{5} x}{\sin x} d x$
Solution:
$I=\frac{\cos ^{5} x}{\sin x} d x$
$=\int \frac{\cos x\left(\cos ^{4} x\right)}{\sin x} d x$
$I=\int \frac{\cos x\left(1-\sin ^{2} x\right)^{2}}{\sin x} d x$
$\left[\because \sin x=t, \frac{d t}{d x}=\cos x, d x=\frac{d t}{\cos x}\right]$
$I=\int \frac{\left(1-t^{2}\right)^{2}}{t} d t$
$I=\int \frac{\left(1-2 t^{2}+t^{4}\right)}{t} d t$
$I=\int \frac{1}{t} d t-\int 2 t d t+\int t^{3} d t$
$I=\log t-t^{2}+\frac{t^{4}}{4}+c$
$I=\log (\sin x)-\sin ^{2} x+\frac{\sin ^{4} x}{4}+c$

Indefinite Integrals Exercise Revision Exercise Question 78

Answer:
$\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c$
Hint:
To solve the given equation we have to split the sin? x term in sin? x.sin² x.
Given:
$\int \frac{\sin ^{6} x}{\cos x} d x$
Solution:
$I=\int \frac{\sin ^{6} x}{\cos x} d x$
$=\int \frac{\sin ^{4} x \sin ^{2} x}{\cos x} d x=\int \frac{\sin ^{4} x}{\cos x}\left(1-\cos ^{2} x\right) d x$
$=\int \frac{\sin ^{4} x}{\cos x} d x-\int \frac{\sin ^{4} x \cos ^{2} x}{\cos x} d x$
$=\int \frac{\sin ^{4} x}{\cos x} d x-\int \sin ^{4} x \cos x d x$
$=\int \frac{\sin ^{2} x\left(1-\cos ^{2} x\right)}{\cos x} d x-\int t^{4} d t$
$=\int \frac{\sin ^{2} x}{\cos x} d x-\int \sin ^{2} x \cos x d x-\frac{t^{5}}{5}$
$=\int \frac{\left(1-\cos ^{2} x\right)}{\cos x} d x-\int t^{2} d t-\frac{\sin ^{5} x}{5}$
$=\int \frac{1}{\cos x} d x-\int \cos x d x-\frac{t^{8}}{3}-\frac{\sin ^{5} x}{5}$
$=\int \sec x d x-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}$
$=\ln |\sec x+\tan x|+\left(-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}\right)+c$
$=\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c$

Indefinite Integrals Exercise Revision Exercise Question 79

Answer:
$\frac{\tan ^{8} x}{3}+\frac{\tan ^{5} x}{5}+c$
Hint:
To solve the given statement divide the numerator and denominator by cos²x.
Given:
$=\int \frac{\sin ^{2} x}{\cos ^{6} x} d x$
Solution:
$=\int \frac{\frac{\sin ^{2} x}{\cos ^{2} x}}{\frac{\cos ^{6} x}{\cos ^{2} x}} d x$
$=\int \tan ^{2} x \sec ^{4} x d x$
$=\int \tan ^{2} x \sec ^{2} x\left(\sec ^{2} x d x\right)$
$=\int t^{2}\left(1+t^{2}\right) d t$
$=\int t^{2} d t+\int t^{4} d t$
$=\frac{t^{3}}{3}+\frac{t^{5}}{5}+c$
$=\frac{\tan ^{8} x}{3}+\frac{\tan ^{5} x}{5}+c$

Indefinite Integrals Exercise Revision Exercise Question 80

Answer:
$I=\frac{\tan ^{5} x}{5}+\frac{2 \tan ^{8} x}{3}+\tan x+c$
Hint:
To solve the given equation we have to split sec? x into sec? x sec² x.
Given:
$\int \sec ^{6} x d x$
Solution:
$I=\int \sec ^{6} x d x$
$=\int \sec ^{4} x \sec ^{2} x d x$
$=\int\left(\sec ^{2} x\right)^{2} \sec ^{2} x d x$
$\sec ^{2} x=\tan ^{2} x+1$
$I=\int\left(\tan ^{2} x+1\right)^{2} \sec ^{2} x d x$
$\left[\because \tan x=t \quad, \sec ^{2} x d x=d t\right.$
$I=\int\left(t^{2}+1\right)^{2} d t$
$I=\int\left(t^{4}+2 t^{2}+1\right) d t$
$I=\frac{t^{5}}{5}+\frac{2 t^{3}}{3}+t+c$
$I=\frac{\tan ^{5} x}{5}+\frac{2 \tan ^{3} x}{3}+\tan x+c$

Indefinite Integrals Exercise Revision Exercise Question 81

Answer:
$\frac{\sec ^{7} x}{7}-\frac{2 \sec ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$
Hint:
You must know about integration of tan x & sec x.
Given:
$\int \tan ^{5} x \sec ^{3} x d x$
Solution:
$\text { let } \sec x=t$
$\sec x \tan x d x=d t$
$\text { Now, } \int \tan ^{4} x \sec ^{2} x(\sec x \tan x d x) \ldots \ldots \ldots . \text { (1) }$
$\int \tan ^{4} x \sec ^{2} x d t$
$\text { put the value of } \sec x=\operatorname{tin}(1)$
$\int \tan ^{2} x \cdot \tan ^{2} x \cdot \sec ^{2} x d t$
$\int\left(\left(\sec ^{2} x-1\right)\left(\sec ^{2} x-1\right) \cdot \sec ^{2} x\right) d t$
$\int\left(t^{2}-1\right)\left(t^{2}-1\right) \cdot t^{2} d t$
$\int\left[t^{4}+(1)^{2}-2\left(t^{2}\right)(1) .\right] t^{2} d t$
$\int\left(t^{4}+1-2 t^{2}\right) t^{2} d t$
$\int t^{6} d t-\int 2 t^{4} d t+\int t^{2} d t$
$=\frac{t^{7}}{7}-\frac{2 t^{5}}{5}+\frac{t^{3}}{3}+c$
$=\frac{\sec ^{7} x}{7}-\frac{2 \sec ^{5} x}{5}+\frac{\sec ^{3} x}{3}+c$

Indefinite Integrals Exercise Revision Exercise Question 82

Answer:
$\frac{\tan ^{4} x}{4}+\frac{\tan ^{6} x}{6}+c$
Hint:
You must know about the integration of tan x & sec x.
Given :
$\int \tan ^{3} x \sec ^{4} x d x$
Solution:
$\operatorname{let} \tan x=t, \sec ^{2} x d x=d t$
$\text { now, } \int \tan ^{3} x \cdot \sec ^{2} x \cdot \sec ^{2} x d x \ldots \ldots \ldots . . \text { (1) }$
$\text { put value of } \tan x=\operatorname{t} \text { (1) }$
$=\int t^{3}\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x$
$=\int t^{3}\left(1+t^{2}\right) d t$
$=\int t^{3} d t+\int t^{5} d t$
$=\frac{t^{4}}{4}+\frac{t^{6}}{6}+c$
$=\frac{\tan ^{4} x}{4}+\frac{\tan ^{6} x}{6}+c$

Indefinite Integrals Exercise Revision Exercise Question 83

Answer:
$-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c$
Hint:
You must know about integration of sec x & cosec x.
Given:
$\int \frac{1}{\sec x+\operatorname{cosec} x} d x$
Soltuion:
$\operatorname{now}, \int \frac{1}{\frac{1}{\cos x}+\frac{1}{\sin x}} d x \quad\left[\because \sec x=\frac{1}{\cos x}, \operatorname{cosec} x=\frac{1}{\sin x}\right]$
$=\int \frac{d x(\cos x \sin x)}{\sin x+\cos x}$
$=\frac{1}{2} \int \frac{2(\cos x \sin x)}{\sin x+\cos x} d x$ $\left[\because(1+2 \sin x \cos x)=(\sin x+\cos x)^{2}\right.$
$=\frac{1}{2} \int \frac{1+2 \sin x \cos x-1}{\sin x+\cos x} d x$
$=\frac{1}{2} \int \frac{(\sin x+\cos x)^{2}-1}{\sin x+\cos x} d x$
$=\frac{1}{2} \int(\sin x+\cos x)=d x-\int \frac{1}{\sin x+\cos x} d x$
$=\frac{1}{2}\left[\int \sin x d x+\int \cos x d x-\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)}\right]$
$=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(x+\left(\frac{\pi}{4}\right)\right.}\right]$
$=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \operatorname{cosec}\left(x+\frac{\pi}{4}\right) d x\right]$
$=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)$
$=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 84

Answer:
$\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
Hint:
You must know about integration of $\int \sqrt{a^{2}+x^{2}}$
Given:
$\int \sqrt{a^{2}+x^{2}} d x$
Solution:
$I=\int 1_{I I} \cdot \sqrt{a_{1}^{2}+x^{2}} d x$
$=\sqrt{a^{2}+x^{2}} \int 1 d x-\int\left(\frac{d}{d x}\left(\sqrt{a^{2}+x^{2}}\right) \int 1 d x\right) d x$
$=\sqrt{a^{2}+x^{2}} \cdot x-\int \frac{1 \times 2 x}{2 \sqrt{a^{2}+x^{2}}} \cdot x d x$
$=\sqrt{a^{2}+x^{2}} \cdot x-\int\left(\frac{x^{2}+a^{2}-a^{2}}{\sqrt{a^{2}+x^{2}}}\right) d x$
$=x \sqrt{a^{2}+x^{2}}-\int \sqrt{a^{2}+x^{2}} d x+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x$
$=x \sqrt{a^{2}+x^{2}}-I+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x$
$\therefore 2 I=x \sqrt{a^{2}+x^{2}}+a^{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|$
$\Rightarrow I=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

Indefinite Integrals Exercise Revision Exercise Question 86

Answer:
$a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\left(\frac{x}{a}\right) \cdot \sqrt{a^{2}-x^{2}}\right)+c$
Hint:
You must know about how to solve integration.
Given:
$\int \sqrt{a^{2}-x^{2}} d x$
Solution:
$\text { let } x=a \sin \theta$
$d x=a \cos \theta d \theta$
$I=\int \sqrt{a^{2}-a^{2} \sin ^{2} \theta} d \theta$
$I=a^{2} \int \sqrt{1-\sin ^{2} \theta} \cdot \cos \theta d \theta \quad \sqrt{1-\sin ^{2} \theta}=\cos ^{2} \theta$
$I=a^{2}\left(\frac{\theta}{2}+\frac{1}{2} \frac{(\sin 2 \theta)}{2}\right)+c$ $\left[\operatorname{let} \theta=\sin ^{-1}\left(\frac{x}{a}\right), \cos \theta=\sqrt{1-\left(\frac{x^{2}}{a^{2}}\right)}\right.$
$I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2} \frac{2(\sin 2 \theta) \cos \theta}{2}\right)+c$
$I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2}\left(\frac{x}{a}\right) \cdot \frac{1}{a} \sqrt{a^{2}-x^{2}}\right)+c$
$I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\left(\frac{x}{a}\right) \cdot \sqrt{a^{2}-x^{2}}\right)+c$

Indefinite Integrals Exercise Revision Exercise Question 87

Answer:
$\frac{1}{6}(3 x+2) \sqrt{3 x^{2}+4 x+1}-\frac{\sqrt{3}}{18} \ln \left|\left(x+\frac{2}{3}\right)+\sqrt{x^{2}+\frac{4 x}{3}+\frac{1}{3}}\right|+c$
Hint:
You must know about how to solve integration.
Given:
$\int \sqrt{3 x^{2}+4 x+1} d x$
Solution:
$\int \sqrt{3} \sqrt{x^{2}+\frac{4 x}{3}+\frac{1}{3}} d x$
$\sqrt{3} \int \sqrt{x^{2}+\frac{4 x}{3}+\frac{4}{9}-\frac{4}{9}+\frac{1}{3}} d x$
$\sqrt{3} \int \sqrt{\left(x+\left(\frac{2}{3}\right)\right)^{2}-\frac{1^{2}}{3}} d x$
$\frac{1}{6}(3 x+2) \sqrt{3 x^{2}+4 x+1}-\frac{\sqrt{3}}{18} \ln \left|\left(x+\frac{2}{3}\right)+\sqrt{x^{2}+\frac{4 x}{3}+\frac{1}{3}}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 88

Answer:
$\left(\frac{\sqrt{3}}{2}\right)\left[\left(\frac{3 x-1}{3}\right) \sqrt{\frac{1}{3}+\frac{2 x}{3}-x^{2}}+\frac{4 \sin ^{-1}(3 x-1)}{9}\right]+c$
Hint:
You must know about formula of $\sqrt{a^{2}-x^{2}}, \sqrt{x^{2}+a^{2}}, \sqrt{x^{2}-a^{2}}$
Given:
$\int \sqrt{1+2 x-3 x^{2}} d x$
Solution:
$\int \sqrt{3} \sqrt{\frac{1}{3}+\frac{2}{3} x-x^{2}} d x$
$\sqrt{3} \int \frac{1}{3}+\frac{1}{9}-\frac{1}{9}+\frac{2 x}{3}-x^{2} d x$
$\int \sqrt{3} \sqrt{\frac{4}{9}-\frac{1}{9}-\frac{2}{3} x+x^{2}} d x$
$\int \sqrt{3} \sqrt{\left(\frac{2}{3}\right)^{2}-\left(x-\frac{1}{3}\right)^{2}} d x$ $\left[\because\left(\sqrt{a^{2}-x^{2}} d x=\frac{1}{2}\left(x \sqrt{a^{2}-x^{2}}\right)+a^{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]\right.$
$\sqrt{3} \cdot\left(\frac{1}{2}\right)\left[\left(x-\frac{1}{3}\right) \sqrt{\left(\frac{2}{3}\right)^{2}-\left(x-\frac{1}{3}\right)^{2}}+\left(\frac{2}{3}\right)^{2} \frac{\sin ^{-1}\left(x-\frac{1}{8}\right)}{\frac{2}{3}}\right]$
$I=\left(\frac{\sqrt{3}}{2}\right)\left[\left(\frac{3 x-1}{3}\right) \sqrt{\frac{1}{3}+\frac{2 x}{3}-x^{2}}+\frac{4}{9} \frac{\sin ^{-1}(3 x-1)}{2}\right]+c$

Indefinite Integrals Exercise Revision Exercise Question 89

Answer:
$\dpi{100} -\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+c$
Hint:
You have to find value of A & B.
Given:
$\int x \sqrt{1+x-x^{2}} d x$
Solution:
$\text { let } x=A\left(\left(\frac{d}{d x}\right)\left(1+x-x^{2}\right)\right)+B$
$x=A(1-2 x)+B$
$x=A+B-2 A x$
A + B = 0 , -2A = 1
$B=\frac{1}{2} \quad A=-\frac{1}{2}$
$\mathrm{I}=\int\left[-\frac{1}{2}(1-2 x)+\frac{1}{2}\right) \sqrt{1+x-x^{2}} d x$
$=-\frac{1}{2} \int(1-2 x) \sqrt{1+x-x^{2}} d x+\frac{1}{2} \sqrt{1+x-x^{2}} d x$
$I=-\frac{1}{2} \int \sqrt{t} d t \ldots .\left(\sqrt{1+x-x^{2}}=\mathrm{t}\right)$
$=-\frac{1}{3} t^{\frac{3}{2}}=-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c$
$I I=\frac{1}{2} \int \sqrt{1+x-x^{2}} d x$
$=\frac{1}{2} \int \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}} d x$ $\left[\because x-\frac{1}{2}=z, d x=d z\right]$
$=\frac{1}{2} \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^{2}-(z)^{2}} d z$
$=\frac{1}{2}\left[\frac{1}{2} z \sqrt{\frac{\sqrt{5}}{4}-z^{2}}+\frac{1}{2} \cdot \frac{\sqrt{5}}{4} \sin ^{-1}\left(\frac{2 z}{\sqrt{5}}\right)+c\right]$
$=\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]$
$\int x \sqrt{1+x-x^{2}}=I+I I$
$I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]$
$I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+C$

Indefinite Integrals Exercise Revision Exercise Question 90

Answer:
$\frac{\left(4 x^{2}+5 x+6\right)^{\frac{3}{2}}}{6}+\frac{7}{8} \int \frac{x}{2} \sqrt{4 x^{2}+5 x+6}-\frac{1}{8} \ln \left|2 x+\sqrt{4 x^{2}+5 x+6}\right|+c$
Hint:
You must have to know about integration method.
Given:
$\int(2 x+3) \sqrt{4 x^{2}+5 x+6} d x$
Solution:
$\int(2 x+3) \sqrt{4 x^{2}+5 x+6} d x$
$\frac{1}{4} \int(8 x+5+7) \sqrt{4 x^{2}+5 x+6} d x$
$\frac{1}{4} \int(8 x+5) \sqrt{4 x^{2}+5 x+6} d x+7 \int \sqrt{4 x^{2}+5 x+6} d x$
$\text { Let } 4 x^{2}+5 x+6=t$
$\frac{\frac{1}{4}\left(t^{\frac{1}{2}}+1\right)}{\frac{3}{2}}+7 \int \sqrt{\left(2 x+\frac{5}{2}\right)^{2}}+\left(6-\frac{25}{4}\right) d x$
$\frac{1}{4}\left(\frac{2\left(\sqrt{4 x^{2}+5 x+6}\right)}{3} \cdot\left(4 x^{2}+5 x+6\right)+7 \int \sqrt{\left(2 x+\frac{5}{2}\right)^{2}}+\frac{1}{4} d x\right.$
$\int \frac{\left(4 x^{2}+5 x+6\right)^{\frac{3}{2}}}{6}+\frac{7}{4} \sqrt{\left(2 x+\frac{5}{2}\right)^{2}}+\left(\frac{1}{2}\right)^{2} d x$
$\int \frac{\left(4 x^{2}+5 x+6\right)^{\frac{3}{2}}}{6}+\frac{7}{4} \sqrt{\left(2 x+\frac{5}{2}\right)^{2}}+\left(\frac{1}{2}\right)^{2} d x$
$\frac{\left(4 x^{2}+5 x+6\right)^{\frac{3}{2}}}{6}+\frac{7}{4} \sqrt{\left(2 x+\frac{5}{2}\right)^{2}\left(-\frac{1}{2}\right)^{2}}-\frac{1}{4} \times \frac{1}{2} \ln \left|2 x+\sqrt{4 x^{2}+5 x+6}\right|+c$
$\frac{\left(4 x^{2}+5 x+6\right)^{\frac{3}{2}}}{6}+\frac{7}{8} \int \frac{x}{2} \sqrt{4 x^{2}+5 x+6}-\frac{1}{8} \ln \left|2 x+\sqrt{4 x^{2}+5 x+6}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 91

Answer:
$\frac{\sin 2 x}{2}\left(x^{2}+\frac{1}{2}\right)+\frac{x \cos 2 x}{2}+c$
Hint:
You must know about integration of cos2x.
Given:
$\int(1+x)^{2} \cos 2 x d x$
Solution:
$\int(1+x)^{2} \cos 2 x d x$
$\int \cos 2 x d x+\int x^{2} \cos 2 x d x$
$\frac{\sin 2 x}{2}+\left[\frac{x^{2} \sin 2 x}{2}-\int 2 x\left(\frac{\sin 2 x}{2}\right) d x\right.$ $\left[1 . I 1 d x=I \int I I d x-\left[\left(\frac{d}{d x}\right) I \int I I d x\right]\right.$
$\frac{\sin 2 x}{2}+\left[\frac{x^{2} \sin 2 x}{2}-\left[x \frac{-\cos 2 x}{2}+-\frac{1}{2} \int \cos 2 x d x\right.\right.$
$\frac{\sin 2 x}{2}+\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{1}{2} \int \cos 2 x d x$
$\frac{\sin 2 x}{2}\left(1+x^{2}\right)+\frac{x \cos 2 x}{2}-\frac{1}{2} \frac{\sin 2 x}{2}+c$
$\sin 2 x\left(1+x^{2}-\frac{1}{2}\right)+\frac{x \cos 2 x}{2}+c$
$\frac{\sin 2 x}{2}\left(x^{2}+\frac{1}{2}\right)+\frac{x \cos 2 x}{2}+c$

Indefinite Integrals Exercise Revision Exercise Question 92

Answer:
$x \log _{10} x-x+c$
Hint: You must know about integration of logx.
Given:
$\int \log _{10} x d x$
Solution:
$\int \log x d x$
$\int \log x \cdot 1 d x$
$x \log x-\int \frac{1}{x} \cdot x d x$ $\left[\because \int 1=x, \int I .11 d x=I \int I I d x-\int \frac{d}{d x} I \int I I d x\right]$
$x \log _{10} x-x+c$

Indefinite Integrals Exercise Revision Exercise Question 93

Answer:
$(\log x) \cdot \log (\log x)-\log x+c$
Hint:
You must know about integration of log.
Given:
$\int \frac{\log (\log x)}{x} d x$
Solution:
$\operatorname{let} \log x=t$
$\frac{d x}{x}=d t$
$\int \log (t) d t$
$t \log t-t+c$ $\int I . I I d x=I \int I I d x-\int \frac{d}{d x} I \int I 1 d x$
put value of t
$(\log x) \cdot \log (\log x)-\log x+c$

Indefinite Integrals Exercise Revision Exercise Question 94

Answer:
$\frac{x \tan 2 x}{2}-\frac{1}{4} \log |\sec 2 x|+c$
Hint:
You must know about integration of sec x & tan x.
Given:
$\int x \sec ^{2} 2 x d x$
Solution:
$\int x \sec ^{2} 2 x d x \quad\left(\int1.11 d x=I \int I 1 d x-\frac{d}{d x} 1 . \int I I d x\right)$
$x\left(\frac{\tan 2 x}{2}\right)-\frac{1}{2} \int \tan 2 x d x$
$\frac{x \tan 2 x}{2}-\frac{1}{2} \int \tan 2 x d x \quad\left(\int \tan 2 x=\log \left|\frac{\sec 2 x}{2}\right|\right)$
$\frac{x \tan 2 x}{2}-\frac{1}{2} \log \left|\frac{\sec 2 x}{2}\right|+c$
$\frac{x \tan 2 x}{2}-\frac{1}{4} \log |\sec 2 x|+c$

Indefinite Integrals Exercise Revision Exercise Question 96

Answer:
$e^{\mathrm{x}}\left[x^{2}+1\right]+c$
Hint:
You must know about integration of I and II.(ILATE)
Given:
$\int(x+1)^{2} e^{x} d x$
Solution:
$\int\left(x^{2}+1+2 x\right) e^{x} d x$
$\int\left(x^{2} e^{x}+e^{x} 1+2 x e^{x}\right) d x$
$e^{x}\left(x^{2}-2 x+2\right)+e^{x}+e^{x}(2 x-2)$ $\int 1.11 d x=I \int 11 d x-\int \frac{d}{d x} I \int \text { IIdx } \quad\left(\int e^{x}=e^{x}\right)$
$e^{x}\left[x^{2}-2 x+2+1+2 x-2\right]+c\left(\int e^{x}=e^{x}\right)$
$e^{x}\left[x^{2}+1\right]+c$

Indefinite Integrals Exercise Revision Exercise Question 97

Answer:
$x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]$
Hint
$\text { Put } x=a \tan \theta$
Given:
$\int \log \left(x+\sqrt{x^{2}+a^{2}}\right) d x$
Solution:
$I=\int 1_{I I} \cdot \log \left(x+\sqrt{x_{I}^{2}+a^{2}}\right) d x$
$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(x+\sqrt{x^{2}+a^{2}}\right)\right\} \int 1 d x\right]$
$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int\left(\frac{1}{x+\sqrt{x^{2}+a^{2}}}\right) \times\left(1+\frac{1 \times 2 x}{2 \sqrt{x^{2}+a^{2}}}\right) \cdot x \cdot d x$
$=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int \frac{x}{\sqrt{x^{2}+a^{2}}} d x$
$\text { putting } x^{2}+a^{2}=\text { tinthe sec o nd inte gral }$
$\Rightarrow 2 x d x=d t$
$\Rightarrow x d x=\frac{d t}{2}$
$I=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t$
$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int t^{-\frac{1}{2}} d t$
$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2}\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]+C$
$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{t}+C$
$=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]$

Indefinite Integrals Exercise Revision Exercise Question 98

Answer:
$-\frac{1}{2} e^{-2 \log x}(2 \log x+1)+C$
Hint:
You must know about integration of e?
Given:
$\int \frac{\log x}{x^{3}} d x$
Solution:
$\int \frac{\log x}{x^{2}} \cdot \frac{1}{x} d x$ $\left(p u t \log x=t, \frac{1}{x} d x=d t\right)$
$\int \frac{t}{e^{2 t}} d t$ $\left(\log x=t, x=e^{t}\right)$
$\int t e^{-2 t} d t$
$t\left(\frac{e^{-2 t}}{-2}\right)-\int 1 \times \frac{e^{-2 t}}{-2} d t \ldots \text { using Byparts }$
$\left(\frac{-t e^{-2 t}}{2}\right)+\frac{1}{2}\left(\frac{e^{-2 t}}{-2}\right)$
$-\frac{1}{2} e^{-2 t}\left(t+\frac{1}{2}\right)+c$
$-\frac{1}{4} e^{-2 t}(2 t+1)+C$
Resubs. t=logx,
$-\frac{1}{2} e^{-2 \log x}(2 \log x+1)+C$

Indefinite Integrals Exercise Revision Exercise Question 99

Answer:
$-\log \frac{(1-x)}{x}-\log |x|+\log |1-x|+c$
Hint:
You must know about ILATE
Given:
$\int\frac{(\log (1-x))}{x^{2}} d x$
Solution:
$\int\frac{(\log (1-x))}{x^{2}} d x$
$\left(\frac{1}{x^{2}}\right) \log \int(1-x) d x \quad \text { (ILATE) }$
$\log (1-x)\left(-\frac{1}{x}\right)\left(-\frac{1}{1-x}\right)\left(-\frac{1}{x}\right) d x$
$-\log \frac{(1-x)}{x}-\int \frac{1}{(1-x) x} d x$
$-\log \frac{(1-x)}{x}-\left(\int \frac{1}{x}+\frac{1}{(1-x)}\right) d x$
$-\log \frac{(1-x)}{x}-\log |x|+\log |1-x|+c$

Indefinite Integrals Exercise Revision Exercise Question 100

Answer:
$(\log x)^{2} \times \frac{x^{4}}{4}-\frac{\log x \cdot x^{4}}{8}+\frac{x^{4}}{32}+C$
Hint:
You must know about the log formula of integration.
Given:
$\int x^{3}(\log x)^{2} d x$
Solution:
$\int x^{3}{ }_{11} \cdot\left(\log _{1} x\right)^{2} \cdot d x$
$=\left(\log x^{2}\right) \int x^{3} d x-\int \frac{2 \log x}{x} \times \frac{x^{4}}{4} d x$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2} \int \log _{1} x \cdot x^{3}{ }_{I I} d x$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \int x^{3} d x-\int\left\{\frac{d}{d x}(\log x) \int x^{3} d x\right\} d x\right]$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\int \frac{1}{x} \times \frac{x^{4}}{4} d x\right]$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\frac{1}{4} \int x^{3} d x\right]$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\frac{x^{4}}{16}\right]+C$
$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{\log x \cdot x^{4}}{8}+\frac{x^{4}}{32}+C$

Indefinite Integrals Exercise Revision Exercise Question 101

Answer: $\frac{1}{n} \log \left|\frac{\sqrt{1+x^{n}}-1}{\sqrt{1+x^{n}}+1}\right|+C$
Hint
Given: $\int \frac{1}{x \sqrt{1+x^{n}}} d x$
Solution:
$I=\int \frac{d x}{x \sqrt{1+x^{n}}}$
$=\int \frac{x^{n-1} d x}{x^{n-1} x^{1} \sqrt{1+x^{n}}}$
$=\int \frac{x^{n-1} d x}{x^{n} \sqrt{1+x^{n}}}$
$\text { putting } x^{n}=t$
$\Rightarrow n x^{n-1} d x=d t$
$\Rightarrow x^{n-1} d x=\frac{d t}{n}$
$I=\frac{1}{n} \int \frac{d t}{t \sqrt{1+t}}$
$\operatorname{let} 1+t=p^{2}$
$d t=2 p d p$

$I=\frac{1}{n} \int \frac{2 p d p}{\left(p^{2}-1\right) p}$
$=\frac{2}{n} \int \frac{d p}{p^{2}-1^{2}}$
$=\frac{2}{n} \times \frac{1}{2} \log \left|\frac{p-1}{p+1}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}\right|+C$
$=\frac{1}{n} \log \left|\frac{\sqrt{1+x^{n}}-1}{\sqrt{1+x^{n}}+1}\right|+C$

Indefinite Integrals Exercise Revision Exercise Question 102

Answer: $-2 \sqrt{1-x}-\frac{2}{5}(1-x)^{5 / 2}+\frac{4}{3}(1-x)^{3 / 2}+c$
Hint: to solve this question we have to use differentiate method
Given:
$\int \frac{x^{2}}{\sqrt{1-x}} d x$
Solution:
$\text { Let } 1-x=t$
$x=1-t$
$\text { differentiating on both sides, }$
$-d x=d t$
$d x=-d t$
$I=-\int \frac{(1-t)^{2}}{\sqrt{t}} d t$
$I=-\int \frac{1+t^{2}-2 t}{\sqrt{t}} d t$
$I=-\int \frac{1}{\sqrt{t}} d t-\int \frac{t^{2}}{\sqrt{t}} d t+\int \frac{2 t}{\sqrt{t}} d t$
$I=-\int t^{-1 / 2} d t-\int \frac{t \sqrt{t} \sqrt{t}}{\sqrt{t}} d t+2 \int \sqrt{t} d t$
$I=-\left(\frac{t^{1 / 2}}{1 / 2}\right)-\int t^{3 / 2} d t+2 \frac{t^{3 / 2}}{3 / 2}$
$I=-2 \sqrt{t}-\frac{t^{3 / 2+1}}{3 / 2+1}+2 x \frac{2}{3}-t^{3 / 2}+c$
$I=-2 \sqrt{t}-\frac{t^{5 / 2}}{5 / 2}+\frac{4}{3} t^{3 / 2}+c$
$I=-2 \sqrt{1-x}-\frac{2}{5}(1-x)^{5 / 2}+\frac{4}{3}(1-x)^{3 / 2}+c$

Indefinite Integrals Exercise Revision Exercise Question 103

Answer: $\frac{2}{9}\left(1+x^{3}\right)^{3 / 2}-\frac{2}{3} \sqrt{1+x^{3}}+c$
Hint: to solve this question we have to use substitute method
Given: $\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x$
Solution:
$\text { Let } I=\int \frac{x^{3} \cdot x^{2}}{\sqrt{1+x^{3}}} d x$
$\text { put } 1+x^{3}=t^{2}, \text { differentiate on both sides, }$
$3 x^{2} d x=2 t d t$
$I=\int \frac{\left(t^{2}-1\right)}{t} \cdot \frac{2 t d t}{3}$
$I=\frac{2}{3} \int\left(t^{2}-1\right) d t$
$I=\frac{2}{3} \int t^{2} d t-\frac{2}{3} \int d t$
$I=\frac{2}{3} \frac{t^{3}}{3}-\frac{2}{3} t+c$
$I=\frac{2}{9}\left(\sqrt{1+x^{3}}\right)^{3}-\frac{2}{3} \sqrt{1+x^{3}}+c$
$I=\frac{2}{9}\left(1+x^{3}\right)^{3 / 2}-\frac{2}{3} \sqrt{1+x^{3}}+c$

Indefinite Integrals Exercise Revision Exercise Question 104

Answer: $\frac{3}{2} \sin ^{-1} x-\frac{x}{2} \sqrt{1-x^{2}}+c$
Hint: to solve this equation we have to solve this by splitting the team
Given:
$I=\int \frac{1+x^{2}}{\sqrt{1-x^{2}}} d x$
Solution:
$\operatorname{Let} I=\int \frac{2-\left(1-x^{2}\right)}{\sqrt{1-x^{2}}} d x$
$I=\int \frac{2}{\sqrt{1-x^{2}}} d x-\int \frac{1-x^{2}}{\sqrt{1-x^{2}}} d x$
$I=\int \frac{2}{\sqrt{1-x^{2}}} d x-\int \sqrt{1-x^{2}} d x$
$I=2 \sin ^{-1} x-\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]+c .$
$\left\{\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\frac{1}{a} \sin ^{-1} \frac{x}{a} \& \int \sqrt{a^{2}-x^{2}} d x=\frac{a}{2} \sqrt{a^{2}-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right\}$
$I=2 \sin ^{-1} x-\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \sin ^{-1} x+c$
$I=\frac{3}{2} \sin ^{-1} x-\frac{x}{2} \sqrt{1-x^{2}}+c$

Indefinite Integrals Exercise Revision Exercise Question 105

Answer: $\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$
Hint: to solve this equation, we have to assume u as $\cos \Theta$
Given: $\int x \sqrt{\frac{1-x}{1+x}} d x$
Solution:
$I=\int x \sqrt{\frac{1-x}{1+x}} d x$
$I=\int x \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} d x$
$I=\int x \frac{(1-x)}{\sqrt{1-x^{2}}} d x$
$I=\int \frac{x-x^{2}}{\sqrt{1-x^{2}}} d x$
$I=\int \frac{x-x^{2}-1+1}{\sqrt{1-x^{2}}} d x$
$I=\int \frac{-x^{2}+1}{\sqrt{1-x^{2}}} d x+\int \frac{x-1}{\sqrt{1-x^{2}}} d x$
$I=\int \sqrt{1-x^{2}} d x+\int \frac{x}{\sqrt{1-x^{2}}} d x-\int \frac{1}{\sqrt{1-x^{2}}} d x$
$I=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1}(x)+C_{1}-\sqrt{1-x^{2}}+C_{2}-\sin ^{-1}(x)$
$+C_{3}\left[\because \int \frac{x}{\sqrt{1-x^{2}}} d x=-\sqrt{1-x^{2}}+C_{2}\right]$
$I=\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$

Indefinite Integrals Exercise Revision Exercise Question 106

Answer: $I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{8}}-1}{\sqrt{1+x^{8}+1}}\right|+c$
Hint: to solve this equation we have to do with differentiate method.
Given: $\int \frac{1}{x \sqrt{1+x^{3}}} d x$
Solution:
$x^{3}=t$
$3 x^{2} d x=d t$
$x^{2} d x=\frac{1}{3} d t$
$I=\int \frac{x^{2}}{x^{3} \sqrt{1+x^{3}}} d x$
$I=\frac{1}{3} \int \frac{d t}{t \sqrt{1+t}}$
$\text { Let } 1+\mathrm{t}=\mathrm{p}^{2} \text { , } \mathrm{dt}=2 \mathrm{pdp}$
$I=\frac{1}{3} \int \frac{2 p d p}{\left.p^{2}-1\right) p}$
$I=\frac{2}{3} \int \frac{d p}{\left(p^{2}-1\right)}$
$\frac{2}{3} \times \frac{1}{3} \times \frac{1}{2} \log \left[\frac{p-1}{p+1}\right]+C \quad\left[\frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left[\frac{x-a}{x+a}\right]+C\right]$
$I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{3}}-1}{\sqrt{1+x^{3}}+1}\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 107

Answer: $\frac{1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}\right]+C$
$\text { wheret }=\sin x-\cos x$
Hint: to solve this we have to put $\sin x+\cos u$ in team of t
Given: $\int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x$
Solution:
$I=\int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x$
$=\int \frac{\sin x+\cos x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} d x$
$=\int \frac{\sin x+\cos x}{1-2 \sin ^{2} \cos ^{2} x} d x$
$=\int \frac{\sin x+\cos x}{1-\frac{1}{2}(2 \sin x \cos x)^{2}} d x$
$=\int \frac{\sin x+ \cos x}{1-\frac{1}{2} \sin ^{2} 2 x} d x$
$\text { putting } \sin x-\cos x=t(i)$
$\Rightarrow(\sin x-\cos x)^{2}=t^{2}$
$\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$
$1-2 \sin x \cos x=t^{2}$
$\sin 2 x=1-t^{2}$
$\text { Differentiating }(i), \text { weget }$
$(\cos x+\sin x) d x=d t$
$I=\int \frac{1}{1-\frac{1}{2}\left(1-t^{2}\right)^{2}} d t$
$=\int \frac{2}{2-\left(1-t^{2}\right)^{2}} d t$
$=\int \frac{2}{(\sqrt{2})^{2}-\left(1-t^{2}\right)^{2}} d t$
$=2 \int \frac{1}{\left(\sqrt{2}+1-t^{2}\right)\left(\sqrt{2}-1+t^{2}\right)} d t$
$=\frac{2}{2 \sqrt{2}} \int\left[\frac{1}{\sqrt{2}+1-t^{2}}+\frac{1}{\sqrt{2}-1+t^{2}}\right] d t$
$=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2}+1-t^{2}} d t+\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2}-1+t^{2}} d t$
$=\frac{1}{\sqrt{2}} \int \frac{1}{(\sqrt{\sqrt{2}+1})^{2}-t^{2}} d t+\frac{1}{\sqrt{2}} \int \frac{1}{(\sqrt{\sqrt{2}-1})^{2}+t^{2}} d t$
$=\frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}+C$
$=\frac{1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}\right]+C$
$\text { wheret }=\sin x-\cos x$

Indefinite Integrals Exercise Revision Exercise Question 108

Answer: $\frac{x^{3}}{3} \tan ^{-1} x-\frac{x^{3}}{6}+\frac{1}{6} \log \left|1+x^{2}\right|+c$
Hint: in this equation we will us ILATE method and then differentiate the terms
Given: $x^{2} \tan ^{-1} x d x$
Solution: Considering $\tan ^{-1} x$ as first function and $x^{2}$ as second

$\begin{aligned} &\tan ^{-1} x \int x^{2} d x-\int\left(\frac{d}{d x} \tan ^{-1} x \int x^{2} d x\right) \\ &=\frac{x^{3}}{3} \tan ^{-1} x \frac{-1}{3} \int \frac{x \cdot x^{3}}{1+u^{2}}-x d u \\ &=\frac{x^{3}}{3} \tan ^{-1} x-\frac{1}{3} \int \frac{x\left(1+x^{2}\right)}{1+x^{2}} d x+\frac{1}{6} \int \frac{2 x}{1+x^{2}} d x \\ &=\frac{x^{3}}{3} \tan ^{-1} x-\frac{x^{2}}{6}+\frac{1}{6} \log \left|1+x^{2}\right|+c \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 109

Answer: $\tan ^{-1} \sqrt{x}(1+x)-\sqrt{x}+c$
Hint: to solve this equation we have to use Byparts method
Given: $\int \tan ^{-1} \sqrt{x} d x$
Solution: $\text { let } \sqrt{x}=t$
$\frac{1}{2 \sqrt{x}} d x=d t$
$d x=2 t d t$
$I=\int \tan ^{-1} t .2 t d t$
$I=\tan ^{-1} t . t^{2}-\int \frac{t^{2}+1-1}{1+t^{2}} d t \ldots . \text { using Byparts }$
$I=\tan ^{-1} t \cdot t^{2}-\int d t+\int \frac{1}{1+t^{2}}$
$I=\tan ^{-1} t \cdot t^{2}-t+\tan ^{-1} t+C$
$I=\tan ^{-1} t\left(1+t^{2}\right)-t+C$
$I=\tan ^{-1} \sqrt{x}(1+x)-\sqrt{x}+C$

Indefinite Integrals Exercise Revision Exercise Question 110

Answer: $\frac{\sqrt{x} \sqrt{1-x}}{2}-\frac{\sin ^{-1} \sqrt{x}(1-2 x)}{2}+C$
Hint: to solve the question we have to use ILATE method
Given: $\int \sin ^{-1} \sqrt{x} d x$
Solution:
$I=\int \sin ^{-1} \sqrt{x} d x$
$\text { Let } \sqrt{x}=\sin t$
$x=\sin ^{2} t \ldots \text { squaring on both sides }$
$d x=\sin t \cos t d t$
$I=\int \sin ^{-1}(\sin t) \sin t \cos t d t$
$I=\frac{1}{2} \int t \sin 2 t d t$
$\left[\int(u . v) d x=u \int v d x-\int\left[\frac{d}{d x} u \cdot \int u d x\right]\right] d x$
$I=\frac{1}{2}\left\{t \frac{\cos 2 t}{2}+\int \frac{\cos 2 t}{2} d t\right\}$
$I=\frac{1}{2}\left\{\frac{-t \cos 2 t}{2}+\frac{\sin 2 t}{4}\right\}+C$
$\sqrt{x}=\sin t$
$t=\sin \sqrt{x}$
$\cos t=\sqrt{1-\sin ^{2} t}$
$\cos t=\sqrt{1-x}$
$I=-\sin ^{-1} \frac{\sqrt{x}(1-2 x)}{2}+\frac{2 \sqrt{x} \sqrt{1-x}}{4}+C$
$I=\frac{\sqrt{x} \sqrt{1-x}}{2}-\frac{\sin ^{-1} \sqrt{x}(1-2 x)}{2}+C$

Indefinite Integrals Exercise Revision Exercise Question 111

Answer: $I=x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+C$
Hint: to solve this equation we have to useByparts equation;
Given: $I=\int \sec ^{-1} \sqrt{x} d x$
Solution: Let $\sec ^{-1} \sqrt{x} \text { be the first function and } 1 \text { as the second function }$
$I=\sec ^{-1} \sqrt{x} \cdot x-\int \frac{1}{\sqrt{x} \sqrt{x-1} \cdot 2 \sqrt{x}} \ldots . . \text { Using Byparts }$
$I=x \sec ^{-1} \sqrt{x}-\frac{1}{2} \int(x-1)^{-1 / 2} d x$
$I=x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+C$

Indefinite Integrals Exercise Revision Exercise Question 112

Answer: $\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^{2}}\right]+C$
Hint: to solve this question we will presume x as $\cos \theta$
Given: $\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$
Solution:
$I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$
$\operatorname{let} x=\cos 2 \theta$
$\frac{d x}{d \theta}=-2 \sin 2 \theta$
$d x=-2 \sin 2 \theta d \theta$
$I=\int \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}-2 \sin 2 \theta d \theta$
$\cos 2 \theta=2 \cos ^{2} \theta-1$
$\cos 2 \theta=1-2 \sin ^{2} \theta$
$I=\int \tan ^{-1} \sqrt{\frac{1-1+2 \sin ^{2} \theta}{1+2 \cos ^{2} \theta-1}} \times-2 \sin 2 \theta d \theta$
$I=\int \tan ^{-1} \sqrt{\tan ^{2} \theta} \times-2 \sin 2 \theta d \theta$
$I=-2 \int \theta \sin 2 \theta d \theta$
$I=-2\left[\theta \int \sin 2 \theta d \theta-\int \frac{d \theta}{d \theta} \int \sin 2 \theta d \theta\right] d x . \text { .applying Byparts }$
$I=-2\left[\theta \times \frac{-\cos 2 \theta}{2}-(-1) \int \frac{\cos 2 \theta}{2}\right] d \theta$
$I=-2\left[-\theta \frac{\cos 2 \theta}{2}-(-1) \int \frac{\cos 2 \theta}{2}\right] d \theta$
$I=-2\left[-\theta \frac{\cos 20}{2}+\int \frac{\cos 20}{2}\right] d \theta$
$I=-2\left[-\frac{\cos 2 \theta}{2}+\frac{\sin 2 \theta}{4}\right]+C$
$I=-2\left[\left(\frac{-1}{2} \cos ^{-1} x\right) \frac{x}{2}+\frac{\sin \left(\cos ^{-1} x\right)}{4}+C\right]$
$I=-2\left[\frac{-x}{4} \cos ^{-1} x+\frac{\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)}{4}\right]+C$
$I=-\frac{2}{4}\left[-x \cos ^{-1} x+\sqrt{1-x^{2}}\right]+C$
$I=\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^{2}}\right]+C$

Indefinite Integrals Exercise Revision Exercise Question 115

Answer: $\left(\sin ^{-1} x\right)^{3} \cdot x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}+6 \sin ^{-1} x-6 \sqrt{1-x^{2}}+C$
Hint: to solve this equation, we have to convert equation into trigonometric formulaes
Given: $\int\left(\sin ^{-1} x\right)^{3} d x$
Solution:
$\text { Let } \sin ^{-1} x=\theta$
$x=\sin \theta=>d x=\cos \theta d \theta$
$I=\int \theta^{3} \cdot \cos \theta d \theta$
$I=\theta^{3} \sin \theta-3 \int \theta^{2} \sin \theta d \theta$
$I=\theta^{3} \sin \theta+3 \theta^{2} \cos \theta-\left(60(-\sin \theta)-\int 6(-\sin \theta)\right.$
$I=\theta^{3} \sin \theta+3 \theta^{2} \cos \theta+60 \sin \theta-6 \cos \theta+C$
$I=\left(\sin ^{-1} x\right)^{3} \cdot x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}+6 \sin ^{-1} x-6 \sqrt{1-x^{2}}+C$

Indefinite Integrals Exercise Revision Exercise Question 113

Answer: $a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}\right]+C$
Hint: to solve this statement we will assume sin as tan
Given: $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
Solution:
$\mathrm{I}=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
$\text { Put } x=a \tan ^{2} t=>\tan t=\sqrt{\frac{x}{a}}$
$d x=a\left(2 \tan (t) \sec ^{2} t d t\right.$
$I=\int \sin ^{-1} \sqrt{\frac{\operatorname{atan}^{2} t}{a+\operatorname{atan}^{2} t}} a\left(2 \tan (t) \sec ^{2} t d t\right.$
$I=2 a \int \operatorname{ttan}(t) \sec ^{2} t d t$
$\text { Let t be the first function and } \tan (t) \sec ^{2} t d t \text { be the} e \text { second, applying Byparts }$
$I=a\left(\operatorname{ttan}^{2} t-\tan t+t\right)+C$
$I=a\left(\tan ^{2} t+t-\tan t\right)+C$
$I=a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}\right]+C$

Indefinite Integrals Exercise Revision Exercise Question 114

Answer: $3 x \sin ^{-1} x+3 \sqrt{1-x^{2}}+C$
To solve this question we will use ILATE Method
Given: $\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x$
Solution:
$\int 3 \sin ^{-1} x d x$
$I=3 \int 1 \cdot \sin ^{-1} x d x$
$I=3 \int 1 \cdot \sin ^{-1} x d x$

$I=3\left[x \sin ^{-1} x-\int \frac{-d t}{2 \sqrt{t}}\right] \ldots \text { applying byparts }$
$I=3\left[x \sin ^{-1} x+\frac{1}{2} \int t^{-\frac{1}{2}} d t\right]$
$I=3\left[x \sin ^{-1} x+\frac{1}{2} \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\right)$
$I=3\left[x \sin ^{-1} x+\frac{1}{2} \frac{\sqrt{t}}{\frac{1}{2}}+C\right]$
$I=\left[3 x \sin ^{-1} x+3 \sqrt{1-x^{2}}+C\right]$

Indefinite Integrals Exercise Revision Exercise Question 116

Answer:
$2\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+C$
Hint: to solve this statement we will suppose x as $\sin \theta$
Given: $\int \cos ^{-1}(1-2 x)^{2} d x$
Solution:
$\int \cos ^{-1}(1-2 x)^{2} d x$
$\operatorname{let} x=\sin \theta, d x=\cos \theta d \theta$
$I=\int \cos ^{-1}\left(1-2 \sin ^{2} \theta\right)^{2} \cos \theta d \theta$
$I=\int \cos ^{-1}(\cos 2 \theta) \cos \theta d \theta$
$I=\int 2 \theta \cos \theta d \theta$
$\int u v d x=u \int v d x-\int\left[\frac{d u}{d v} \int v d x\right] d x$
$I=2\left[\theta \sin \theta-\int \sin \theta d \theta\right]$
$I=2[\theta \sin \theta+\cos \theta]+c$
$I=2\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+C$

Indefinite Integrals Exercise Revision Exercise Question 117

Answer: $\left(\sin ^{-1} x\right) \cdot \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2} \log \frac{1+x}{1-x}+C$

Hint: to solve this equation, we have to assume x as $\sin \theta$ and then do byparts method
Given: $\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}} d x$
Soluton: Let $x=\sin \theta$
$d x=\cos \theta d \theta$
$I=\int \frac{\sin \theta(\theta \cos \theta d \theta)}{\left(1-\sin ^{2} \theta\right)^{\frac{3}{2}}}$
$\int \frac{(\sin \theta)(\theta) \cos \theta d \theta}{\cos ^{3} \theta}$
$\int \frac{\sin \theta}{\cos ^{2} \theta} \theta d \theta$
$I=\int \theta \cdot \sec \theta \tan \theta d \theta$
$\left[\int \sec \theta \tan \theta d \theta=\int d \sec \theta=\sec \theta\right]$
$I=\theta \sec \theta \int 1 \cdot \sec \theta d \theta$
$I=\theta \sec \theta-\ln |\sec \theta+\tan \theta|+C$
$I=\left(\sin ^{-1} x\right) \cdot \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2} \log \frac{1+x}{1-x}+C$

Indefinite Integrals Exercise Revision Exercise Question 118

Answer: $\frac{e^{2 x} \tan x}{2}+C$
Hint: to solve this equation, we have to assume 2x as t
Given: $\int e^{2} x\left(\frac{1+\sin 2 x}{1+\cos 2 x}\right) d x$
Solution:
$\int e^{2 x}\left(\frac{1+\sin 2 x}{1+\cos 2 x}\right) d x$
$\int e^{x} f(x)+f^{\prime}(x) d x=e^{x}(x)+C$
$\begin{aligned} &2 x=t \\ &2 d x=d t \\ &d x=\frac{d t}{2} \end{aligned}$ $\left[\begin{array}{l} \cos ^{2} \theta=2 \cos ^{2} \theta-1 \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \\ 1+\cos 2 x=2 \cos ^{2} x \\ \sin ^{2} x=2 \sin x \cos x \end{array}\right]$
$I=\int e^{t}\left(\frac{1+\sin t}{1+\cos t}\right) \frac{d t}{2}$
$=\int e^{t}\left(\frac{1+\sin t}{1+\cos t}\right) \cdot \frac{d t}{2}$
$=\frac{1}{2} \int e^{t}\left(\frac{1+2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \cos ^{2} \frac{t}{2}}\right) d t$
$=\frac{1}{2} \int e^{t}\left(\frac{1}{2} \sec ^{2} \frac{t}{2}+\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^{2} \frac{t}{2}}\right) d t$
$=\frac{1}{2} \int e^{t}\left(\frac{1}{2} \sec ^{2} \frac{t}{2}+\tan \frac{t}{2}\right) d t$
$=\frac{1}{2} e^{t} \tan \frac{t}{2}+C$
$=\frac{e^{t} \tan \frac{t}{2}}{2}+C$
$=\frac{e^{2 x} \tan x}{2}+C . \text { resubs. } t=2 x$

Indefinite Integrals Exercise Revision Exercise Question 119

Answer: $-e^{-x / 2} \sec \frac{x}{2}+C$
Hint : To solve this equation we have to use differentiation method
Given: $\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x$
Solution:
$\int e^{-x}\left(f^{\prime}(x)-f(x) d x=e^{-x} f(x)+C\right.$
$\operatorname{Let} \frac{x}{2}=t$
$x=2 t$
$d x=2 d t$
$I=\int \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} \cdot e^{-t} \cdot 2 d t$
$I=2 \int e^{-t} \cdot \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} \cdot d t$
$\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=\frac{\sqrt{\sin ^{2} t+\cos ^{2} t-2 \sin t \cdot \cos t}}{2 \cos ^{2} t}$
$I=\frac{\sqrt{(\sin t-\cos t)^{2}}}{2 \cos ^{2} t}=\frac{\sin t-\cos t}{2 \cos ^{2} t}$
$=\frac{1}{2}\left(\frac{\sin t}{\cos ^{2} t}-\frac{\cos t}{\cos ^{2} t}\right)$
$=\frac{1}{2}\left(\frac{\tan t}{\cos t}-\frac{1}{\cos t}\right)$
$=\frac{1}{2}(\tan t \cdot \sec t-\sec t)$
$f(t)=\frac{1}{2} \sec t ; f^{\prime}(t)=\frac{1}{2} \tan t \sec t$
$\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=f^{\prime}(t)-f(t)$
$I=-2 e^{-t} f(t)+C$
$I=-2 e^{-t} \cdot \frac{1}{2} \sec t+C$
$I=-e^{-x / 2} \sec \frac{x}{2}+C$

Indefinite Integrals Exercise Revision Exercise Question 12

Answer: $I=e^{x} \frac{1}{\tan ^{2} x}+C$
Hint: to solve this equation, we have to break the $\left(1+x^{2}\right)$
Given: $\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x$
Solution:
$I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x$
$I=\int e^{x} \frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}} d x \ldots \int e^{x}\left(f^{\prime}(x)+f(x) d x=e^{x} f(x)+C\right.$
$I=e^{x} \frac{1}{1+x^{2}}+C$

Indefinite Integrals Exercise Revision Exercise Question 121

Answer: $\frac{e^{m \tan ^{-1} x}}{m^{2}+1}\left(m \cos \left(\tan ^{-1} x\right)+\sin \left(\tan ^{-1} x\right)+C\right.$
Given: $\int \frac{e^{m \operatorname{tan}^{-1} x}}{\left(1+x^{2}\right)^{3 / 2}} d x$
Hint: using integration by parts
Explanation
$\text { Let } I=\int \frac{e^{m \tan ^{-1} x}}{\left(1+x^{2}\right)^{3 / 2}} d x$
$I=\int \frac{e^{m \theta}}{\left(1+\tan ^{2} \theta\right)^{3 / 2}} \cdot \sec ^{2} \theta d \theta$ $x=\tan \theta \cdot \theta=\tan ^{-1} x$
Put $d x=\sec ^{2} \theta d \theta$
$I=\int \frac{e^{m \theta}}{\left(\sec ^{2} \theta\right)^{3 / 2}} \cdot \sec ^{2} \theta d \theta$
$I=\int \frac{e^{m \theta}}{\sec ^{3} \theta} \sec ^{2} \theta d \theta$
$I=\int \frac{e^{m \theta} d \theta}{\sec \theta}=\int e^{m \theta} \cos \theta d \theta\left[\because \frac{1}{\cos \theta}=\sec \theta\right]$
$I=\cos \theta \frac{e^{m \theta}}{m}-\int(-\sin \theta) \frac{e^{m \theta}}{m} d \theta \ldots u s i n g \text { Byparts }$
$I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m} \int \sin \theta e^{m \theta} d \theta$
$I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m}\left[\sin \theta \frac{e^{m \theta}}{m}-\int \cos \theta \frac{e^{m \theta}}{m} d \theta\right]$
$I=\frac{1}{m} \cos \theta e^{m \theta}+\frac{1}{m^{2}} \sin \theta e^{m \theta}-\frac{1}{m^{2}} \int \cos \theta e^{m \theta} d \theta$
$I=\frac{1}{m} \cos \theta \cdot e^{m \theta}+\frac{1}{m^{2}} \int \sin \theta e^{m \theta}-\frac{1}{m^{2}} I \ldots \text { w?ere } I=\int \cos \theta \cdot e^{m \theta} d \theta$
$\left(1+\frac{1}{m^{2}}\right) I=\frac{1}{m} \cos \theta \cdot e^{m \theta}+\frac{1}{m} \sin \theta e^{m \theta}+C$
$\left(\frac{m^{2}+1}{m^{2}}\right) I=e^{m \theta}\left(\frac{\cos \theta}{m}+\frac{\sin \theta}{m^{2}}\right)+C$
$\frac{\left(m^{2}+1\right)}{m^{2}} I=e^{m \theta}\left(\frac{m \cos \theta+\sin \theta}{m^{2}}\right)+m^{2} C$
$\left(m^{2}+1\right) I=e^{m \theta}(m \cos \theta+\sin \theta)+m^{2} C$
$I=\frac{e^{m \theta}}{m^{2}+1}(m \cos \theta+\sin \theta)+m^{2} C$
$=\frac{e^{m \tan ^{-1} x}}{m^{2}+1}\left(m \cos \left(\tan ^{-1} x\right)+\sin \left(\tan ^{-1} x\right)+C\right.$

Indefinite Integrals Exercise Revision Exercise Question 122

Answer: $\frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\left(\frac{3 x-4}{4(x-1)^{2}}\right)+C$
Given: $\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$
Hint: you must know the steps to integrate by partial function
Explanation: let $I=\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$
$\frac{x^{2}}{(x-1)^{3}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}+\frac{D}{(x+1)}$
Multiplying by $(x-1)^{3}(x+1)$
$x^{2}=A(x-1)^{3}(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^{3}$
Putting x=1

$1=A(0)+B(0)+C(2)+D(0) \Rightarrow c=\frac{1}{2}$
Putting x=-1
$1=A(0)+B(0)+C(2)+D(-8) \Rightarrow c=\frac{-1}{8}$

Putting x=0

$\begin{aligned} &1=A(1)+B(-1)+C(1)+D(-1)\\ &A-B+C-D=0\\ &A-B+\frac{1}{2}-\left(-\frac{1}{8}\right)=0 \Rightarrow A-B+\frac{1}{2}+\frac{1}{8}=0\\ &A-B+\frac{5}{8}=0 \Rightarrow A-B=-\frac{5}{8} \end{aligned}$
Putting x=2
$\begin{aligned} &4=A(1)(3)+B(1)(3)+C(3)+D(1) \\ &4=3 A+3 B+3\left(\frac{1}{2}\right)+\left(-\frac{1}{8}\right) \\ &\Rightarrow 3 A+3 B=4-\frac{3}{2}+\frac{1}{8}=>3(A+B)=\frac{32-12+1}{8}=\frac{21}{8} \\ &A+B=\frac{3}{8} \end{aligned}$

Adding (i) and (ii)

$\begin{aligned} &A+B=\frac{-5}{8}+\frac{7}{8}=\frac{2}{8} \\ &2 A-\frac{2}{8} \Rightarrow A=\frac{1}{8} \end{aligned}$
Put in (1)
$\begin{aligned} &\frac{1}{8}-B=\frac{5}{8} \Rightarrow B=\frac{1}{8}+\frac{5}{8}=\frac{6}{8}=>B=\frac{3}{4} \\ &\frac{x^{2}}{(x-1)^{3}}=\frac{1}{8(x-1)}+\frac{3}{4(x-1)^{2}}=\frac{1}{2(x-1)^{3}}=\frac{1}{8(x+1)} \\ &\frac{1}{8} \int \frac{d x}{x-1}+\frac{3}{4} \int \frac{(x-1)^{-2+1}}{-2+1}+\frac{1}{2} \frac{(x-1)^{-3+1}}{-3+1}-\frac{1}{8} \log |x+1|+C \\ &=\frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\frac{3 x-4}{4(x+1)^{2}}+C \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 123

Answer: $=\frac{1}{3} \log |x+1|-\frac{1}{6} \log \left|x^{2}+x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}+C$
Given: $\int \frac{x}{x^{3}-1} d x$
Hint: using partial function
Explanation let $I=\int \frac{x}{x^{3}-1} d x$
$\frac{x}{x^{3}-1}=\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+c}{x^{2}+x+1}$
Multiplying by $(x-1)\left(x^{2}+x+1\right)$
$x=A\left(x^{2}+x+1\right)+(B x+C)(x-1)$
Putting x=1

$\begin{aligned} &1=A(1+1+1)+(B(1)+C)(1-1) \\ &1=3 A+0 \Rightarrow A=\frac{1}{3} \end{aligned}$
Putting x=-1

$\begin{aligned} &-1=A+(-2)(-B+C)\\ &-1=\frac{1}{3}+2 B-2 C\\ &2(B-C)=-1 \frac{-1}{3} \Rightarrow B-C=\frac{-2}{3}\\ &B-C=>\frac{-2}{3} \end{aligned}$
Putting x=0
$\begin{aligned} &0=A(0+0+1)+(0+C)(0-1) \\ &0=A(1)+C(-1) \\ &0=\frac{1}{3}-C=>C=\frac{1}{3} \end{aligned}$
Put in (1)
$\begin{aligned} &B-\frac{1}{3}=\frac{-2}{3}=>B=\frac{1}{3} \frac{-2}{3}=>B=\frac{-1}{3} \\ &\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3(x-1)}+\frac{\frac{-1}{3} x+\frac{1}{3}}{x^{2}+x+1} \\ &\int \frac{x d x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x-1}{x^{2}+x+1} d x \\ &=\frac{1}{3} \log |x-1|-\frac{1}{3} I_{1} \quad \text { (2) } \end{aligned}$
Where $I_{1}=\int \frac{x-1}{x^{2}+x+1} d x$
$\begin{aligned} &=\frac{1}{2} \int \frac{2(x-1)+3-3}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+1} d x-\frac{3}{2} \int \frac{1}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x \quad\left[\begin{array}{l} x^{2}+x+1=t \\ (2 x+1) d x=d t \end{array}\right] \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &=\frac{1}{2} \log |t|-\frac{3}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \\ &=\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}$

By (2), we get

$\begin{aligned} &=\frac{1}{3} \log |x-1|-\frac{1}{3}\left[\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]+C \\ &=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^{2}+x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 124

Answer: $\frac{1}{2} \log \left|\frac{1+x}{\sqrt{1+x^{2}}}\right|+\frac{1}{2} \tan ^{-1} x+C$
Given: $\int \frac{1}{1+x+x^{2}+x^{3}} d x$
Hint: using partial function
Explanation: let $I=\int \frac{1}{1+x+x^{2}+x^{2}} d x$
$\begin{aligned} &\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{(1+x)+x^{2}(1+x)}=\frac{1}{\left(1+x^{2}\right)(1+x)} \\ &\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{\left(1+x^{2}\right)(1+x)} \\ &\frac{1}{(1+x)\left(1+x^{2}\right)}=\frac{A}{(1+x)}+\frac{B x+C}{\left(1+x^{2}\right)} \end{aligned}$
Multiplying by $(1+x)\left(1+x^{2}\right)$
$1=A\left(1+x^{2}\right)+(B x+C)(1+x)$
$\begin{aligned} &\text { put } x=-1 \\ &1=A(1+1)+(B(-1)+C)(1-1) \\ &1=A(2)+0=>A=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\text { put } x=0 \\ &1=A(1+0)+(B(0)+C)(1+0) \\ &1=A+C=>1-\frac{1}{2}=C \Rightarrow \frac{1}{2} \end{aligned}$
$\begin{aligned} &\text { putting } x=1 \\ &1=A(1+1)+(B+C)(1+1) \\ &1=2 A+2 B+2 C \\ &1=2\left(\frac{1}{2}\right)+2 B+2\left(\frac{1}{2}\right) \\ &1=2 B+1=>2 B=1-2=>B=-\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{(1+x)\left(1+x^{2}\right)}=\frac{1}{2(1+x)}+\frac{1}{2}\left(\frac{-x+1}{\left(1+x^{2}\right)}\right) \\ &\int \frac{1}{(1+x)\left(1+x^{2}\right)} d x=\frac{1}{2} \int \frac{1}{1+x} d x-\frac{1}{2} \int \frac{x-1}{1+x^{2}} d x \\ &=\frac{1}{2} \int \frac{1}{1+x^{2}} d x-\frac{1}{2^{2}} \int \frac{2 x}{1+x^{2}} d x+\frac{1}{2} \int \frac{1}{1+x^{2}} d x \\ &=\frac{1}{2} \log |1+x|-\frac{1}{4} \log \left|1+x^{2}\right|+\frac{1}{2} \tan ^{-1} x+C \\ &=\frac{1}{2}\left[\log |1+x|-\log \left|\sqrt{1+x^{2}}\right|\right]+\frac{1}{2} \tan ^{-1} x+C \\ &=\frac{1}{2} \log \left|\frac{1+x}{\sqrt{1+x^{2}}}\right|+\frac{1}{2} \tan ^{-1} x+C \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 125

Answer: $\frac{1}{3}\left(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{1}{\sqrt{5}} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+C$
Given: $\int \frac{1}{\left(x^{2}+2\right)\left(x^{2}+5\right)} d x$
Hint: Using $\int \frac{1}{1+x^{2}} d x$
Explanation: let $I=\int \frac{1}{\left(x^{2}+2\right)\left(x^{2}+5\right)} d x$
$=\frac{1}{3} \int \frac{\left(x^{2}+5\right)-\left(x^{2}+2\right)}{\left(x^{2}+5\right)\left(x^{2}+2\right)} d x$
$=\frac{1}{3} \int\left(\frac{1}{x^{2}+2}-\frac{1}{x^{2}+5}\right) d x$
$=\frac{1}{3} \int \frac{1}{x^{2}+2} d x-\frac{1}{3} \int \frac{1}{x^{2}+5} d x$
$=\frac{1}{3} \int \frac{1}{x^{2}+(\sqrt{2})^{2}} d x-\frac{1}{3} \int \frac{1}{x^{2}+(\sqrt{5})^{2}} d x$
$=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{1}{3 \sqrt{5}} \tan ^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$
$=\frac{1}{3}\left(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{1}{\sqrt{5}} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+C$

Indefinite Integrals Exercise Revision Exercise Question 126

Answer: $2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c$
Given: $\int \frac{x^{2}-2}{x^{5}-x} d x$
Hint: using partial fraction
Explanation let $I=\int \frac{x^{2}-2}{x^{5}-x} d x$
$\begin{aligned} &=\int \frac{x^{2}-2}{x\left(x^{4}-1\right)} d x=\int \frac{x^{2}-2}{x\left(x^{2}-1\right)\left(x^{2}+1\right)} \\ &I=\int \frac{x^{2}-2}{x(x+1)(x-1)\left(x^{2}+1\right)} d x \\ &\frac{x^{2}-2}{x(x+1)(x-1)\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}+\frac{D x+E}{x^{2}+1} \end{aligned}$
Multiply by $x(x+1)(x-1)\left(x^{2}+1\right)$
$x^{2}-2=A(x+1)(x-1)\left(x^{2}+1\right)+B(u)(u-1)\left(x^{2}+1\right)+C(x)(x+1)\left(x^{2}+1\right)+(D x+E)(x)(x+1)(x-1)$
$\begin{aligned} &\text { putting } x=0 \\ &0-2=A(1)(-1)(1)+B(0)+C(0)+(D x+6)(10) \\ &-2=-A \Rightarrow A=2 \end{aligned}$
$\begin{aligned} &\text { putting } x=1 \\ &1-2=A(2)(0)(2)+B(0)+C(1)(2)(2)+(D x+6)(0) \\ &-1=4 C=>C=\frac{-1}{4} \end{aligned}$
$\begin{aligned} &\text { putting } x=-1\\ &1-2=A(0)+B(-1)(-2)(2)+C(0)+D(0)\\ &-1=>4 B \Rightarrow B=\frac{-1}{4} \end{aligned}$
$\text { On solving }, D=-\frac{3}{2} \& E=0$
$\int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x$
$\int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x$
$=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1}-\frac{3}{4} \int \frac{2 x d x}{x^{2}+1}$
$=2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c$

Indefinite Integrals Exercise Revision Exercise Question 127

Answer: $-2 \sqrt{1-x}+\cos ^{-1} x+\sqrt{x(1-x)}+C$
Given: $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$
Hint: You must know the derivatives and integration of sinx and cosx
Explanation:
Let $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$
$\begin{aligned} &=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta \\ &=\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \times 2 \sin \theta \cos \theta d \theta \\ &=-2 \int \tan \frac{\theta}{2} \sin \theta \cos \theta d \theta \end{aligned}$ $\left[\begin{array}{l} \text { put } \sqrt{x}=\cos \theta \\ x=\cos ^{2} \theta \\ d x=2 \cos \theta(-\sin \theta) d \theta \\ d x=-2 \sin \theta \cos \theta d \theta \end{array}\right]$
$\begin{aligned} &=-2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos \theta d \theta \\ &=-4 \int \sin ^{2} \frac{\theta}{2} \cos \theta d \theta \end{aligned}$ $\left[\begin{array}{l} \because 1-\cos 2 \theta=2 \sin ^{2} \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$
$\begin{aligned} &=-4 \int\left(\frac{1-\cos \theta}{2}\right) \cos \theta d \theta \\ &=-2 \int\left(\cos \theta-\cos ^{2} \theta\right) d \theta \\ &=-2 \int \cos \theta d \theta+2 \int \cos ^{2} \theta d \theta \end{aligned}$
$\begin{aligned} &=-2 \int \cos \theta d \theta+2 \int \frac{1+\cos 2 \theta}{2} d \theta \\ &=-2 \int \cos \theta d \theta+\int 1 d \theta+\int \cos 2 \theta d \theta \end{aligned}$
$\begin{aligned} &=-2 \sin \theta+\theta+\frac{\sin 2 \theta}{\theta}+C \\ &=-2 \sin \theta+\theta+\frac{2 \sin \theta \cos \theta}{2}+C \\ &=-2 \sin \theta+\theta+\sin \theta \cos \theta+C \\ &=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}+C \quad\left[\because \cos \theta=\sqrt{1-\sin ^{2} \theta}\right] \\ &=-2 \sqrt{1-x}+\cos ^{-1} x+\sqrt{x(1-x)}+C \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 128

Answer: $-2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C$
Given: $\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$
Hint: using partial fraction
Explanation: let $I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$
$\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}$
Multiplying by : $(x+1)^{2}(x+2)$ $x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}$

Putting x=-1

$x^{2}+x+1=A(x+1)(u+2)+B(x+2)+C(x+1)^{2}$

$\begin{aligned} &\text { putting } x=-1\\ &1-1+1=A(0)(1)+B(1)+C(0)\\ &=0+B+0=>B=1 \end{aligned}$

$\begin{aligned} &\text { putting } x=-2 \\ &4+(-2)+1=A(-1)(0)+B(0)+C(1) \\ &3=0+0+C=>3 \end{aligned}$

$\begin{aligned} &\text { putting } x=0\\ &0+0+1=A(1)(2)+B(2)+C(1)\\ &1=2 A+2 B+C\\ &1=2 A+2(1)+3\\ &2 A=-4 \Rightarrow A=-2 \end{aligned}$

$\frac{x^{2}+x+1}{(x+1)^{2}+x+2}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{x+2}$

$\int \frac{x^{2}+x+1}{(x+1)^{2}+x+2}=-2 \int \frac{1}{x+1} d x+1 \int \frac{1}{(x+1)^{2}} d x+3 \int \frac{d x}{x+2}$

$=-2 \log |x+1|+1 \int(x+1)^{-2} d x+3 \log |x+2|+C$

$=-2 \log |x+1|+1\left[\frac{(x+1)^{2+1}}{-2+1}\right]+3 \log |x+2|+C$

$=-2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C$

Indefinite Integrals Exercise Revision Exercise Question 129

Answer: $\frac{1}{2} \cot 2 x \cdot e^{2 x}+C$
Hint:Using $\int e^{x}\left(f(x)-f^{\prime}(x)\right) d x$
Given: $\int \frac{\sin 4 x-2}{1-\cos 4 x} \cdot e^{2 x} d x$
Explanation:
Let
$\begin{aligned} &I=\int \frac{\sin 4 x-2}{1-\cos 4 x} \cdot e^{2 x} d x \\ &=\frac{1}{2} \int\left(\frac{\sin 2 t-2}{1-\cos 2 t}\right) e^{t} d t \\ &=\frac{1}{2} \int e^{t}\left(\frac{\sin t \cos t-1}{\sin ^{2} t}\right) d t \end{aligned}$ $\left[\begin{array}{l} \text { put } 2 x=t \\ 2 d x=d t \\ d x=\frac{1}{2} d t \end{array}\right]$
$\begin{aligned} &=\frac{1}{2} \int e^{t}\left(\frac{\cos t}{\sin t}-\frac{1}{\sin ^{2} t}\right) d t \\ &=\frac{1}{2} \int e^{t}\left(\cot t-\cos e c^{2} t\right) d t \\ &=\frac{1}{2} \int e^{t} \cot d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t \\ &=\frac{1}{2}\left[\cot \cdot e^{t}-\int\left(-\cos e c^{2} t\right) \cdot e^{t} d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t\right] \\ &=\frac{1}{2} \cot t \cdot e^{t}+\frac{1}{2} \int \cos e c^{2} t \cdot e^{t} d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t \\ &=\frac{1}{2} \cot t \cdot e^{t}+C \\ &=\frac{1}{2} \cot 2 x e^{2 x}+C \end{aligned}$

Indefinite Integrals Exercise Revision Exercise Question 130

Answer:
$\frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C$
Given: $\int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x$
Hint: using partial fraction and $\int \frac{1}{x} d x, \int \frac{1}{1+x^{2}} d x$
Explanation:
$\begin{aligned} &I=\int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x\left(1+\cot ^{2} x\right)}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x \cos e c^{2} x}{1+\cot ^{3} x} d x \\ &=-\int \frac{t}{1+t^{3}} d t \end{aligned}$ $\left[\begin{array}{l} p u t \cot x=t \\ -\cos e c^{2} x=d t \\ \cos e c^{2} x d x=-d t \end{array}\right]$
$=-\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}$
$\begin{aligned} &\text { now: } \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{A}{1+t}+\frac{B}{t^{2}-t+1} \\ &\text { multiplying by }(1+t)\left(t^{2}-t+1\right) \\ &t=A\left(t^{2}-t+1\right)+(B t+C)(t+1) \end{aligned}$
$\begin{aligned} &\text { putting\: t }=-1 \\ &-1=A(1+1+1)+(B(-1)+C)(0) \\ &=3 A \Rightarrow A=-\frac{1}{3} \end{aligned}$
$\begin{aligned} &\text { putting } t=0 \\ &0=A(0-0+1)+(B(0)+C)(0+1) \\ &0=A(1)+C(1) \\ &0=A+C \Rightarrow 0=\frac{-1}{3}+C \Rightarrow C=\frac{1}{3} \end{aligned}$
$\begin{aligned} &\text { putting } t=+1 \\ &1=A(1-1+1)+(B+C)(2) \\ &1=A(1)+2 B+2 C \\ &1=\frac{-1}{3}+2 B+\frac{2}{3} \end{aligned}$
$\begin{aligned} &1+\frac{1}{3}-\frac{2}{3}=2 B \\ &2 B=\frac{3+1-2}{3} \\ &2 B=\frac{2}{3} \Rightarrow B=\frac{1}{3} \end{aligned}$
$\frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{\frac{-1}{3}}{t+1}+\frac{\frac{1}{3}(t+1)}{t^{2}-t+1}$
$\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3} \int \frac{t+1}{t^{2}-t+1} d t$
$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3 \times 2} \int \frac{2 t+2-3+3}{t^{2}-t+1} d t$
$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{t^{2}-t+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}$
$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C$
$=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C$
$=\frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C$

Revising the proper set of questions is essential. And class 12, mathematics chapter 18 consists of around 32 exercises. This brings to the point that students might forget what they have learned in the first exercise when they reach the 32nd one. Therefore, it becomes essential to revise the previous portion before jumping to the next chapter.

The RE portion consists of 130 questions with subparts to be answered. These questions come under the concept of evaluating the integrals, Solutions on indefinite integral, Integration by parts, Questions on Fundamental integration formulas, Integration of trigonometric functions.

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RD Sharma Chapter wise Solutions

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