RD Sharma Class 12 Exercise 18.9 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.9 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:58 AM IST

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RD Sharma Class 12th Exercise 18.9 deals with the chapter Indefinite Integrals. This exercise contains 72 questions that are divided into Level 1 and Level 2 based on their complexity. It consists of 62 Level 1 questions and 10 Level 2 questions. RD Sharma solutions The Level 1 questions cover concepts like the integration of trigonometric and logarithmic equations, and Level 2 questions contain more complex functions that can be solved using the given theorems.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.9
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.9

Indefinite Integrals exercise 18.9 question 1

Solution: We have ,
$I=\int \frac{\log (\log x)}{x}dx$
Put $\log x=t\Rightarrow \frac{1}{x}dx=dt$
$\\ \\ I=\int \log t dt=t[\log t-1]+c\\ \\ I=\log x[\log (\log x)-1]+c$

Indefinite Integrals exercise 18 .9 question 2

Answer: $-\frac{1}{2}\left[\log \left(1+\frac{1}{x}\right)\right]^{2}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\log \left(1+\frac{1}{x}\right)}{x(1+x)} d x$
Solution:
Let $I=\int \frac{\log \left(1+\frac{1}{x}\right)}{x(1+x)} d x$
Put $\log \left(1+\frac{1}{x}\right)=t \Rightarrow \frac{1}{1+\frac{1}{x}}\left(\frac{-1}{x^{2}}\right) d x=d t$
$\begin{aligned} &\Rightarrow \frac{1}{\frac{x+1}{x}}\left(\frac{-1}{x^{2}}\right) d x=d t \\ &\Rightarrow \frac{x}{x+1}\left(\frac{-1}{x^{2}}\right) d x=d t \\ &\Rightarrow \frac{-1}{x(x+1)} d x=d t \Rightarrow d x=-(x+1) x \; d t \text { then } \end{aligned}$
$I=\int-\frac{t}{x(1+x)}(x+1) x \; d t=-\int t d t \quad\left[\because \frac{1}{x} d x=d t \Rightarrow d x=x\; d t\right]$
$\begin{aligned} &=-\frac{t^{1+1}}{1+1}+c \\ &=\frac{-t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{-1}{2}\left\{\log \left(1+\frac{1}{x}\right)\right\}^{2}+c \quad\left[\because t=\log \left(1+\frac{1}{x}\right)\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 3

Answer: $\frac{2}{3}(1+\sqrt{x})^{3}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{(1+\sqrt{x})^{2}}{\sqrt{x}} d x$
Solution:
Let $I=\int \frac{(1+\sqrt{x})^{2}}{\sqrt{x}} d x$
Put $1+\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t$
$\begin{aligned} &\Rightarrow d x=2 \sqrt{x} d t \text { then } \\ &I=\int \frac{t^{2}}{\sqrt{x}} 2 \sqrt{x} d t=\int 2 t^{2} d t=2 \int t^{2} d t \end{aligned}$
$\begin{aligned} &=2 \frac{t^{2+1}}{2+1}+c=2 \frac{t^{3}}{3}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{2}{3}(1+\sqrt{x})^{3}+c \quad[\because t=1+\sqrt{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 4

Answer: $\frac{2}{3}\left(1+e^{x}\right)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \sqrt{1+e^{x}} \cdot e^{x} d x$
Solution:
Let $I=\int \sqrt{1+e^{x}} \cdot e^{x} d x$
Put $1+e^{x}=t \Rightarrow e^{x} d x=d t \text { then }$
$\begin{aligned} I &=\int \sqrt{t} d t=\int t^{\frac{1}{2}} d t \\ &=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{2}{3}\left(1+e^{x}\right)^{\frac{3}{2}}+c \quad\left[\because t=1+e^{x}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 5

Answer: $-\frac{3}{5}(\cos x)^{\frac{5}{3}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \sqrt[3]{\cos ^{2} x} \cdot \sin x\; d x$
Solution:
Let $I=\int \sqrt[3]{\cos ^{2} x} \cdot \sin x \; d x$
$\begin{aligned} &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ &\Rightarrow \sin x d x=-d t \text { then } \\ &I=\int \sqrt[3]{t^{2}}(-d t)=-\int t^{\frac{2}{3}} d t \end{aligned}$
$\begin{aligned} &=-\frac{t^{\frac{2}{3}+1}}{\frac{2}{3}+1}+c=-\frac{t^{\frac{5}{3}}}{\frac{5}{3}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-\frac{3}{5}(\cos x)^{\frac{5}{3}}+c \quad[\because t=\cos x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 6

Answer: $\frac{-1}{\left(1+e^{x}\right)}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{e^{x}}{\left(1+e^{x}\right)^{2}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{e^{x}}{\left(1+e^{x}\right)^{2}} d x \\ &\text { Put } 1+e^{x}=t \Rightarrow e^{x} d x=d t \text { then } \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{t^{2}} d t=\int t^{-2} d t \\ &=-\frac{t^{-2+1}}{-2+1}+c=\frac{t^{-1}}{-1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$\begin{aligned} &=-\frac{1}{t}+c \\ &=\frac{-1}{\left(1+e^{x}\right)}+c \quad\left[\because t=1+e^{x}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 7

Answer: $\frac{-1}{4} \cot ^{4} x+c$
Hint: Use substitution method to solve this integral.
Given: $\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x \; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x \; d x \\ &\text { Put } \cot \mathrm{x}=t \Rightarrow-\operatorname{cosec}^{2} x\; d x=d t \end{aligned}$
$\begin{aligned} &\Rightarrow \operatorname{cosec}^{2} x \; d x=-d t \text { then } \\ &I=\int t^{3}(-d t)=-\int t^{3} d t \end{aligned}$
$\begin{aligned} &=-\frac{t^{3+1}}{3+1}+c=-\frac{t^{4}}{4}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{-1}{4} \cot ^{4} x+c\; \; \; \; \; [\because t=\cot x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 8

Answer: $\frac{e^{2 \sin ^{-1} x}}{2}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\left\{e^{\sin ^{-1} x}\right\}^{2}}{\sqrt{1-x^{2}}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\left\{e^{\sin ^{-1} x}\right\}^{2}}{\sqrt{1-x^{2}}} d x \\ &\text { Put } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow d x=\sqrt{1-x^{2}} d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{\left(e^{t}\right)^{2}}{\sqrt{1-x^{2}}} \sqrt{1-x^{2}} d t=\int\left(e^{t}\right)^{2} d t \\ &=\int e^{2 t} d t=\frac{e^{2 t}}{2}+c \quad\left[\because \int e^{a x} d x=\frac{e^{a x}}{a}+c\right] \end{aligned}$
$=\frac{e^{2 \sin ^{-1} x}}{2}+c\; \; \; \left[\because t=\sin ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 9

Answer: $2 \sqrt{x-\cos x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{1+\sin x}{\sqrt{x-\cos x}} d x$
Solution:
$\text { Let } I=\int \frac{1+\sin x}{\sqrt{x-\cos x}} d x$
$\begin{aligned} &\text { Put } x-\cos x=t \Rightarrow(1+\sin x) d x=d t \\ &\Rightarrow(1+\sin x) d x=d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{1}{\sqrt{t}} d t=\int t^{\frac{-1}{2}} d t \\ &=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$=2 \sqrt{t}+c$
$=2 \sqrt{x-\cos x}+c\; \; \; \; \; [\because t=x-\cos x]$

Indefinite Integrals exercise 18.9 question 10

Answer: $\frac{-1}{\sin ^{-1} x}+C$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{1}{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)^{2}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1}{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)^{2}} d x \\ &\text { Put } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow d x=\sqrt{1-x^{2}} d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{1}{\sqrt{1-x^{2}} \cdot t^{2}} \sqrt{1-x^{2}} d t=\int \frac{1}{t^{2}} d t \\ &=\frac{t^{-2+1}}{-2+1}+c=\frac{t^{-1}}{-1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$=\frac{-1}{t}+c=\frac{-1}{\sin ^{-1} x}+c\; \; \; \left[\because t=\sin ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 11

Answer: $\frac{-2}{\sqrt{\sin x}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\cot x}{\sqrt{\sin x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\cot x}{\sqrt{\sin x}} d x \\ &\text { Put } \sin x=t \Rightarrow \cos x d x=d t \\ &\Rightarrow d x=\frac{d t}{\cos x} \text { then } \end{aligned}$
$I=\int \frac{\cot x}{\sqrt{t}} \frac{d t}{\cos x}=\int \frac{\cos x}{\sin x} \cdot \frac{1}{\sqrt{t}} \cdot \frac{d t}{\cos x} d t \quad\left[\because \cot x=\frac{\cos x}{\sin x}\right]$
$\begin{aligned} &=\int \frac{1}{t . t^{\frac{1}{2}}} d t \quad[\because t=\sin x] \\ &=\int \frac{1}{t^{1+\frac{1}{2}}} d t=\int \frac{1}{t^{\frac{3}{2}}} \mathrm{dt} \end{aligned}$
$I=\int t^{\frac{-3}{2}} d t$
$\begin{aligned} &=\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+c=\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-2 \frac{1}{\sqrt{t}}+c=\frac{-2}{\sqrt{\sin x}}+c\; \; \; \; [\because t=\sin x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 12

Answer: $\frac{2}{\sqrt{\cos x}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\tan x}{\sqrt{\cos x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\tan x}{\sqrt{\cos x}} d x \\ &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ &\Rightarrow d x=\frac{-d t}{\sin x} \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{\sin x}{t \sqrt{t}} \frac{-d t}{\sin x} \\ &=-\int \frac{1}{t^{1+\frac{1}{2}}} d t=-\int \frac{1}{t^{\frac{3}{2}}} d t \\ I &=-\int t^{\frac{-3}{2}} d t \end{aligned}$
$\begin{aligned} &=-\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+c=-\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-(-2) \frac{1}{\sqrt{t}}+c=\frac{2}{\sqrt{t}}+c \\ &=\frac{2}{\sqrt{\cos x}}+c\; \; \; \; \; \; [\because t=\cos x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 13

Answer: $2 \sqrt{\sin x}-\frac{2}{5}(\sin x)^{\frac{5}{2}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x=\int \frac{\cos ^{2} x \cdot \cos x}{\sqrt{\sin x}} d x \\ &=\int \frac{\left(1-\sin ^{2} x\right) \cdot \cos x}{\sqrt{\sin x}} d x \quad\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow \cos ^{2} x=1-\sin ^{2} x \end{array}\right] \\ &\operatorname{Put} \sin x=t \Rightarrow \cos x d x=d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{\left(1-t^{2}\right)}{\sqrt{t}} d t=\int\left\{\frac{1}{\sqrt{t}}-\frac{t^{2}}{\sqrt{t}}\right\} d t \\ &=\int t^{\frac{-1}{2}} d t-\int t^{2-\frac{1}{2}} d t=\int t^{\frac{-1}{2}} d t-\int t^{\frac{4-1}{2}} d t \\ &=\int t^{\frac{-1}{2}} d t-\int t^{\frac{3}{2}} d t \end{aligned}$
$=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}-\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\mathrm{c}$
$\begin{aligned} &=2 t^{\frac{1}{2}}-\frac{2}{5} t^{\frac{5}{2}}+c \\ &=2 \sqrt{\sin x}-\frac{2}{5}(\sin x)^{\frac{5}{2}}+c\; \; \; \; [\because t=\sin x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 14

Answer: $-2 \sqrt{\cos x}+\frac{2}{5}(\cos x)^{\frac{5}{2}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\sin ^{3} x}{\sqrt{\cos x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\sin ^{3} x}{\sqrt{\cos x}} d x=\int \frac{\sin ^{2} x \cdot \sin x}{\sqrt{\cos x}} d x \\ &=\int \frac{\left(1-\cos ^{2} x\right) \cdot \sin x}{\sqrt{\cos x}} d x \quad\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow \sin ^{2} x=1-\cos ^{2} x \end{array}\right] \\ &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{\left(1-t^{2}\right)}{\sqrt{t}}(-d t)=-\int\left\{\frac{1-t^{2}}{\sqrt{t}}\right\} d t \\ &=-\int\left\{t^{\frac{-1}{2}}-t^{2-\frac{1}{2}}\right\} d t=-\int\left\{t^{\frac{-1}{2}}-t^{\frac{4-1}{2}}\right\} d t \\ &=-\int\left\{t^{\frac{-1}{2}}-t^{\frac{3}{2}}\right\} d t \end{aligned}$
$=-\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\mathrm{c}$
$\begin{aligned} &=-2 t^{\frac{1}{2}}+\frac{2}{5} t^{\frac{5}{2}}+c \\ &=-2 \sqrt{\cos x}+\frac{2}{5}(\cos x)^{\frac{5}{2}}+c\; \; \; \; \; [\because t=\cos x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 15

Answer: $2 \sqrt{\tan ^{-1} x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{1}{\sqrt{\tan ^{-1} x}\left(1+x^{2}\right)} d x$
Solution:
$\text { Let } I=\int \frac{1}{\sqrt{\tan ^{-1} x}\left(1+x^{2}\right)} d x$
$\text { Put } \tan ^{-1} x=t \Rightarrow \frac{1}{1+x^{2}} d x=d t \Rightarrow d x=\left(1+x^{2}\right) d t \text { then }$
$\begin{aligned} I &=\int \frac{1}{\sqrt{t}\left(1+x^{2}\right)}\left(1+x^{2}\right) d t=\int \frac{1}{\sqrt{t}} d t \\ &=\int t^{\frac{-1}{2}} d t=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{c}=2 \sqrt{t}+c \\ &=2 \sqrt{\tan ^{-1} x}+c\left[\because t=\tan ^{-1} x\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 16

Answer: $2 \sqrt{\tan x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$
Solution:
$\begin{aligned} &I=\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x \\ &=\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos x \cdot \cos x} d x \end{aligned}$
$=\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos ^{2} x} d x=\int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x}} \cdot \frac{1}{\cos ^{2} x} d x$
$\begin{aligned} &=\int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec ^{2} x\; d x=\int(\tan x)^{\frac{1}{2}-1} \cdot \sec ^{2} x\; d x \\ &=\int(\tan x)^{\frac{-1}{2}} \sec ^{2} x \; d x \end{aligned}$
$\text { Put } \tan x=t \Rightarrow \sec ^{2} x d x=d t \text { then }$
$I=\int t^{\frac{-1}{2}} d t=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{c}=2 \sqrt{t}+c \\ &=2 \sqrt{\tan x}+c\; \; \; \; [\because t=\tan x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 17

Answer: $\frac{1}{3}(\log x)^{3}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{1}{x}(\log x)^{2} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1}{x}(\log x)^{2} d x \\ &\text { Put } \log x=t \Rightarrow \frac{1}{x} d x=d t \\ &\Rightarrow d x=x \; d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{1}{x} t^{2} \cdot x d t=\int t^{2} d t \\ &=\frac{t^{2+1}}{2+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{3}}{3}+c \\ &=\frac{1}{3}(\log x)^{3}+c \quad[\because t=\log x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 18

Answer: $\frac{1}{6} \sin ^{6} x+c$
Hint: Use substitution method to solve this integral.
Given: $\int \sin ^{5} x \cos x\; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \sin ^{5} x \cos x d x \\ &\text { Put } \sin x=t \Rightarrow \cos x d x=d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int t^{5} d t \\ &=\frac{t^{5+1}}{5+1}+\mathrm{c} \quad \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{6}}{6}+c \\ &=\frac{1}{6} \sin ^{6} x+c \quad \quad[\because t=\sin x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 21

Answer: $\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int(4 x+2) \sqrt{x^{2}+x+1} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int(4 x+2) \sqrt{x^{2}+x+1} d x \\ &\Rightarrow I=2 \int(2 x+1) \sqrt{x^{2}+x+1} d x \\ &\text { Put } x^{2}+x+1=t \Rightarrow(2 x+1) d x=d t \text { then } \end{aligned}$
$I=2 \int(2 x+1) \sqrt{t} \cdot \frac{d t}{(2 x+1)}=2 \int \sqrt{t} d t$
$=2 \int t^{\frac{1}{2}} d t=2\left[\frac{t_{2}^{\frac{1}{+1}}}{\frac{1}{2}+1}\right]+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=2\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]+\mathrm{C}=2 \cdot \frac{2}{3} t^{\frac{3}{2}}+c$
$=\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+c \quad\left[\because t=x^{2}+x+1\right]$

Indefinite Integrals exercise 18.9 question 22

Answer: $2 \sqrt{2 x^{2}+3 x+1}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{(4 x+3)}{\sqrt{2 x^{2}+3 x+1}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{(4 x+3)}{\sqrt{2 x^{2}+3 x+1}} d x \\ &\text { Put } 2 x^{2}+3 x+1=t \Rightarrow(4 x+3) d x=d t \text { then } \\ &I=\int \frac{1}{\sqrt{t}}=\int t^{\frac{-1}{2}} d t \end{aligned}$
$=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c=2 \sqrt{t}+c \\ &=2 \sqrt{2 x^{2}+3 x+1}+c \quad \quad\left[\because t=2 x^{2}+3 x+1\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 23

Answer: $2 \sqrt{x}-2 \log |\sqrt{x}+1|+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{1}{1+\sqrt{x}} d x$
Solution:
$\begin{aligned} &\text { Let }\; I=\int \frac{1}{1+\sqrt{x}} d x \\ &\text { Put } x=t^{2} \Rightarrow d x=2 t \; d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{1}{1+\sqrt{t^{2}}} 2 t d t=\int \frac{2 t}{1+t} d t \\ &=2 \int \frac{t}{1+t} d t=2 \int \frac{1+t-1}{1+t} d t \end{aligned}$
$\begin{aligned} &=2 \int \frac{(1+t)-1}{1+t} d t=2 \int\left\{\frac{1+t}{1+t}-\frac{1}{1+t}\right\} d t \\ &=2 \int\left\{1-\frac{1}{1+t}\right\} d t=2 \int 1 . d t-2 \int \frac{1}{1+t} d t \end{aligned}$
$=2 \int 1 . d t-2 \int \frac{1}{1+t} d t$ ...... $(i)$
$\text { Now } 2 \int 1 . d t=2 \frac{t^{0+1}}{0+1}+c_{1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=2 t+c_{1}$ ........ $(ii)$
$\begin{aligned} &\text { and } 2 \int \frac{1}{1+t} d t \\ &\text { Put } 1+t=p \Rightarrow d t=d p \text { then } \end{aligned}$
$2 \int \frac{1}{1+t} d t=2 \int \frac{1}{p} d p=2 \log |p|+c_{2}$
$=2 \log |t+1|+c_{2}$ ......... $(iii)$
Putting the values of equation (ii) and (iii) in (i) then
$\begin{aligned} &I=2 t+c_{1}-\left(2 \log |t+1|+c_{2}\right) \\ &\Rightarrow I=2 t+c_{1}-2 \log |t+1|-c_{2} \\ &\therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c_{1}-c_{2} \end{aligned}$
$\therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c \quad\left[\begin{array}{c} \because t^{2}=x \Rightarrow t=\sqrt{x} \\ \text { and } c=c_{1}-c_{2} \end{array}\right]$

Indefinite Integrals exercise 18.9 question 24

Answer: $-e^{\cos ^{2} x}+c$
Hint:Use substitution method to solve this integral.
Given: $\int e^{\cos ^{2} x} \sin 2 x\; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int e^{\cos ^{2} x} \sin 2 x d x \\ &\text { Put } \cos ^{2} x=t \Rightarrow 2 \cos x(-\sin x) d x=d t \end{aligned}$
$\begin{aligned} &\Rightarrow-(2 \cos x \sin x) d x=d t\\ &\Rightarrow-\sin 2 x\; d x=d t \end{aligned}$ $[\because \sin 2 x=2 \sin x \; \cos x]$
$\begin{aligned} &\text { Then }\\ &I=\int e^{t}(-d t)=-\int e^{t} d t=-e^{t}+c \quad\left[\because \int e^{x} d x=e^{x}+c\right]\\ &=-e^{\cos ^{2} x}+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because t=\cos ^{2} x\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 25

Answer: $\frac{-1}{2(x+\sin x)^{2}}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{1+\cos x}{(x+\sin x)^{3}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1+\cos x}{(x+\sin x)^{2}} d x \\ &\text { Put } x+\sin x=t \Rightarrow(1+\cos x) d x=d t \text { then } \end{aligned}$
$I=\int \frac{1}{t^{3}} d t=\int t^{-3} d t=\frac{t^{-3+1}}{-3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{t^{-2}}{-2}+c=\frac{-1}{2} \frac{1}{t^{2}}+c$
$=\frac{-1}{2(x+\sin x)^{2}}+c \quad[\because t=x+\sin x]$

Indefinite Integrals exercise 18.9 question 26

Answer: $\frac{-1}{\sin x+\cos x}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{\cos x-\sin x}{1+\sin 2 x} d x$
Solution:
$\text { Let } I=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x$
$=\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+\sin 2 x} d x \quad\left[\because 1=\sin ^{2} x+\cos ^{2} x\right]$
$=\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} d x \quad[\because \sin 2 x=2 \sin x \cos x]$
$=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$
$\text { Put } \sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t \text { then }$
$I=\int \frac{(\cos x-\sin x)}{t^{2}} \frac{d t}{(\cos x-\sin x)}$
$=\int \frac{1}{t^{2}}=\int t^{-2} d t=\frac{t^{-2+1}}{-2+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c$
$=\frac{-1}{\sin x+\cos x}+c \quad[\because t=\sin x+\cos x]$

Indefinite Integrals exercise 18.9 question 27

Answer: $\frac{1}{2 b(a+b \cos 2 x)}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{\sin 2 x}{(a+b \cos 2 x)^{2}} d x$
Solution:
$\text { Let } I=\int \frac{\sin 2 x}{(a+b \cos 2 x)^{2}} d x$
$\begin{aligned} &\text { Put } a+b \cos 2 x=t \\ &\Rightarrow b(-\sin 2 x) \cdot 2 d x=d t \\ &\Rightarrow-2 b \sin 2 x \; d x=d t \\ &\Rightarrow \sin 2 x \; d x=\frac{d t}{-2 b} \end{aligned}$
$I=\int \frac{1}{t^{2}} \cdot \frac{d t}{-2 b}=\frac{-1}{2 b} \int \frac{1}{t^{2}} d t$
$=\frac{-1}{2 b} \int t^{-2} d t=\frac{-1}{2 b} \frac{t^{-2+1}}{-2+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{-1}{2 b} \cdot \frac{t^{-1}}{-1}+c=\frac{1}{2 b} \cdot \frac{1}{t}+c$
$=\frac{1}{2 b(a+b \cos 2 x)}+c \quad \quad[\because t=a+b \cos 2 x]$

Indefinite Integrals exercise 18.9 question 28

Answer: $(\log x)^{2}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{\log x^{2}}{x} d x$
Solution:
$\text { Let } I=\int \frac{\log x^{2}}{x} d x$
$=\int \frac{2 \log x}{x} d x\left[\because \log x^{m}=m \log x\right]$
$\operatorname{Put} \log x=t \Rightarrow \frac{1}{x} d x=d t, \text { then }$
$I=\int 2 \cdot \frac{t}{x} \cdot x \; d t=2 \int t\; d t=2 \int \frac{t^{1+1}}{1+1}+c$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=2 \frac{t^{1+1}}{2}+c=t^{2}+c=(\log x)^{2}+c$ $[\because t=\log x]$

Indefinite Integrals exercise 18.9 question 29

Answer: $\frac{1}{1+\cos x}+C$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{\sin x}{(1+\cos x)^{2}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\sin x}{(1+\cos x)^{2}} d x \\ &\operatorname{Put} 1+\cos x=t \Rightarrow-\sin x d x=d t, \text { then } \end{aligned}$
$I=\int \frac{1}{t^{2}}(-d t)=-\int \frac{1}{t^{2}} d t=-\int t^{-2} d t$
$=-\left[\frac{t^{-2+1}}{-2+1}\right]+c=-\frac{-t^{-1}}{-1}+c$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{1}{t}+c=\frac{1}{1+\cos x}+c$ $[\because t=1+\cos x]$

Indefinite Integrals exercise 18.9 question 30

Answer: $\frac{1}{2}\{\log (\sin x)\}^{2}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \cot x \cdot \log (\sin x) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \cot x \cdot \log (\sin x) d x \\ &\text { Put } \log (\sin x)=t \Rightarrow \frac{1}{\sin x} \cos x\; d x=d t \\ &\Rightarrow \cot x\; d x=d t \Rightarrow d x=\frac{d t}{\cot x} \text { then } \end{aligned}$
$\Rightarrow I=\int \cot x \cdot t \cdot \frac{d t}{\cot x}=\int t \; d t$
$=\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]$
$=\frac{1}{2}\{\log (\sin x)\}^{2}+c \quad[\because t=\log (\sin x)]$

Indefinite Integrals exercise 18.9 question 31

Answer: $\frac{1}{2}\{\log (\sec x+\tan x)\}^{2}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \sec x \cdot \log (\sec x+\tan x) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \sec x \cdot \log (\sec x+\tan x) d x \\ &\text { Put } \log (\sec x+\tan x)=t \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{(\sec x+\tan x)}\left(\sec x \tan x+\sec ^{2} x\right) d x=d t \\ &\Rightarrow \frac{1}{(\sec x+\tan x)} \sec x(\tan x+\sec x) d x=d t \end{aligned}$
$\begin{aligned} &\Rightarrow \sec x\; d x=d t \Rightarrow d x=\frac{d t}{\sec x} \text { then } \\ &\Rightarrow I=\int \sec x \cdot t \cdot \frac{d t}{\sec x}=\int t \; d t \end{aligned}$
$=\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]$
$=\frac{1}{2}\{\log (\sec x+\tan x)\}^{2}+c \quad[\because t=\log (\sec x+\tan x)]$

Indefinite Integrals exercise 18.9 question 32

Answer: $\frac{1}{2}\{\log (\operatorname{cosec} x-\cot x)\}^{2}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x \\ &\text { Put } \log (\operatorname{cosec} x-\cot x)=t \end{aligned}$
$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left(-\operatorname{cosec} x \cdot \cot x-\left(-\operatorname{cosec}^{2} x\right)\right) d x=d t$
$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left\{\operatorname{cosec}^{2} x-\operatorname{cosec} x \cdot \cot x\right\} d x=d t$
$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)} \operatorname{cosec} x\{\operatorname{cosec} x-\cot x\} d x=d t$
$\Rightarrow \operatorname{cosec} x d x=d t \Rightarrow d x=\frac{d t}{\operatorname{cosec} x} \text { then }$
$\Rightarrow I=\int \operatorname{cosec} x \cdot t \cdot \frac{d t}{\operatorname{cosec} x}=\int t \; d t$
$=\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]$
$=\frac{1}{2}\{\log (\operatorname{cosec} x-\cot x)\}^{2}+c \quad \quad[\because t=\log (\operatorname{cosec} x-\cot x)]$

Indefinite Integrals exercise 18.9 question 33

Answer: $\frac{1}{4} \sin \left(x^{4}\right)+c$
Hint:Use substitution method to solve this integral.
Given: $\int x^{3} \cos x^{4} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int x^{3} \cos x^{4} d x \\ &\text { Put } x^{4}=t \Rightarrow 4 x^{3} d x=d t \Rightarrow d x=\frac{d t}{4 x^{3}} \text { then } \end{aligned}$
$\Rightarrow I=\int x^{3} \cos t \frac{d t}{4 x^{3}}=\frac{1}{4} \int \cos t \; d t$
$=\frac{1}{4} \int \sin t \; d t \quad\left[\because \int \cos x \; d x=\sin x+c\right]$
$=\frac{1}{4} \sin \left(x^{4}\right)+c \quad\left[\because t=x^{4}\right]$

Indefinite Integrals exercise 18.9 question 34

Answer: $\frac{3}{2}\left(x^{2}-1\right)^{\frac{2}{3}}+c$
Hint:Use substitution method to solve this integral.
Given: $\int \frac{2 x}{\sqrt[3]{x^{2}-1}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{2 x}{\sqrt[3]{x^{2}-1}} d x \\ &\text { Put } x^{2}-1=t \Rightarrow 2 x d x=d t \text { then } \end{aligned}$
$\Rightarrow I=\int \frac{1}{\sqrt[3]{t}} d t=\int \frac{1}{t^{\frac{1}{3}}} d t=\int t^{-\frac{1}{3}} d t$
$=\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c=\frac{t^{\frac{2}{3}}}{\frac{2}{3}}+c$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{3}{2} t^{\frac{2}{3}} \\ &=\frac{3}{2}\left(x^{2}-1\right)^{\frac{2}{3}}+c \quad\left[\because t=x^{2}-1\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 36

Answer: $-\frac{\cos \left(x^{4}+1\right)}{4}+c$
Hint: Use substitution method to solve this integral.
Given: $\int x^{3} \sin \left(x^{4}+1\right) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int x^{3} \sin \left(x^{4}+1\right) d x \\ &\text { Put } x^{4}+1=t \Rightarrow 4 x^{3} d x=d t \\ &\Rightarrow d x=\frac{d t}{4 x^{3}} \text { then } \end{aligned}$
$\begin{aligned} \Rightarrow I &=\int x^{3} \sin t \frac{d t}{4 x^{3}}=\frac{1}{4} \int \sin t\; d t \\ &=-\frac{\cos t}{4}+c \end{aligned}$ $\left[\because \int \sin x\; d x=-\cos x+c\right]$
$=-\frac{\cos \left(x^{4}+1\right)}{4}+c \quad\left[\because t=x^{4}+1\right]$

Indefinite Integrals exercise 18.9 question 37

Answer: $\tan \left(x e^{x}\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{(x+1) e^{x}}{\cos ^{2}\left(x e^{x}\right)} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{(x+1) e^{x}}{\cos ^{2}\left(x e^{x}\right)} d x \\ &\text { Put } x e^{x}=t \Rightarrow\left(x e^{x}+1 . e^{x}\right) d x=d t \\ &\Rightarrow(x+1) e^{x} d x=d t \quad \text { then } \end{aligned}$
$\Rightarrow I=\int \frac{1}{\cos ^{2} t} d t$
$=\int \sec ^{2} t d t \quad\left[\because \frac{1}{\cos x}=\sec x\right]$
$\begin{aligned} &=\tan t+c \\ &=\tan \left(x e^{x}\right)+c \quad\left[\because t=x e^{x}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 38

Answer: $\frac{1}{3} \sin \left(e^{x^{3}}\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int x^{2} e^{x^{3}} \cos \left(e^{x^{3}}\right) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int x^{2} e^{x^{3}} \cos \left(e^{x^{3}}\right) d x \\ &\text { Put } e^{x^{3}}=t \Rightarrow e^{x^{3}} \cdot 3 x^{2} d x=d t \\ &\Rightarrow d x=\frac{d t}{e^{x^{3}} \cdot 3 x^{2}} \text { then } \end{aligned}$
$\Rightarrow I=\int e^{x^{3}} \cdot x^{2} \cdot \cos t \frac{d t}{e^{x^{3}} .3 x^{2}}=\frac{1}{3} \int \cos t\; d t$
$=\frac{1}{3} \sin t \; d t \quad\left[\because \int \cos x \; d x=\sin x+c\right]$
$=\frac{1}{3} \sin \left(e^{x^{3}}\right)+c \quad\left[\because t=e^{x^{3}}\right]$

Indefinite Integrals exercise 18.9 question 39

Answer: $\frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x \\ &\text { Put } x^{2}+3=t \Rightarrow 2 x d x=d t \\ &\Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}$
$\begin{aligned} \Rightarrow I &=\int 2 x \sec ^{3} t \cdot \tan t \frac{d t}{2 x}=\int \sec ^{3}(t) \cdot \tan (t) d t \\ &=\int \sec ^{2}(t) \sec (t) \tan (t) d t \end{aligned}$
$\text { Again Put } \sec t=u \Rightarrow \sec t \tan t \; d t=d u \text { then }$
$I=\int u^{2} d u=\left[\frac{u^{2+1}}{2+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{array}{ll} =\frac{u^{3}}{3}+c=\frac{\sec ^{3} t}{3}+c & {[\because \sec t=u]} \end{array}$
$=\frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c \quad\left[\because t=x^{2}+3\right]$

Indefinite Integrals exercise 18.9 question 40

Answer: $\frac{1}{3}(\log x+x)^{3}+c$
Hint: Use substitution method to solve this integral.
Given: $\int\left(\frac{x+1}{x}\right)(\log x+x)^{2} d x$
Solution:
$\text { Let } I=\int\left(\frac{x+1}{x}\right)(\log x+x)^{2} d x$
$\begin{aligned} &=\int\left(\frac{x}{x}+\frac{1}{x}\right)(\log x+x)^{2} d x \\ &=\int\left(1+\frac{1}{x}\right)(\log x+x)^{2} d x \end{aligned}$
$\operatorname{Put} \log x+x=t \Rightarrow\left(\frac{1}{x}+1\right) d x=d t$
$\Rightarrow d x=\frac{1}{\left(1+\frac{1}{x}\right)} \text { dt then }$
$I=\int\left(1+\frac{1}{x}\right) t^{2} \frac{1}{\left(1+\frac{1}{x}\right)} d t=\int t^{2} d t$
$I=\left[\frac{t^{2+1}}{2+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{t^{3}}{3}+c=\frac{1}{3}(\log x+x)^{3}+c \quad[\because t=\log x+x]$

Indefinite Integrals exercise 18.9 question 41

Answer: $-\frac{1}{3}\left(1-\tan ^{2} x\right)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \tan x \cdot \sec ^{2} x \sqrt{1-\tan ^{2} x}\; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \tan x \cdot \sec ^{2} x \sqrt{1-\tan ^{2} x} d x \\ &\text { Put } \quad 1-\tan ^{2} x=t \Rightarrow-2 \tan x \cdot \sec ^{2} x d x=d t \\ &\Rightarrow d x=\frac{1}{-2 \tan x \cdot \sec ^{2} x} \text { dt then } \end{aligned}$
$\Rightarrow I=\int \tan x \cdot \sec ^{2} x \cdot \sqrt{t} \frac{1}{-2 \tan x \cdot \sec ^{2} x} \mathrm{dt}$
$=-\frac{1}{2} \int \sqrt{t} d t=-\frac{1}{2} \int t^{\frac{1}{2}} d t$
$=-\frac{1}{2}\left[\frac{t_{}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=-\frac{1}{2}\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]+c=-\frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}}+c$
$=-\frac{1}{3}\left(1-\tan ^{2} x\right)^{\frac{3}{2}}+c \quad\left[\because t=1-\tan ^{2} x\right]$

Indefinite Integrals exercise 18 .9 question 42

Answer: $-\frac{1}{2} \cos \left\{1+(\log x)^{2}\right\}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \log x \cdot \frac{\sin \left\{1+(\log x)^{2}\right\}}{x} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \log x \cdot \frac{\sin \left\{1+(\log x)^{2}\right\}}{x} d x \\ &\text { Put } 1+(\log x)^{2}=t \Rightarrow 2 \log x \cdot \frac{1}{x} \cdot d x=d t \\ &\Rightarrow d x=\frac{x \cdot d t}{2 \log x} \text { then } \end{aligned}$
$\begin{aligned} \Rightarrow I &=\int \log x \cdot \frac{\sin t}{x} \cdot \frac{x \; d t}{2 \log x} d t \\ &=\frac{1}{2} \int \sin t\; d t \end{aligned}$
$\begin{aligned} &=\frac{1}{2}(-\cos t)+c \\ &=-\frac{1}{2} \cos \left\{1+(\log x)^{2}\right\}+c \quad\left[\because t=1+(\log x)^{2}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 43

Answer: $-\frac{1}{2 x}-\frac{1}{4} \sin \left(\frac{2}{x}\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{1}{x^{2}} \cos ^{2}\left(\frac{1}{x}\right) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1}{x^{2}} \cos ^{2}\left(\frac{1}{x}\right) d x \\ &\text { Put } \frac{1}{x}=t \Rightarrow-\frac{1}{x^{2}} d x=d t \\ &\Rightarrow d x=-x^{2} \text { dt then } \end{aligned}$
$I=\int \frac{1}{x^{2}} \cdot \cos ^{2} t \cdot\left(-x^{2}\right) d t=-\int \cos ^{2} t \; d t$
$=-\int\left\{\frac{1+\cos 2 t}{2}\right\} d t$ $\left[\begin{array}{l} \because 2 \cos ^{2} A-1=\cos 2 A \\ \Rightarrow 2 \cos ^{2} A=1+\cos 2 A \\ \Rightarrow \cos ^{2} A=\frac{1+\cos 2 A}{2} \end{array}\right]$
$\begin{aligned} &=-\int\left\{\frac{1}{2}+\frac{\cos 2 t}{2}\right\} d t=-\int \frac{1}{2} d t-\frac{1}{2} \int \cos 2 t d t \\ &=-\frac{1}{2} \int 1 . d t-\frac{1}{2} \int \cos 2 t d t=-\frac{1}{2} \int t^{0} d t-\frac{1}{2} \int \cos 2 t d t \end{aligned}$
$=-\frac{1}{2} t-\frac{1}{4} \sin 2\; t$ $\left[\begin{array}{c} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \cos a\; x \; d x=\frac{\sin a x}{a}+c \end{array}\right]$
$=-\frac{1}{2 x}-\frac{1}{4} \sin \left(\frac{2}{x}\right)+c$

Indefinite Integrals exercise 18.9 question 44

Answer: $\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+c$
Hint: Use substitution method to solve this integral.
Given: $\int \sec ^{4} x \tan x\; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \sec ^{4} x \tan x \; d x \\ &\text { Put } \tan x=t \Rightarrow \sec ^{2} x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{\sec ^{2} x} \text { then } \end{aligned}$
$I=\int \sec ^{4} x \cdot t \cdot \frac{d t}{\sec ^{2} x}=\int \sec ^{2} x \cdot t\; d t$
$=\int\left\{1+\tan ^{2} x\right\} . t \; d t \quad\left[\begin{array}{l} \because \sec ^{2} x-\tan ^{2} x=1 \\ \Rightarrow \sec ^{2} x=1+\tan ^{2} x \end{array}\right]$
$\begin{aligned} &=\int\left(1+t^{2}\right) . t\; d t \quad[\because \tan x=t] \\ &=\int\left(t+t^{2} t\right) d t=\int\left(t+t^{3}\right) d t \\ &=\int t \; d t+\int t^{3} d t \end{aligned}$
$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{t^{2}}{2}+\frac{t^{4}}{4}+c \\ &=\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+c \quad[\because t=\tan x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 45

Answer: $2 \sin \left(e^{\sqrt{x}}\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{e^{\sqrt{x}} \cos \left(e^{\sqrt{x}}\right)}{\sqrt{x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{e^{\sqrt{x}} \cos \left(e^{\sqrt{x}}\right)}{\sqrt{x}} d x \\ &\text { Put } e^{\sqrt{x}}=t \Rightarrow e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} d x=d t \\ &\Rightarrow d x=\frac{2 \sqrt{x}}{e^{\sqrt{x}}} \mathrm{dt} \text { then } \end{aligned}$
$\begin{aligned} &I=\int \frac{e^{\sqrt{x}} \cos t}{\sqrt{x}} \frac{2 \sqrt{x}}{e^{\sqrt{x}}} d t \\ &=2 \int \cos t \; d t \end{aligned}$
$=2 \sin t+c \quad\left[\because \int \cos x\; d x=\sin x+c\right]$
$=2 \sin \left(e^{\sqrt{x}}\right)+c \quad\left[\because t=e^{\sqrt{x}}\right]$

Indefinite Integrals exercise 18.9 question 46

Answer: $\frac{1}{4} \sin ^{4} x-\sin ^{2} x+\log |\sin x|+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\cos ^{5} x}{\sin x} d x$
Solution:
$\text { Let } I=\int \frac{\cos ^{5} x}{\sin x} d x$
$\begin{aligned} &\text { Put } \sin x=t \Rightarrow \cos x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{\cos x} \text { then } \end{aligned}$
$I=\int \frac{\cos ^{5} x}{t} \frac{d t}{\cos x}=\int \frac{\cos ^{4} x}{t} d t$
$=\int \frac{\left(\cos ^{2} x\right)^{2}}{t} d t=\int \frac{\left(1-\sin ^{2} x\right)^{2}}{t} d t$ $\left[\begin{array}{l} \because \cos ^{2} x+\sin ^{2} x=1 \\ \Rightarrow \cos ^{2} x=1-\sin ^{2} x \end{array}\right]$
$=\int \frac{\left(1-t^{2}\right)^{2}}{t} d t$ $[\because \sin x=t]$
$=\int\left\{\frac{1+\left(t^{2}\right)^{2}-2 t^{2}}{t}\right\} d t$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &=\int\left\{\frac{1+\left(t^{4}\right)-2 t^{2}}{t}\right\} d t \\ &=\int\left\{\frac{1}{t}+\frac{t^{4}}{t}-\frac{2 t^{2}}{t}\right\} d t \\ &=\int\left\{\frac{1}{t}+t^{3}-2 t\right\} d t \end{aligned}$
$=\int \frac{1}{t} d t+\int t^{3} d t-2 \int t\; d t$
$=\log |t|+\frac{t^{3+1}}{3+1}-2 \frac{t^{1+1}}{1+1}+c$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\log |t|+\frac{t^{4}}{4}-2 \frac{t^{2}}{2}+c \\ &=\frac{1}{4} \sin ^{4} x-\sin ^{2} x+\log |\sin x|+c \quad[\because t=\sin x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 47

Answer: $-2 \cos \sqrt{x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}}\; d x=d t \\ &\Rightarrow d x=2 \sqrt{x} \; d t \text { then } \end{aligned}$
$I=\int \frac{\sin t}{\sqrt{x}} 2 \sqrt{x} d t=2 \int \sin t\; d t$
$=2[-\cos t]+\mathrm{c} \quad\left[\because \int \sin x \; d x=-\cos x+c\right]$
$\begin{aligned} &=-2 \cos t+c \\ &=-2 \cos \sqrt{x}+c \quad[\because t=\sqrt{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 48

Answer: $-\cot \left(x e^{x}\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{(x+1) e^{x}}{\sin ^{2}\left(x e^{x}\right)} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{(x+1) e^{x}}{\sin ^{2}\left(x e^{x}\right)} d x \\ &\text { Put } x e^{x}=t \Rightarrow\left(x e^{x}+1 . e^{x}\right) d x=d t \\ &\Rightarrow e^{x}(x+1) d x=d t \quad \text { then } \end{aligned}$
$I=\int \frac{1}{\sin ^{2}(t)} d t=\int \operatorname{cosec}^{2} t \; d t \quad\left[\because \frac{1}{\sin x}=\operatorname{cosec\; x}\right]$
$=[-\cot t]+\mathrm{c} \quad\left[\because \int \operatorname{cosec}^{2} x \; d x=-\cot x+c\right]$
$=-\cot \left(x e^{x}\right)+c \quad\left[\because t=x e^{x}\right]$

Indefinite Integrals exercise 18.9 question 49

Answer: $\frac{5^{x+\tan ^{-1} x}}{\log 5}+C$
Hint: Use substitution method to solve this integral.
Given: $\int 5^{x+\tan ^{-1} x} \cdot\left(\frac{x^{2}+2}{1+x^{2}}\right) d x$
Solution:
$\begin{aligned} &\text { Let } I=\int 5^{x+\tan ^{-1} x} \cdot\left(\frac{x^{2}+2}{1+x^{2}}\right) d x \\ &\text { Put } x+\tan ^{-1} x=t \\ &\Rightarrow\left(1+\frac{1}{1+x^{2}}\right) d x=d t \Rightarrow\left(\frac{\left(1+x^{2}\right)+1}{1+x^{2}}\right) d x=d t \\ &\Rightarrow\left(\frac{1+x^{2}+1}{1+x^{2}}\right) d x=d t \Rightarrow\left(\frac{2+x^{2}}{1+x^{2}}\right) d x=d t \text { then } \end{aligned}$
$I=\int 5^{t} d t=\frac{5^{t}}{\log 5}+c \quad\left[\because \int a^{x} d x=\frac{a^{x}}{\log a}+c\right]$
$=\frac{5^{x+\tan ^{-1} x}}{\log 5}+c \quad\left[\because t=x+\tan ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 50

Answer: $\frac{1}{m} e^{m \sin ^{-1} x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{e^{m \sin ^{-1} x}}{\sqrt{1-x^{2}}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{e^{m \sin ^{-1} x}}{\sqrt{1-x^{2}}} d x \\ &\text { Put } m \sin ^{-1} x=t \Rightarrow m \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow \mathrm{d} \mathrm{x}=\frac{\sqrt{1-x^{2}}}{m} \mathrm{dt} \text { then } \end{aligned}$
$I=\int \frac{e^{t}}{\sqrt{1-x^{2}}} \cdot \frac{\sqrt{1-x^{2}}}{m} d t \Rightarrow \frac{1}{m} \int e^{t} d t$
$=\frac{1}{m} e^{t}+c \quad\left[\because \int e^{x} d x=e^{x}+c\right]$
$=\frac{1}{m} e^{m \sin ^{-1} x}+c \quad\left[\because t=m \sin ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 51

Answer: $2 \sin \sqrt{x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ &\Rightarrow d x=2 \sqrt{x}\; d t \text { then } \end{aligned}$
$\begin{aligned} I &=\int \frac{\cos t}{\sqrt{x}} \cdot 2 \sqrt{x} \; d t \Rightarrow 2 \int \cos t \; d t \\ &=2 \sin t+\mathrm{c} \quad\left[\because \int \cos x\; d x=\sin x+c\right] \\ &=2 \sin \sqrt{x}+c \quad[\because t=\sqrt{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 52

Answer: $-\cos \left(\tan ^{-1} x\right)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x$
Solution:
$\text { Let } I=\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x$
$\begin{aligned} \text { put } \tan ^{-1} x=t & \Rightarrow \frac{1}{1+x^{2}} d x=d t \\ & \Rightarrow d x=\left(1+x^{2}\right) d t \end{aligned}$
$I=\int \frac{\sin t}{1+x^{2}}\left(1+x^{2}\right) d t=\int \sin t \; d t$
$\begin{array}{ll} =-\cos t+c & {\left[\because \int \sin x \; d x=-\cos x+c\right]} \\ =-\cos \left(\tan ^{-1} x\right)+c & {\left[\because t=\tan ^{-1} x\right]} \end{array}$

Indefinite Integrals exercise 18.9 question 53

Answer: $-\cos (\log x)+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\sin (\log x)}{x} d x$
Solution:
$\text { Let } I=\int \frac{\sin (\log x)}{x} d x$
$\begin{aligned} &\text { Put } \log x=t \Rightarrow \frac{1}{x} d x=d t \Rightarrow d x=x \; d t \\ &\text { Then } I=\int \frac{\sin t}{x} \cdot x d t=\int \sin t \; d t \end{aligned}$
$\begin{array}{ll} =-\cos t+c & \ \because \sin x \; d x=-\cos x+c] \\ =-\cos (\log x)+c & {[\because t=\log x]} \end{array}$

Indefinite Integrals exercise 18 .9 question 54

Answer: $\frac{1}{m} e^{m \tan ^{-1} x}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{e^{m \tan ^{-1} x}}{1+x^{2}} d x$
Solution:
$\text { Let } I=\int \frac{e^{m \tan ^{-1} x}}{1+x^{2}} d x$
$\begin{aligned} &\text { Put } m \tan ^{-1} x=t \Rightarrow m \frac{1}{1+x^{2}} d x=d t \\ &\Rightarrow d x=\frac{1+x^{2}}{m} d t \end{aligned}$
$\begin{aligned} &I=\int \frac{e^{t}}{1+x^{2}} \cdot \frac{1}{m} \cdot\left(1+x^{2}\right) d t \\ &=\int \frac{e^{t}}{m} d t \end{aligned}$
$=\frac{1}{m} \int e^{t} d t=\frac{1}{m}\left(e^{t}\right)+c \quad\left[\because \int e^{x} d x=e^{x}+c\right]$
$=\frac{1}{m} e^{m \tan ^{-1} x}+c \quad\left[\because t=m \tan ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 55

Answer: $\frac{1}{6 a^{2}}\left[\left(x^{2}+a^{2}\right)^{\frac{3}{2}}-\left(x^{2}-a^{2}\right)^{\frac{3}{2}}\right]+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} d x$
Solution:
$\text { Let } I=\int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} d x$
On Rationalising we get
$I=\int\left(\frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}} \times \frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}}\right) d x$
$=\int\left(\frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}\right)\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}\right) d x$
$=\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(\sqrt{x^{2}+a^{2}}\right)^{2}-\left(\sqrt{x^{2}-a^{2}}\right)^{2}} d x$
$=\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{\left(x^{2}+a^{2}\right)-\left(x^{2}-a^{2}\right)} d x$
$=\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{x^{2}+a^{2}-x^{2}+a^{2}} d x$
$=\int \frac{x\left(\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}-a^{2}}\right)}{2 a^{2}} d x$
$=\frac{1}{2 a^{2}} \int\left(x \sqrt{x^{2}+a^{2}}-x \sqrt{x^{2}-a^{2}}\right) d x$
$=\frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x-\frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x$ .......... $(i)$
$\text { Now } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x$
$\begin{aligned} &\text { Put } x^{2}+a^{2}=t \Rightarrow 2 x \; d x=d t \Rightarrow d x=\frac{a \imath}{2 x} \\ &\text { Then, } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}+a^{2}} d x=\frac{1}{2 a^{2}} \int x \sqrt{t} \cdot \frac{d t}{2 x}=\frac{1}{4 a^{2}} \int t^{\frac{1}{2}} d t \end{aligned}$
$=\frac{1}{4 a^{2}} \cdot \frac{t_{2}^{\frac{1}{1}+1}}{\frac{1}{2}+1}+c_{1}=\frac{1}{4 a^{2}} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}}+c_{1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{1}{6 a^{2}}\left(x^{2}+a^{2}\right)^{\frac{3}{2}}+c_{1}$ ...... $(ii)$ $\left(\because t=x^{2}+a^{2}\right)$
$\text { and, } \frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x$
$\begin{aligned} &\text { Put, } x^{2}-a^{2}=u \Rightarrow 2 x d x=d u \Rightarrow d x=\frac{d u}{2 x} \\ &\frac{1}{2 a^{2}} \int x \sqrt{x^{2}-a^{2}} d x=\frac{1}{2 a^{2}} \int x \cdot \sqrt{u} \cdot \frac{d u}{2 x}=\frac{1}{4 a^{2}} \int u_{}^{\frac{1}{2}} d u \text { then, } \end{aligned}$
$=\frac{1}{4 a^{2}}\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c_{2} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{1}{4 a^{2}}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]+c_{2} \\ &=\frac{1}{4 a^{2}} \cdot \frac{2}{3} u_{}^{\frac{3}{2}}+c_{2} \end{aligned}$
$=\frac{1}{6 a^{2}}\left(x^{2}-a^{2}\right)^{\frac{3}{2}}+c_{2}$ .... $(iii)$ $\left(\because u=x^{2}-a^{2}\right)$
Putting the values of eqn $(ii)$ and eqn $(iii)$ in $(i)$ then,
$\begin{aligned} &I=\frac{1}{6 a^{2}}\left(x^{2}+a^{2}\right)^{\frac{3}{2}}+c_{1}-\frac{1}{6 a^{2}}\left(x^{2}-a^{2}\right)^{\frac{3}{2}}-c_{2} \\ &=\frac{1}{6 a^{2}}\left[\left(x^{2}+a^{2}\right)^{\frac{3}{2}}-\left(x^{2}-a^{2}\right)^{\frac{3}{2}}\right]+c \quad\left(\because c=c_{1}-c_{2}\right) \end{aligned}$

Indefinite Integrals exercise 18.9 question 56

Answer: $\int \frac{x \tan ^{-1} x^{2}}{1+x^{4}} d x$
Hint: Use substitution method to solve this integral.
Given: $\frac{1}{4}\left(\tan ^{-1} x^{2}\right)^{2}+c$
Solution:
$I=\int \frac{x \tan ^{-1} x^{2}}{1+x^{4}} d x$
$\begin{aligned} &\text { Put } \tan ^{-1} x^{2}=t \Rightarrow \frac{1}{1+\left(x^{2}\right)^{2}} 2 x\; d x=d t \\ &\Rightarrow \frac{2 x}{1+x^{4}}\; d x=d t \Rightarrow d x=\frac{1+x^{4}}{2 x} d t \\ &\text { Then, } I=\int \frac{x t}{1+x^{4}} \cdot \frac{1+x^{4}}{2 x} d t=\frac{1}{2} \int t\; d t \end{aligned}$
$=\frac{1}{2} \frac{t^{1+1}}{1+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{1}{2} \cdot \frac{t^{2}}{2}+c=\frac{1}{4} t^{2}+c \\ &=\frac{1}{4}\left(\tan ^{-1} x^{2}\right)^{2}+c \quad\left[\because t=\tan ^{-1} x^{2}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 57

Answer: $\frac{1}{4}\left(\sin ^{-1} x\right)^{4}+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x$
Solution:
$\text { Let } I=\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x$
$\begin{aligned} &\text { put } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow d x=\sqrt{1-x^{2}} d t \end{aligned}$
$I=\int \frac{t^{3}}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}} d t=\int t^{3} d t$
$=\frac{t^{3+1}}{3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{1}{4} t^{4}+c=\frac{1}{4}\left(\sin ^{-1} x\right)^{4}+c \quad\left[\because t=\sin ^{-1} x\right]$

Indefinite Integrals exercise 18.9 question 58

Answer: $-\frac{1}{3} \cos (2+3 \log x)+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{\sin (2+3 \log x)}{x} d x$
Solution:
$\text { Let } I=\int \frac{\sin (2+3 \log x)}{x} d x$
$\begin{aligned} &\text { Put } 2+3 \log x=t \Rightarrow 3 \cdot \frac{1}{x} d x=d t \Rightarrow d x=\frac{x}{3} d t \\ &\text { Then, } I=\int \frac{\sin t}{x} \cdot \frac{x}{3} d t=\frac{1}{3} \int \sin t \; d t \end{aligned}$
$=\frac{1}{3}(-\cos t)+c \quad\left[\because \int \sin x\; d x=-\cos x+c\right]$
$\begin{aligned} &=\frac{-1}{3} \cos t+c \\ &=\frac{-1}{3} \cos (2+3 \log x)+c \end{aligned}$ $[\because t=2+3 \log x]$

Indefinite Integrals exercise 18.9 question 59

Answer: $\frac{1}{2} e^{x^{2}}+c$
Hint: Use substitution method to solve this integral

Given: $\int x\: e^{x^{2}} d x$
Solution:
$\text { Let } I=\int x\: e^{x^{2}} d x$
$\begin{aligned} &\text { Put } x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow d x=\frac{1}{2 x} d t \\ &\text { Then, } I=\int x \cdot e^{t} \cdot \frac{d t}{2 x}=\frac{1}{2} \int e^{t} d t=\frac{1}{2} e^{t}+c \end{aligned}$ $\left[\because \int e^{x} d x=e^{x}+c\right]$
$=\frac{1}{2} e^{x^{2}}+c$ $\left[\because t=x^{2}\right]$

Indefinite Integrals exercise 18.9 question 60

Answer: $1+e^{x}-\log \left|1+e^{x}\right|+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{e^{2 x}}{1+e^{x}} d x$
Solution:
$I=\int \frac{e^{2 x}}{1+e^{x}} d x$
$\begin{aligned} &\text { Put } 1+e^{x}=t \Rightarrow e^{x} d x=d t \Rightarrow d x=\frac{d t}{e^{x}} \\ &\text { Then, } I=\int \frac{e^{2 x}}{t} \cdot \frac{d t}{e^{x}}=\int \frac{\left(e^{x}\right)^{2}}{t} \cdot \frac{d t}{e^{x}} \end{aligned}$
$=\int \frac{e^{x}}{t} d t=\int \frac{t-1}{t} d t \quad\left[\because 1+e^{x} \Rightarrow e^{x}=t-1\right]$
$\begin{aligned} &=\int\left(\frac{t}{t}-\frac{1}{t}\right) d t=\int\left(1-\frac{1}{t}\right) d t \\ &=\int 1 . d t-\int \frac{1}{t} d t=\int t^{0} d t-\int \frac{1}{t} d t \end{aligned}$
$=\frac{t^{0+1}}{0+1}-\log |t|+c$ $\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]$
$\begin{aligned} &=t-\log |t|+c \\ &=1+e^{x}-\log \left|1+e^{x}\right|+c\; \; \; \; \; \; \; \; \; \left[\because t=1+e^{x}\right] \end{aligned}$

Indefinite Integrals exercise 18.9 question 61

Answer: $2 \tan (\sqrt{x})+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} d x$
Solution:
$\text { Let } I=\int \frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} d x$
$\begin{aligned} &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 \sqrt{x}\; d t \\ &\text { Then, } I=\int \frac{\sec ^{2} t}{\sqrt{x}} \cdot 2 \sqrt{x}\; d t=2 \int \sec ^{2} t \; d t \end{aligned}$
$=2 \tan t+c \quad\left[\because \int \sec ^{2} x d x=\tan x+c\right]$
$=2 \tan \sqrt{x}+c \quad[\because t=\sqrt{x}]$

Indefinite Integrals exercise 18.9 question 63

Answer: $(x+1)+2 \sqrt{x+1}-2 \tan ^{-1}(\sqrt{x+1})-2 \log |x+2|+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{x+\sqrt{x+1}}{x+2} d x$
Solution:
$\text { Let } \mathrm{I}=\int \frac{x+\sqrt{x+1}}{x+2}$
$\text { Put } x+1=t^{2} \Rightarrow d x=2 t\; d t \text { then }$
$I=\int \frac{\left(t^{2}-1\right)+\sqrt{t^{2}}}{t^{2}+1} 2 t \; d t=2 \int \frac{\left(t^{2}-1\right)+t}{t^{2}+1} t\; d t \quad\left[\begin{array}{l} \because x+1=t^{2} \\ \Rightarrow x=t^{2}-1 \end{array}\right]$
$\Rightarrow I=2 \int\left(\frac{t^{2}+t-1}{t^{2}+1}\right) t \; d t=2 \int\left(\frac{t^{2} \cdot t+t . t-t}{t^{2}+1}\right) d t$
$\begin{aligned} &\Rightarrow I=2 \int\left(\frac{t^{3}+t^{2}-t}{t^{2}+1}\right) d t=2 \int\left\{\frac{t^{3}}{t^{2}+1}+\frac{t^{2}}{t^{2}+1}-\frac{t}{t^{2}+1}\right\} d t \\ &\Rightarrow I=2\left[\int \frac{t^{3}}{t^{2}+1} d t+\int \frac{t^{2}}{t^{2}+1} d t-\int \frac{t}{t^{2}+1} d t\right] \end{aligned}$
We can write
$I=2\left(I_{1}+I_{2}-I_{3}\right)$ $.........(i)$
$\begin{aligned} \text { where } I_{1}=\int \frac{t^{3}}{t^{2}+1} d t \end{aligned}$
$I_{2}=\int \frac{t^{2}}{t^{2}+1} d t$
$\text { and } I_{3}=\int \frac{t}{t^{2}+1} d t$
$\text { Now } I_{1}=\int \frac{t^{3}}{t^{2}+1} d t=\int\left(\frac{t^{3}+t-t}{t^{2}+1}\right) d t$
$=\int\left(\frac{\left(t^{3}+t\right)-t}{t^{2}+1}\right) d t=\int\left(\frac{t^{3}+t}{t^{2}+1}-\frac{t}{t^{2}+1}\right) d t$
$=\int\left(\frac{t\left(t^{2}+1\right)}{t^{2}+1}-\frac{t}{t^{2}+1}\right) d t=\int\left(t-\frac{t}{t^{2}+1}\right) d t$
$=\int t\; d t-\int \frac{t}{t^{2}+1} d t$
$\begin{aligned} &\text { put } t^{2}+1=u \Rightarrow 2 t \; d t=d u \Rightarrow t \; d t=\frac{d u}{2} \text { then } \\ &I_{1}=\int t \; d t-\int \frac{1}{u} \frac{d u}{2}=\int t\; d t-\frac{1}{2} \int \frac{d u}{u} \end{aligned}$
$=\frac{t^{1+1}}{1+1}-\frac{1}{2} \log |u|+c_{1}\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]$
$=\frac{t^{2}}{2}-\frac{1}{2} \log \left|1+t^{2}\right|+c_{1}$ $.......(ii)$ $\left[\because u=t^{2}+1\right]$
$\text { And } I_{2}=\int \frac{t^{2}}{t^{2}+1} d t=\int\left(\frac{t^{2}+t-t}{t^{2}+1}\right) d t$
$\begin{aligned} &=\int\left(\frac{\left(t^{2}+1\right)-1}{t^{2}+1}\right) d t=\int\left(\frac{t^{2}+1}{t^{2}+1}-\frac{1}{t^{2}+1}\right) d t \\ &=\int\left(1-\frac{1}{t^{2}+1}\right) d t=\int t^{0} d t-\int \frac{1}{t^{2}+1} d t \end{aligned}$
$=\frac{t^{0+1}}{0+1}-\tan ^{-1}(t)+c_{2}\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]$
$=t-\tan ^{-1}(t)+c_{2}$ $.......(iii)$
$\begin{aligned} &\text { Also } I_{3}=\int \frac{t}{t^{2}+1} d t \\ &\text { put } t^{2}+1=p \Rightarrow 2 t d t=d p \Rightarrow t d t=\frac{d p}{2} \text { then } \\ &I_{3}=\int \frac{1}{p} \frac{d p}{2}=\frac{1}{2} \int \frac{1}{p} d p=\frac{1}{2} \log |p|+c_{3} \end{aligned}$
$=\frac{1}{2} \log \left|1+t^{2}\right|+c_{3}$ $......(iv)$
$\begin{aligned} &\text { Substituting the values of } I_{1}, I_{2}, I_{3} \text { from eqn(ii), (iii) and (iv) in }(i) \text { then }\\ &I=2\left[\frac{t^{2}}{2}-\frac{1}{2} \log \left|1+t^{2}\right|+c_{1}+t-\tan ^{-1}(t)+c_{2}-\frac{1}{2} \log \left|1+t^{2}\right|-c_{3}\right] \end{aligned}$
$=2\left[\frac{t^{2}}{2}+t-\tan ^{-1}(t)-\left(\frac{1}{2}+\frac{1}{2}\right) \log \left|1+t^{2}\right|+c_{1}+c_{2}-c_{3}\right]$
$=2\left[\frac{t^{2}}{2}+t-\tan ^{-1}(t)-\log \left|1+t^{2}\right|+c_{4}\right] \quad\left[\because c_{4}=c_{1}+c_{2}-c_{3}\right]$
$\begin{aligned} &=2 \cdot \frac{t^{2}}{2}+2 t-2 \tan ^{-1}(t)-2 \log \left|1+t^{2}\right|+2 c_{4} \\ &=t^{2}+2 t-2 \tan ^{-1}(t)-2 \log \left|1+t^{2}\right|+c \end{aligned}$
$\begin{aligned} &\text { [ since } x+1=t^{2} \text { ] } \\ &=(x+1)+2 \sqrt{x+1}-2 \tan ^{-1} \sqrt{x+1}-2 \log |1+x+1|+c \\ &\mathrm{I}=(x+1)+2 \sqrt{x+1}-2 \tan ^{-1}(\sqrt{x+1})-2 \log |x+2|+c \end{aligned}$

Indefinite Integrals exercise 18.9 question 64

Answer: $\frac{1}{(\log 5)^{3}} \cdot 5^{5^{5^{x}}}+c$
Hint: Use substitution method to solve this integral

Given: $\int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x$
Solution:
$\begin{aligned} &\text { let } I=\int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x \\ &\text { Putting } 5^{5^{5^{x}}}=t \end{aligned}$
$\Rightarrow\left(5^{5^{5^{x}}} \cdot \log 5.5^{5^{x}} \cdot \log 5.5^{x} \log 5\right) d x=d t$
$\begin{aligned} &\Rightarrow 5^{5^{5^{5}}} 5^{5^{x}} 5^{x}(\log 5)^{3} d x=d t \\ &\Rightarrow\left(5^{5^{5^{x}}} 5^{5^{x}} 5^{x}\right) d x=\frac{d t}{(\log 5)^{3}} \text { then } \end{aligned}$
$I=\int \frac{d t}{(\log 5)^{3}}=\frac{1}{(\log 5)^{3}} \int 1 d t$
$=\frac{1}{(\log 5)^{3}} \int t^{0} d t=\frac{1}{(\log 5)^{3}} \frac{t^{0+1}}{0+1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{1}{(\log 5)^{3}} \cdot t+c=\frac{1}{(\log 5)^{3}} \cdot 5^{5^{5^{x}}}+c\left[\because 5^{5^{5^{x}}}=t\right]$

Indefinite Integrals exercise 18.9 question 65

Answer: $\frac{1}{2} \sec ^{-1}\left(x^{2}\right)+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{x \sqrt{x^{4}-1}} d x$
Solution:
$\begin{aligned} &\operatorname{let} I=\int \frac{1}{x \sqrt{x^{4}-1}} d x=\int \frac{1}{x \sqrt{\left(x^{2}\right)^{2}-1}} d x \\ &\text { Putting } x^{2}=t \Rightarrow 2 x \; d x=d t \Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}$
$I=\int \frac{1}{x \sqrt{t^{2}-1}} \frac{d t}{2 x}=\frac{1}{2} \int \frac{1}{x^{2} \sqrt{t^{2}-1}} d t$
$=\frac{1}{2} \int \frac{1}{t \sqrt{t^{2}-1}} d t \quad\left[\because x^{2}=t\right]$
$=\frac{1}{2} \sec ^{-1} \mathrm{t}+c \quad\left[\because \int \frac{1}{x \sqrt{x^{2}-1}} d x=\sec ^{-1} x+c\right]$
$=\frac{1}{2} \sec ^{-1}\left(x^{2}\right)+c \quad\left[\because t=x^{2}\right]$

Indefinite Integrals exercise 18.9 question 66

Answer: $2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+c$
Hint: Use substitution method to solve this integral

Given: $\int \sqrt{e^{x}-1} \; d x$
Solution:
$\begin{aligned} &\operatorname{let} I=\int \sqrt{e^{x}-1} d x \\ &\text { Putting } \mathrm{e}^{x}-1=t^{2} \Rightarrow e^{x} d x=2 t d t \Rightarrow d x=\frac{2 t \cdot d t}{e^{x}} \text { then } \end{aligned}$
$I=\int \sqrt{t^{2}} \frac{2 t \; d t}{e^{x}}$
$=2 \int \frac{t \cdot t}{t^{2}+1} d t \quad\left[\because e^{x}-1=t^{2} \Rightarrow t^{2}+1=e^{x}\right]$
$\begin{aligned} &=2 \int \frac{t^{2}}{t^{2}+1} d t \\ &=2 \int\left(\frac{t^{2}+1-1}{t^{2}+1}\right) d t \end{aligned}$
$\begin{aligned} &=2 \int\left(\frac{\left(t^{2}+1\right)-1}{t^{2}+1}\right) d t \\ &=2 \int\left(\frac{\left(t^{2}+1\right)}{t^{2}+1}-\frac{1}{t^{2}+1}\right) d t=2 \int\left(1-\frac{1}{t^{2}+1}\right) d t \end{aligned}$
$=2 \int 1 d t-2 \int \frac{1}{t^{2}+1} d t=2 \int t^{0} d t-2 \int \frac{1}{t^{2}+1} d t$
$=2 \frac{t^{0+1}}{0+1}-2 \tan ^{-1}(t)+c$ $\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c \end{array}\right]$
$\begin{aligned} &=2 t-2 \tan ^{-1}(t)+c \\ &=2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+\mathrm{c} \end{aligned}$ $\left[\because t^{2}=e^{x}-1 \Rightarrow t=\sqrt{e^{x}-1}\right]$

Indefinite Integrals exercise 18.9 question 67

Answer: $\log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x$
Solution:
$\operatorname{let} I=\int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x$
$\begin{aligned} &=\int \frac{1}{(x+1)\left(x^{2}+2 x+1+1\right)} d x \\ &=\int \frac{1}{(x+1)\left\{\left(x^{2}+2 x+1\right)+1\right\}} d x \end{aligned}$
$=\int \frac{1}{(x+1)\left\{(x+1)^{2}+1\right\}} d x$ $\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\text { Putting } x+1=\tan u$ $......(i)$
$\Rightarrow d x=\sec ^{2} u \; d u \text { then }$
$I=\int \frac{1}{\tan u\left\{\tan ^{2} u+1\right\}} \sec ^{2} u\; d u$
$=\int \frac{1}{\tan u\left\{\sec ^{2} u\right\}} \sec ^{2} u \; d u\left[\because 1+\tan ^{2} u=\sec ^{2} u\right]$
$=\int \frac{1}{\tan u} d u=\int \cot u\; d u$
$=\log |\sin u|+c$ $........(ii)$ $\left[\because \int \cot x \; d x=\log |\sin x|+c\right]$
$\text { Also, from (i) } x+1=\tan u=\frac{\sin u}{\cos u} \quad\left[\because \tan x=\frac{\sin x}{\cos x}\right]$
$\begin{aligned} &\Rightarrow(x+1) \cos u=\sin u \\ &\Rightarrow(x+1)^{2} \cos ^{2} u=\sin ^{2} u \quad[\text { Squaring on both sides }] \\ &\Rightarrow(x+1)^{2}\left(1-\sin ^{2} u\right)=\sin ^{2} u \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &\Rightarrow(x+1)^{2}-(x+1)^{2} \sin ^{2} u=\sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\sin ^{2} u+(x+1)^{2} \sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\left[1+(x+1)^{2}\right] \sin ^{2} u \end{aligned}$
$\Rightarrow \sin ^{2} u=\frac{(x+1)^{2}}{1+(x+1)^{2}}=\frac{(x+1)^{2}}{x^{2}+2 x+1+1}=\frac{(x+1)^{2}}{x^{2}+2 x+2} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\Rightarrow \sin u=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+2}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+2}}$ $........(iii)$
$\begin{aligned} &\text { From (ii) and (iii) we get }\\ &I=\log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c \end{aligned}$

Indefinite Integrals exercise 18.9 question 68

Answer: $\frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x \\ &\text { Put } 1+x^{3}=t^{2} \Rightarrow 3 x^{2} d x=2 t \; d t \\ &\Rightarrow d x=\frac{2 t}{3 x^{2}} d t \text { then } \end{aligned}$
$I=\int \frac{x^{5}}{\sqrt{t^{2}}} \frac{2 t \; d t}{3 x^{2}}=\int \frac{2}{3} \frac{x^{3} \cdot t}{t} d t=\frac{2}{3} \int x^{3} d t$
$=\frac{2}{3} \int\left(t^{2}-1\right) \mathrm{dt} \quad\left[\because 1+x^{3}=t^{2} \Rightarrow t^{2}-1=x^{3}\right]$
$=\frac{2}{3} \int t^{2} d t-\frac{2}{3} \int 1 d t=\frac{2}{3} \int t^{2} d t-\frac{2}{3} \int t^{0} d t$
$=\frac{2}{3} \frac{t^{2+1}}{2+1}-\frac{2}{3} \frac{t^{0+1}}{0+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=\frac{2}{3} \cdot \frac{t^{3}}{3}-\frac{2}{3} t+c$
$=\frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+\mathrm{c} \quad\left[\because t^{2}=1+x^{3} \Rightarrow t=\sqrt{1+x^{3}}\right]$

Indefinite Integrals exercise 18.9 question 69

Answer: $-\frac{20}{3}\left(5-x^{2}\right)^{\frac{3}{2}}+\frac{4}{5}\left(5-x^{2}\right)^{\frac{5}{2}}+c$
Hint: Use substitution method to solve this integral

Given: $\int 4 x^{3} \sqrt{5-x^{2}} \; d x$
Solution:
$\begin{aligned} &\text { Let } I=\int 4 x^{3} \sqrt{5-x^{2}} \; d x \\ &\text { Put } 5-x^{2}=t^{2} \Rightarrow-2 x \; d x=2 t\; d t \\ &\Rightarrow d x=\frac{t}{-x} d t \text { then } \end{aligned}$
$I=\int 4 x^{3} \sqrt{t^{2}} \frac{t \cdot d t}{-x}=-\int 4 x^{2} \cdot t . t \; d t$
$=-4 \int\left(5-t^{2}\right) t^{2} d t=-4 \int\left(5 t^{2}-t^{2} \cdot t^{2}\right) d t \quad\left[\because 5-x^{2}=t^{2} \Rightarrow x^{2}=5-t^{2}\right]$
$=-4 \int\left(5 t^{2}-t^{4}\right) d t$
$=-20 \int t^{2} d t+4 \int t^{4} d t$
$=-20 \cdot \frac{t^{2+1}}{2+1}+4 \cdot \frac{t^{4+1}}{4+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$=-20 \cdot \frac{t^{3}}{3}+4 \frac{t^{5}}{5}+c$
$=-\frac{20}{3}\left(\sqrt{5-x^{2}}\right)^{3}+\frac{4}{5}\left(\sqrt{5-x^{2}}\right)^{5}+c \quad\left[\because t^{2}=5-x^{2} \Rightarrow t=\sqrt{5-x^{2}}\right]$
$=-\frac{20}{3}\left(5-x^{2}\right)^{\frac{3}{2}}+\frac{4}{5}\left(5-x^{2}\right)^{\frac{5}{2}}+c$

Indefinite Integrals exercise 18.9 question 70

Answer: $\frac{1}{2} \log |1+\sqrt{x}|+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{\sqrt{x}+x} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1}{\sqrt{x}+x} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ &\Rightarrow d x=2 \sqrt{x} \; d t \text { then } \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{t+t^{2}} 2 \sqrt{x} \; d t \\ &=\int \frac{1}{t(1+t)} 2 \sqrt{x} \; d t=2 \int \frac{1}{1+t} d t \end{aligned}$ $\left[\begin{array}{l} \because \sqrt{x}=t \\ \Rightarrow x=t^{2} \end{array}\right]$
$\begin{aligned} &\text { Again put } 1+t=u \Rightarrow d t=d u \text { then }\\ &I=2 \int \frac{1}{u} d u=\frac{1}{2} \log |u|+c \end{aligned}$ $\left[\because \int \frac{1}{x} d x=\log |x|+c\right]$
$=\frac{1}{2} \log |t+1|+c \quad[\because u=1+t]$
$=\frac{1}{2} \log |1+\sqrt{x}|+c \quad[\because t=\sqrt{x}]$

Indefinite Integrals exercise 18.9 question 71

Answer: $-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} d x$
Solution:
$\text { Let } I=\int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} d x$
$=\int \frac{\frac{1}{x^{3}}}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}} \cdot \frac{1}{x^{3}}} d x \quad\left[\text { Dividing numerator and denominator by } \frac{1}{x^{3}}\right]$
$I=\int \frac{x^{-3}}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}} \cdot x^{-3}} d x=\int \frac{\left(x^{4}+1\right)^{-\frac{3}{4}}}{x^{3} x^{2} x^{-3}} d x$
$=\int \frac{\left(x^{4}+1\right)^{-\frac{3}{4}}}{x^{5} x^{-3}} d x=\int \frac{\left(x^{4}+1\right)^{-\frac{3}{4}}}{x^{5}\left(x^{4}\right)^{\frac{-3}{4}}} d x$
$=\int \frac{1}{x^{5}}\left(\frac{x^{4}+1}{x^{4}}\right)^{\frac{-3}{4}} d x=\int \frac{1}{x^{5}}\left(\frac{x^{4}}{x^{4}}+\frac{1}{x^{4}}\right)^{\frac{-3}{4}} d x$
$=\int \frac{1}{x^{5}}\left(1+\frac{1}{x^{4}}\right)^{\frac{-3}{4}} d x$
$\begin{aligned} &\text { Put } \frac{1}{x^{4}}=\mathrm{t} \Rightarrow \frac{-4 d x}{x^{5}}=d t \\ &\Rightarrow d x=\frac{x^{5}}{-4} d t \text { then } \end{aligned}$
$I=\int \frac{1}{x^{5}}(1+t)^{\frac{-3}{4}} \frac{x^{5}}{-4} d t=\frac{-1}{4} \int(1+t)^{\frac{-3}{4}} d t$
$\begin{aligned} &\text { Again put } 1+t=u \Rightarrow d t=d u \text { then }\\ &I=\frac{-1}{4} \int(u)^{\frac{-3}{4}} d u=\frac{-1}{4} \frac{u^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}$
$=\frac{-1}{4} \frac{u^{\frac{1}{4}}}{\frac{1}{4}}+c=\frac{-1}{4} \times 4 \cdot u^{\frac{1}{4}}+c$
$=-u^{\frac{1}{4}}+c=-(1+t)^{\frac{1}{4}}+c \quad[\because u=1+t]$
$=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c \quad\left[\because t=\frac{1}{x^{4}}\right]$

Indefinite Integrals exercise 18.9 question 72

Answer: $\frac{1}{3 \cos ^{2} x}-\cos x-\frac{2}{\cos x}+c$
Hint: Use substitution method to solve this integral

Given: $\int \frac{\sin ^{5} x}{\cos ^{4} x} d x$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{\sin ^{5} x}{\cos ^{4} x} d x \\ &\text { Put } \cos \mathrm{x}=t \Rightarrow-\sin x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{-\sin x} \text { then } \end{aligned}$
$I=\int \frac{\sin ^{5} x}{t^{4} } \frac{d t}{-\sin x}=-\int \frac{\sin ^{4} x}{t^{4}} d t$
$=-\int \frac{\left(\sin ^{2} x\right)^{2}}{t^{4}} d t=-\int \frac{\left(1-\cos ^{2} x\right)^{2}}{t^{4}} d t$
$I=-\int \frac{\left(1+\cos ^{4} x-2 \cos ^{2} x\right)}{t^{4}} d t \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$=-\int\left[\frac{1+t^{4}-2 t^{2}}{t^{4}}\right] d t \quad[\because \cos x=t]$
$\begin{aligned} &=-\int\left[\frac{1}{t^{4}}+\frac{t^{4}}{t^{4}}-\frac{2 t^{2}}{t^{4}}\right] d t \\ &=-\int\left[\frac{1}{t^{4}}+1-\frac{2}{t^{2}}\right] d t \end{aligned}$
$\begin{aligned} &=-\int\left(t^{-4}+1-2 t^{-2}\right) d t \\ &=-\int t^{-4} d t-\int t^{0} d t+2 \int t^{-2} d t \end{aligned}$
$=-\frac{t^{-4+1}}{-4+1}-\frac{t^{0+1}}{0+1}+2 \frac{t^{-2+1}}{-2+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=-\frac{t^{-3}}{-3}-t+2 \frac{t^{-1}}{-1}+c \\ &=\frac{1}{3 t^{3}}-t-2 \cdot \frac{t^{-1}}{1}+c \end{aligned}$
$=\frac{1}{3 \cos ^{2} x}-\cos x-\frac{2}{\cos x}+c \quad[\because t=\cos x]$

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