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RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 12:06 PM IST

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.11
RD Sharma Chapter wise Solutions

RD Sharma Class 12th Exercise 18.11 explains the chapter ‘Indefinite Integrals.’ The exercise contains 12 sums that are of Level 1 difficulty and concept-oriented. However, they are pretty easy to understand how to integrate a function w.r.t to x and also cover trigonometric integrals of values like Tan, Sec, Cosec, Cot. Students should refer to the theorems of those trigonometric values to solve these questions.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.11

Indefinite Integrals Exercise 18.11 Question 2

Answer: $\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \tan x \cdot \sec ^{4} x d x$
Sol : - Let $I=\int \tan x \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \tan x \sec ^{2} x \cdot \sec ^{2} x d x$
$\mathrm{I}=\int \tan x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$ $\text { (if, } \left.\sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan x+\tan ^{3} x\right) \sec ^{2} x d x$
Substitute $\tan x=t \rightarrow \sec ^{2} x d x=d t$ then
$\mathrm{I}=\int\left(\mathrm{t}+t^{3}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t} \text { ) }$
$=\int\left(\mathrm{t}+t^{3}\right) d t=\int \mathrm{t} d t+\int t^{3} d t$
$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{C}\right)$ (i)
$I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+C$
$I=\frac{\tan ^{2} x}{2}+\frac{\tan ^{4} x}{4}+\mathrm{C}$ $\text { (if, } t=\tan x)$

Indefinite Integrals Exercise 18.11 Question 3

Answer: $\frac{1}{6} \tan ^{6} x+\frac{1}{8} \tan ^{8} x+C$
Hint: Use substitution method to solve this integral.
Given: $\int \tan ^{5} x \cdot \sec ^{4} x d x$
Solution: Let,$\mathrm{I}=\int \tan ^{5} x \sec ^{4} x d x$
Re-Write $I=\int \tan ^{5} x \sec ^{2} \cdot \sec ^{2} x d x$
$\left.I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt} \text { , then }$
$\left.I=\int\left(\mathrm{t}^{5}+t^{7}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t}\right)$
$=\int\left(\mathrm{t}^{5}+t^{7}\right) d t=\int \mathrm{t}^{5} d t+\int t^{7} d t$
$=\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right)$......(i)
$\mathrm{I}=\frac{t^{6}}{6}+\frac{t^{8}}{8}+\mathrm{C}$
$\left.\mathrm{I}=\frac{\tan ^{6} x}{6}+\frac{\tan ^{8} x}{8}+\mathrm{C}\: \: \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\tan \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 4

Answer:$\frac{1}{6} \sec ^{6} x+C$
Hint:- Use substitution method to solve this integral.
Given:$\int \sec ^{6} x \cdot \tan x d x$
Solution:Let,$\mathrm{I}=\int \sec ^{6} x \cdot \tan x d x$
Re-write,$I=\int \sec ^{5} x \sec x \cdot \tan x d x$$\mathrm{I}=\int \sec ^{5} x(\sec x \cdot \tan x) d x$

Substitute, $\sec x=t \Rightarrow \sec x \cdot \tan x d x=d t$, then,

$I=\int t^{5} \cdot \mathrm{dt} \quad(\text { if }, \sec x=t)$

$=\frac{t^{5+1}}{5+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$

$\left.=\frac{t^{6}}{6}+c=\frac{\sec ^{6} x}{6}+C \quad \text { (if, } \sec x=t\right)$

Indefinite Integrals Exercise 18.11 Question 5

Answer:$\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec \times| C$
Hint:- Use substitution method to solve this integral.
Given:$\int \tan ^{5} x d x$
Solution: Let,$\mathrm{I}=\int \tan ^{5} x d x$
Re-Write,$\mathrm{I}=\int \tan ^{3} x \cdot \tan ^{2} x \cdot d x$
$I=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ $\text { (if, } \left.\tan ^{2} x=\sec ^{2} x-1\right)$
$\mathrm{I}=\int\left(\tan ^{3} x \cdot \sec ^{2} x-\tan ^{3}\right) d x$
$I=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{3} x d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{2} x \cdot \tan x d x$
$\left.=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x-1\right) \cdot \tan x d x \quad \text { (if, } \sec ^{2} x-\tan ^{2} x=1\right)$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x \tan x-\tan x\right) d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \sec ^{2} x \tan x d x+\int \tan x d x$
Put $tan x = t \Rightarrow \sec^{2}x d x = d t , then$
$\mathrm{I}=\int \mathrm{t}^{3} d t-\int t d t+\int \tan x d x$
$=\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sec \times| C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \tan x d x=\log |\sec x|+c\right)$
$=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x| C$
$\left.=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x| C\: \: \: \: \: \: \quad \text { (if, } t=\tan x\right)$

Indefinite Integrals Exercise 18.11 Question 6

Answer: $\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+C$
Hint: Use substitution method to solve this integral.
Given:$\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Solution:Let, $I=\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \sqrt{\tan x} \cdot \sec ^{2} x \cdot \sec ^{2} x d x$
$I=\int \sqrt{\tan x}\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad\left(\because \sec ^{2} x-\tan ^{2}=1\right)$
$I=\int\left(\sqrt{\tan x}+\sqrt{\tan x} \cdot \tan ^{2} x\right) \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{1}{2}+2} x\right) \cdot \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \cdot \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{t} \text { , then }$
$\begin{array}{ll} \mathrm{I}=\int\left(t^{\frac{1}{2}}+t^{5 / 2}\right) d t \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & (\because \tan \mathrm{x}=\mathrm{t}) \end{array}$
$=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{5}{2}}+1}{\frac{5}{2}+1}+C \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$=\frac{\tan ^{3 / 2} x}{3 / 2}+\frac{\tan ^{7 / 2} x}{7 / 2}+\mathrm{C} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because \mathrm{t}=\tan \mathrm{x})$
$=\frac{2}{3} \tan ^{3 / 2} x+\frac{2}{7} \tan ^{7 / 2} x+C$

Indefinite Integrals Exercise 18.11 Question 7

$\begin{aligned} &\text { Answer }:-\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+C\\ &\text { Hint :- Use substitution method to solve this integral. }\\ &\text { Given :- } \int \sec ^{4} 2 x d x\\ &\text { Sol: - Let } I=\int \sec ^{4} 2 x d x\\ &\text { Re-writing, } I=\int \sec ^{2} 2 x \cdot \sec ^{2} 2 x d x \end{aligned}$
$\begin{aligned} &\left.I=\int\left(1+\tan ^{2} 2 x\right) \sec ^{2} 2 x d x\: \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right) \\ &I=\int\left(\sec ^{2} 2 x+\tan ^{2} 2 x \cdot \sec ^{2} 2 x\right) d x \end{aligned}$
$\begin{aligned} &\text { Substitute } \tan 2 \mathrm{x}=\mathrm{t}\\ &\sec ^{2} 2 x .2 d x=d t, \text { then } \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \sec ^{2} 2 x \mathrm{~d} x+\int \tan ^{2} 2 x \cdot \sec ^{2} 2 x \mathrm{dx} \\ &=\int \sec ^{2} 2 x \mathrm{~d} x+\int t^{2} \cdot \frac{d t}{2} \: \: \: \: \: \: \: \: \: \: \quad(\text { since }, \tan 2 x=t) \end{aligned}$
$=\frac{\tan 2 x}{2}+\frac{1}{2} \frac{t^{2+1}}{2+1}+C$ $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \sec ^{2} d x=\tan x+\mathrm{c}$
$\begin{aligned} &=\frac{\tan 2 x}{2}+\frac{1}{2} \cdot \frac{t^{3}}{3}+C \\ &\left.=\frac{\tan 2 x}{2}+\frac{1}{6} \cdot \tan ^{3} 2 x+C \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } t=\tan 2 x\right) \end{aligned}$

Indefinite Integrals Exercise 18.11 Question 8

Answer: $-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cot 3 x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \operatorname{cosec}^{4} 3 x d x$
Solution:Let $\mathrm{I}=\int \operatorname{cosec}^{4} 3 x d x$
Re-Write $\mathrm{I}=\int \operatorname{cosec}^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x d x$
$\left.I=\int\left(1+\cot ^{2} 3 x\right) \cdot \operatorname{cosec}^{2} 3 x d x \: \: \: \: \: \: \: \: \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\begin{aligned} &I=\int\left(\operatorname{cosec}^{2} 3 x+\cot ^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x\right) d x \\ &I=\int \operatorname{cosec}^{2} 3 x \cdot \cot ^{2} 3 x d x+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}$
Substitute, cot3x = t
$\begin{aligned} &\operatorname{cosec}^{2} 3 x \cdot 3 d x=d t, \text { then } \\ &I=\int t^{2} \frac{d t}{3}+\int \operatorname{cosec}^{2} 3 x d x \\ &=-\frac{1}{3} \int t^{2} d t+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}$
$=-\frac{1}{3} \frac{t^{2+1}}{2+1}+\frac{-\cot 3 x}{3}+C$ $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \operatorname{cosec}^{2} x d x=-\cot x+c$
$=-\frac{1}{3} \cdot \frac{t^{3}}{3}-\frac{\cot 3 x}{3}+C$
$=-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cdot \cot 3 x+\mathrm{C} \: \: \: \: \: \: \: \: \quad(\text { if, } \cot 3 \mathrm{x}=\mathrm{t})$

Indefinite Integrals Exercise 18.11 Question 9

Answer: $-\frac{\cot ^{n+1} x}{n+1}+\mathrm{C}$
Hint:Use substitution method to solve this integral.
Given:$\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x, \mathrm{n} \neq-1$
Solution:let,$\mathrm{I}=\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x$
Substitute, cot x = t
$\operatorname{cosec}^{2} x d x=\text { dt then }$
$\begin{array}{ll} \mathrm{I}=\int t^{n} \cdot \operatorname{cosec}^{2} \mathrm{x} \cdot \frac{d t}{-\operatorname{cosec}^{2} x} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text { (if, } \cot \mathrm{x}=\mathrm{t}) \end{array}$
$=-\int t^{n}$
$=-\left[\frac{t^{n+1}}{n+1}\right]+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$\left.=-\frac{\cot ^{n+1} x}{n+1}+C \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\cot \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 10

Answer: $-\frac{1}{6} \cot ^{6} x-\frac{1}{8} \cot ^{8} x+C$
Hint :- Use substitution method to solve this integral.
Given:$\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Solution:Let, $\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x$ $\text { (if, } \left.\operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{5} x \cdot \cot ^{2}\right) \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x$

Substitute, cotx = t

$-\operatorname{cosec}^{2} x d x=d t, \text { then }$

$I=\int\left(t^{5}+t^{7}\right) \operatorname{cosec}^{2} x \cdot \frac{d t}{-\operatorname{cosec}^{2} x} \mathrm{dx} \: \: \: \: \: \: \: \: \quad(\text { as } \cot x=t)$

$\begin{aligned} &=-\int\left(t^{5}+t^{7}\right) \mathrm{dt} \\ &=-\int t^{5} d t-\int t^{7} \mathrm{dt} \end{aligned}$

$=-\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c}\right)$

$=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+C$

$=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+\mathrm{C}$ $(\text { if, } t=\cot x)$

Indefinite Integrals Exercise 18.11 Question 11

Answer: $-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{5} x d x$
Solution: Let $I=\int \cot ^{5} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{5} x d x$
$=\int \cot ^{3} x \cdot \cot ^{2} x d x$
$\mathrm{I}=\int \cot ^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
$\begin{aligned} &I=\int\left(\cot ^{3} x \cdot \operatorname{cosec}^{2} x-\cot ^{3} x\right) d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{3} x d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{2} x \cot x d x \end{aligned}$
$\begin{aligned} &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) \cot x d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cdot \cot x-\cot x\right) d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \operatorname{cosec}^{2} x \cdot \cot x d x+\int \cot x d x \end{aligned}$

Substitute, cotx = t

$\begin{aligned} &-\operatorname{cosec}^{2} x d x=d t, \text { then } \\ &I=\int t^{3}(-d t)-\int t(-d t)+\int \cot x d x \\ &=-\int t^{3} d t+\int t d t+\int \cot x d x \end{aligned}$

$=-\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sin x|+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c} \& \int \cot x d x=\log |\sin x|+c\right)$

$\begin{aligned} &=-\frac{1}{4} t^{4}+\frac{1}{2} t^{2}+\log |\sin x|+C \\ &=-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C \end{aligned}$ $\left ( as,t=\cot x \right )$

Indefinite Integrals Exercise 18.11 Question 12

Answer: $-\frac{1}{5} \cot ^{5} x+\frac{1}{3} \cot ^{3} x-\cot x-x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{6} x d x$
Solution: Let,$I=\int \cot ^{6} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{6} x d x$
$\mathrm{I}=\int \cot ^{2} x \cdot \cot ^{4} x d x$
$\mathrm{I}=\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{4} x d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
$\begin{aligned} &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{4} x d x \\ &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \cot ^{2} x d x \end{aligned}$
$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x\left(\operatorname{cosec}^{2} x-1\right) d x \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\begin{aligned} &1=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int\left(\cot ^{2} x \cdot \operatorname{cosec}^{2} x-\cot ^{2} x\right) d x \\ &\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \cot ^{2} x d x \end{aligned}$
$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int\left(\operatorname{cosec}^{2} x-1\right) d x \text { (if, } \cot ^{2} x=\operatorname{cosec}^{2} x-1\right)$
$\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \operatorname{cosec}^{2} x d x-\int 1 d x$
Substitute, cotx = t
$-\operatorname{cosec}^{2} x d x=d t, \text { then }$
$\begin{aligned} &\mathrm{I}=\int t^{4}(-d t)-\int t^{2}(-d t)+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int 1 \mathrm{dx} \\ &=-\int t^{4} d t+\int t^{2} d t+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int x^{0} \mathrm{dx} \end{aligned}$
$=-\frac{t^{4+1}}{4+1}+\frac{t^{2+1}}{2+1}-\cot x-\frac{x^{0+1}}{0+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \operatorname{cosec}^{2} x d x=-\cot x+c\right)$
$\begin{aligned} &=-\frac{t^{5}}{5}+\frac{t^{3}}{3}-\cot x-x+C \\ &=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{8} x}{3}-\cot x-x+C \end{aligned}$ $\left ( if,\cot x=t \right )$

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