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

RD Sharma books are well known for their comprehensive and knowledgeable materials. CBSE schools and teachers widely use them to set up question papers and homework assignments. This is why RD Sharma books are the best option for students when it comes to.

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 ,
$I = \int \frac{\log (\log x)}{x} d x$
Put $\log x = t \Rightarrow \frac{1}{x} d x = d t$
$I = \int \log t d t = t [\log t - 1] + c I = \log x [\log (\log x) - 1] + c$

Indefinite Integrals exercise 18 .9 question 2

Answer: $- \frac{1}{2} {[\log (1 + \frac{1}{x})]}^{2} + c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x$
Solution:
Let $I = \int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x$
Put

\log (1 + \frac{1}{x}) = t \Rightarrow \frac{1}{1 + \frac{1}{x}} (\frac{- 1}{x^{2}}) d x = d t

\begin{aligned} \Rightarrow \frac{1}{\frac{x + 1}{x}} (\frac{- 1}{x^{2}}) d x = d t \\ \Rightarrow \frac{x}{x + 1} (\frac{- 1}{x^{2}}) d x = d t \\ \Rightarrow \frac{- 1}{x (x + 1)} d x = d t \Rightarrow d x = - (x + 1) x d t then \end{aligned}

I = \int - \frac{t}{x (1 + x)} (x + 1) x d t = - \int t d t [∵ \frac{1}{x} d x = d t \Rightarrow d x = x d t]

\begin{aligned} = - \frac{t^{1 + 1}}{1 + 1} + c \\ = \frac{- t^{2}}{2} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = \frac{- 1}{2} {\log (1 + \frac{1}{x})}^{2} + c [∵ t = \log (1 + \frac{1}{x})] \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 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 [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = \frac{2}{3} (1 + \sqrt{x})^{3} + c [∵ t = 1 + \sqrt{x}] \end{aligned}

Indefinite Integrals exercise 18.9 question 4

Answer: $\frac{2}{3} {(1 + e^{x})}^{\frac{3}{2}} + c$
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 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}} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = \frac{2}{3} {(1 + e^{x})}^{\frac{3}{2}} + c [∵ t = 1 + e^{x}] \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} Put \cos x = t \Rightarrow - \sin x d x = d t \\ \Rightarrow \sin x d x = - d t 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}} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = - \frac{3}{5} (\cos x)^{\frac{5}{3}} + c [∵ t = \cos x] \end{aligned}

Indefinite Integrals exercise 18.9 question 6

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

Indefinite Integrals exercise 18.9 question 7

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

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: $\int \frac{{e^{\sin^{- 1} x}}^{2}}{\sqrt{1 - x^{2}}} d x$
Solution:
$\begin{aligned} Let I = \int \frac{{e^{\sin^{- 1} x}}^{2}}{\sqrt{1 - x^{2}}} d x \\ Put \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t \\ \Rightarrow d x = \sqrt{1 - x^{2}} d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{{(e^{t})}^{2}}{\sqrt{1 - x^{2}}} \sqrt{1 - x^{2}} d t = \int {(e^{t})}^{2} d t \\ = \int e^{2 t} d t = \frac{e^{2 t}}{2} + c [∵ \int e^{a x} d x = \frac{e^{a x}}{a} + c] \end{aligned}$
$= \frac{e^{2 \sin^{- 1} x}}{2} + c [∵ t = \sin^{- 1} x]$

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:
$Let I = \int \frac{1 + \sin x}{\sqrt{x - \cos x}} d x$
$\begin{aligned} Put x - \cos x = t \Rightarrow (1 + \sin x) d x = d t \\ \Rightarrow (1 + \sin x) d x = d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{1}{\sqrt{t}} d t = \int t^{\frac{- 1}{2}} d t \\ = \frac{t^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + c = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \end{aligned}$
$= 2 \sqrt{t} + c$

= 2 \sqrt{x - \cos x} + c [∵ 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: $\int \frac{1}{\sqrt{1 - x^{2}} {(\sin^{- 1} x)}^{2}} d x$
Solution:
$\begin{aligned} Let I = \int \frac{1}{\sqrt{1 - x^{2}} {(\sin^{- 1} x)}^{2}} d x \\ Put \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t \\ \Rightarrow d x = \sqrt{1 - x^{2}} d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{1}{\sqrt{1 - x^{2}} \cdot t^{2}} \sqrt{1 - x^{2}} d t = \int \frac{1}{t^{2}} d t \\ = \frac{t^{- 2 + 1}}{- 2 + 1} + c = \frac{t^{- 1}}{- 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \end{aligned}$
$= \frac{- 1}{t} + c = \frac{- 1}{\sin^{- 1} x} + c [∵ t = \sin^{- 1} x]$

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:
$\begin{aligned} Let I = \int \frac{\cot x}{\sqrt{\sin x}} d x \\ Put \sin x = t \Rightarrow \cos x d x = d t \\ \Rightarrow d x = \frac{d t}{\cos x} then \end{aligned}$
$I = \int \frac{\cot x}{\sqrt{t}} \frac{d t}{\cos x} = \int \frac{\cos x}{\sin x} \cdot \frac{1}{\sqrt{t}} \cdot \frac{d t}{\cos x} d t [∵ \cot x = \frac{\cos x}{\sin x}]$
$\begin{aligned} = \int \frac{1}{t . t^{\frac{1}{2}}} d t [∵ t = \sin x] \\ = \int \frac{1}{t^{1 + \frac{1}{2}}} d t = \int \frac{1}{t^{\frac{3}{2}}} dt \end{aligned}$
$I = \int t^{\frac{- 3}{2}} d t$
$\begin{aligned} = \frac{t^{\frac{- 3}{2} + 1}}{\frac{- 3}{2} + 1} + c = \frac{t^{\frac{- 1}{2}}}{\frac{- 1}{2}} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = - 2 \frac{1}{\sqrt{t}} + c = \frac{- 2}{\sqrt{\sin x}} + c [∵ t = \sin x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{\tan x}{\sqrt{\cos x}} d x \\ Put \cos x = t \Rightarrow - \sin x d x = d t \\ \Rightarrow d x = \frac{- d t}{\sin x} then \end{aligned}$
$\begin{aligned} I & = \int \frac{\sin x}{t \sqrt{t}} \frac{- d t}{\sin x} \\ = - \int \frac{1}{t^{1 + \frac{1}{2}}} d t = - \int \frac{1}{t^{\frac{3}{2}}} d t \\ I & = - \int t^{\frac{- 3}{2}} d t \end{aligned}$
$\begin{aligned} = - \frac{t^{\frac{- 3}{2} + 1}}{\frac{- 3}{2} + 1} + c = - \frac{t^{\frac{- 1}{2}}}{\frac{- 1}{2}} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = - (- 2) \frac{1}{\sqrt{t}} + c = \frac{2}{\sqrt{t}} + c \\ = \frac{2}{\sqrt{\cos x}} + c [∵ t = \cos x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{\cos^{3} x}{\sqrt{\sin x}} d x = \int \frac{\cos^{2} x \cdot \cos x}{\sqrt{\sin x}} d x \\ = \int \frac{(1 - \sin^{2} x) \cdot \cos x}{\sqrt{\sin x}} d x [\begin{matrix} ∵ \sin^{2} x + \cos^{2} x = 1 \\ \Rightarrow \cos^{2} x = 1 - \sin^{2} x \end{matrix}] \\ Put \sin x = t \Rightarrow \cos x d x = d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{(1 - t^{2})}{\sqrt{t}} d t = \int {\frac{1}{\sqrt{t}} - \frac{t^{2}}{\sqrt{t}}} d t \\ = \int t^{\frac{- 1}{2}} d t - \int t^{2 - \frac{1}{2}} d t = \int t^{\frac{- 1}{2}} d t - \int t^{\frac{4 - 1}{2}} d t \\ = \int t^{\frac{- 1}{2}} d t - \int t^{\frac{3}{2}} d t \end{aligned}$
$= \frac{t^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} - \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{t^{\frac{1}{2}}}{\frac{1}{2}} - \frac{t^{\frac{5}{2}}}{\frac{5}{2}} + c$
$\begin{aligned} = 2 t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{5}{2}} + c \\ = 2 \sqrt{\sin x} - \frac{2}{5} (\sin x)^{\frac{5}{2}} + c [∵ t = \sin x] \end{aligned}$

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} 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{(1 - \cos^{2} x) \cdot \sin x}{\sqrt{\cos x}} d x [\begin{matrix} ∵ \sin^{2} x + \cos^{2} x = 1 \\ \Rightarrow \sin^{2} x = 1 - \cos^{2} x \end{matrix}] \\ Put \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t then \end{aligned}

\begin{aligned} I & = \int \frac{(1 - t^{2})}{\sqrt{t}} (- d t) = - \int {\frac{1 - t^{2}}{\sqrt{t}}} d t \\ = - \int {t^{\frac{- 1}{2}} - t^{2 - \frac{1}{2}}} d t = - \int {t^{\frac{- 1}{2}} - t^{\frac{4 - 1}{2}}} d t \\ = - \int {t^{\frac{- 1}{2}} - 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} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]

= - \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + \frac{t^{\frac{5}{2}}}{\frac{5}{2}} + 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 [∵ 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: $\int \frac{1}{\sqrt{\tan^{- 1} x} (1 + x^{2})} d x$
Solution:
$Let I = \int \frac{1}{\sqrt{\tan^{- 1} x} (1 + x^{2})} d x$

Put \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t \Rightarrow d x = (1 + x^{2}) d t then

\begin{aligned} I & = \int \frac{1}{\sqrt{t} (1 + x^{2})} (1 + x^{2}) 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} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \end{aligned}

\begin{aligned} = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c = 2 \sqrt{t} + c \\ = 2 \sqrt{\tan^{- 1} x} + c [∵ t = \tan^{- 1} x] \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:
$\begin{aligned} I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x \\ = \int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos x \cdot \cos x} d x \end{aligned}$
$= \int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cdot \cos^{2} x} d x = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x}} \cdot \frac{1}{\cos^{2} x} d x$
$\begin{aligned} = \int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec^{2} x d x = \int (\tan x)^{\frac{1}{2} - 1} \cdot \sec^{2} x d x \\ = \int (\tan x)^{\frac{- 1}{2}} \sec^{2} x d x \end{aligned}$
$Put \tan x = t \Rightarrow \sec^{2} x d x = d t then$
$I = \int t^{\frac{- 1}{2}} d t = \frac{t^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c = 2 \sqrt{t} + c \\ = 2 \sqrt{\tan x} + c [∵ t = \tan x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{1}{x} (\log x)^{2} d x \\ Put \log x = t \Rightarrow \frac{1}{x} d x = d t \\ \Rightarrow d x = x d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{1}{x} t^{2} \cdot x d t = \int t^{2} d t \\ = \frac{t^{2 + 1}}{2 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \end{aligned}$
$\begin{aligned} = \frac{t^{3}}{3} + c \\ = \frac{1}{3} (\log x)^{3} + c [∵ t = \log x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \sin^{5} x \cos x d x \\ Put \sin x = t \Rightarrow \cos x d x = d t then \end{aligned}$
$\begin{aligned} I & = \int t^{5} d t \\ = \frac{t^{5 + 1}}{5 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \end{aligned}$
$\begin{aligned} = \frac{t^{6}}{6} + c \\ = \frac{1}{6} \sin^{6} x + c [∵ t = \sin x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 21

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

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:
$\begin{aligned} Let I = \int \frac{(4 x + 3)}{\sqrt{2 x^{2} + 3 x + 1}} d x \\ Put 2 x^{2} + 3 x + 1 = t \Rightarrow (4 x + 3) d x = d t then \\ I = \int \frac{1}{\sqrt{t}} = \int t^{\frac{- 1}{2}} d t \end{aligned}$
$= \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c = 2 \sqrt{t} + c \\ = 2 \sqrt{2 x^{2} + 3 x + 1} + c [∵ t = 2 x^{2} + 3 x + 1] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{1}{1 + \sqrt{x}} d x \\ Put x = t^{2} \Rightarrow d x = 2 t d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{1}{1 + \sqrt{t^{2}}} 2 t d t = \int \frac{2 t}{1 + t} d t \\ = 2 \int \frac{t}{1 + t} d t = 2 \int \frac{1 + t - 1}{1 + t} d t \end{aligned}$
$\begin{aligned} = 2 \int \frac{(1 + t) - 1}{1 + t} d t = 2 \int {\frac{1 + t}{1 + t} - \frac{1}{1 + t}} d t \\ = 2 \int {1 - \frac{1}{1 + t}} d t = 2 \int 1 . d t - 2 \int \frac{1}{1 + t} d t \end{aligned}$
$= 2 \int 1 . d t - 2 \int \frac{1}{1 + t} d t$ ...... $(i)$
$Now 2 \int 1 . d t = 2 \frac{t^{0 + 1}}{0 + 1} + c_{1} [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= 2 t + c_{1}$ ........ $(i i)$
$\begin{aligned} and 2 \int \frac{1}{1 + t} d t \\ Put 1 + t = p \Rightarrow d t = d p then \end{aligned}$
$2 \int \frac{1}{1 + t} d t = 2 \int \frac{1}{p} d p = 2 \log | p | + c_{2}$
$= 2 \log | t + 1 | + c_{2}$ ......... $(i i i)$
Putting the values of equation (ii) and (iii) in (i) then

\begin{aligned} I = 2 t + c_{1} - (2 \log | t + 1 | + c_{2}) \\ \Rightarrow I = 2 t + c_{1} - 2 \log | t + 1 | - c_{2} \\ ∴ I = 2 \sqrt{x} - 2 \log | \sqrt{x} + 1 | + c_{1} - c_{2} \end{aligned}

∴ I = 2 \sqrt{x} - 2 \log | \sqrt{x} + 1 | + c [\begin{matrix} ∵ t^{2} = x \Rightarrow t = \sqrt{x} \\ and c = c_{1} - c_{2} \end{matrix}]

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:
$\begin{aligned} Let I = \int e^{\cos^{2} x} \sin 2 x d x \\ Put \cos^{2} x = t \Rightarrow 2 \cos x (- \sin x) d x = d t \end{aligned}$
$\begin{aligned} \Rightarrow - (2 \cos x \sin x) d x = d t \\ \Rightarrow - \sin 2 x d x = d t \end{aligned}$ $[∵ \sin 2 x = 2 \sin x \cos x]$
$\begin{aligned} Then \\ I = \int e^{t} (- d t) = - \int e^{t} d t = - e^{t} + c [∵ \int e^{x} d x = e^{x} + c] \\ = - e^{\cos^{2} x} + c [∵ t = \cos^{2} x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{1 + \cos x}{(x + \sin x)^{2}} d x \\ Put x + \sin x = t \Rightarrow (1 + \cos x) d x = d t then \end{aligned}$
$I = \int \frac{1}{t^{3}} d t = \int t^{- 3} d t = \frac{t^{- 3 + 1}}{- 3 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{t^{- 2}}{- 2} + c = \frac{- 1}{2} \frac{1}{t^{2}} + c$
$= \frac{- 1}{2 (x + \sin x)^{2}} + c [∵ t = x + \sin x]$

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:
$Let I = \int \frac{\cos x - \sin x}{1 + \sin 2 x} d x$
$= \int \frac{\cos x - \sin x}{\sin^{2} x + \cos^{2} x + \sin 2 x} d x [∵ 1 = \sin^{2} x + \cos^{2} x]$
$= \int \frac{\cos x - \sin x}{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x} d x [∵ \sin 2 x = 2 \sin x \cos x]$
$= \int \frac{\cos x - \sin x}{(\sin x + \cos x)^{2}} d x [∵ a^{2} + b^{2} + 2 a b = (a + b)^{2}]$
$Put \sin x + \cos x = t \Rightarrow (\cos x - \sin x) d x = d t then$
$I = \int \frac{(\cos x - \sin x)}{t^{2}} \frac{d t}{(\cos x - \sin x)}$
$= \int \frac{1}{t^{2}} = \int t^{- 2} d t = \frac{t^{- 2 + 1}}{- 2 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{t^{- 1}}{- 1} + c = \frac{- 1}{t} + c$
$= \frac{- 1}{\sin x + \cos x} + c [∵ t = \sin x + \cos x]$

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:
$Let I = \int \frac{\sin 2 x}{(a + b \cos 2 x)^{2}} d x$
$\begin{aligned} Put a + b \cos 2 x = t \\ \Rightarrow b (- \sin 2 x) \cdot 2 d x = d t \\ \Rightarrow - 2 b \sin 2 x d x = d t \\ \Rightarrow \sin 2 x d x = \frac{d t}{- 2 b} \end{aligned}$
$I = \int \frac{1}{t^{2}} \cdot \frac{d t}{- 2 b} = \frac{- 1}{2 b} \int \frac{1}{t^{2}} d t$
$= \frac{- 1}{2 b} \int t^{- 2} d t = \frac{- 1}{2 b} \frac{t^{- 2 + 1}}{- 2 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{- 1}{2 b} \cdot \frac{t^{- 1}}{- 1} + c = \frac{1}{2 b} \cdot \frac{1}{t} + c$
$= \frac{1}{2 b (a + b \cos 2 x)} + c [∵ t = a + b \cos 2 x]$

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:
$Let I = \int \frac{\log x^{2}}{x} d x$
$= \int \frac{2 \log x}{x} d x [∵ \log x^{m} = m \log x]$
$Put \log x = t \Rightarrow \frac{1}{x} d x = d t, then$
$I = \int 2 \cdot \frac{t}{x} \cdot x d t = 2 \int t d t = 2 \int \frac{t^{1 + 1}}{1 + 1} + c$ $[∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= 2 \frac{t^{1 + 1}}{2} + c = t^{2} + c = (\log x)^{2} + c$ $[∵ t = \log x]$

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:
$\begin{aligned} Let I = \int \frac{\sin x}{(1 + \cos x)^{2}} d x \\ Put 1 + \cos x = t \Rightarrow - \sin x d x = d t, then \end{aligned}$
$I = \int \frac{1}{t^{2}} (- d t) = - \int \frac{1}{t^{2}} d t = - \int t^{- 2} d t$
$= - [\frac{t^{- 2 + 1}}{- 2 + 1}] + c = - \frac{- t^{- 1}}{- 1} + c$ $[∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{1}{t} + c = \frac{1}{1 + \cos x} + c$ $[∵ t = 1 + \cos x]$

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:
$\begin{aligned} Let I = \int \cot x \cdot \log (\sin x) d x \\ Put \log (\sin x) = t \Rightarrow \frac{1}{\sin x} \cos x d x = d t \\ \Rightarrow \cot x d x = d t \Rightarrow d x = \frac{d t}{\cot x} then \end{aligned}$
$\Rightarrow I = \int \cot x \cdot t \cdot \frac{d t}{\cot x} = \int t d t$
$= \frac{t^{1 + 1}}{1 + 1} + c = \frac{t^{2}}{2} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c,]$
$= \frac{1}{2} {\log (\sin x)}^{2} + c [∵ t = \log (\sin x)]$

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:
$\begin{aligned} Let I = \int \sec x \cdot \log (\sec x + \tan x) d x \\ Put \log (\sec x + \tan x) = t \end{aligned}$
$\begin{aligned} \Rightarrow \frac{1}{(\sec x + \tan x)} (\sec x \tan x + \sec^{2} x) d x = d t \\ \Rightarrow \frac{1}{(\sec x + \tan x)} \sec x (\tan x + \sec x) d x = d t \end{aligned}$
$\begin{aligned} \Rightarrow \sec x d x = d t \Rightarrow d x = \frac{d t}{\sec x} then \\ \Rightarrow I = \int \sec x \cdot t \cdot \frac{d t}{\sec x} = \int t d t \end{aligned}$
$= \frac{t^{1 + 1}}{1 + 1} + c = \frac{t^{2}}{2} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c,]$
$= \frac{1}{2} {\log (\sec x + \tan x)}^{2} + c [∵ t = \log (\sec x + \tan x)]$

Indefinite Integrals exercise 18.9 question 32

Answer: $\frac{1}{2} {\log (cosec x - \cot x)}^{2} + c$
Hint:Use substitution method to solve this integral.
Given:

\int cosec x \cdot \log (cosec x - \cot x) d x

Solution:
$\begin{aligned} Let I = \int cosec x \cdot \log (cosec x - \cot x) d x \\ Put \log (cosec x - \cot x) = t \end{aligned}$
$\Rightarrow \frac{1}{(cosec x - \cot x)} (- cosec x \cdot \cot x - (- {cosec}^{2} x)) d x = d t$
$\Rightarrow \frac{1}{(cosec x - \cot x)} {{cosec}^{2} x - cosec x \cdot \cot x} d x = d t$
$\Rightarrow \frac{1}{(cosec x - \cot x)} cosec x {cosec x - \cot x} d x = d t$
$\Rightarrow cosec x d x = d t \Rightarrow d x = \frac{d t}{cosec x} then$
$\Rightarrow I = \int cosec x \cdot t \cdot \frac{d t}{cosec x} = \int t d t$
$= \frac{t^{1 + 1}}{1 + 1} + c = \frac{t^{2}}{2} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c,]$
$= \frac{1}{2} {\log (cosec x - \cot x)}^{2} + c [∵ t = \log (cosec x - \cot x)]$

Indefinite Integrals exercise 18.9 question 33

Answer:

\frac{1}{4} \sin (x^{4}) + c

Hint:Use substitution method to solve this integral.
Given:

\int x^{3} \cos x^{4} d x

Solution:
$\begin{aligned} Let I = \int x^{3} \cos x^{4} d x \\ Put x^{4} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow d x = \frac{d t}{4 x^{3}} then \end{aligned}$
$\Rightarrow I = \int x^{3} \cos t \frac{d t}{4 x^{3}} = \frac{1}{4} \int \cos t d t$
$= \frac{1}{4} \int \sin t d t [∵ \int \cos x d x = \sin x + c]$
$= \frac{1}{4} \sin (x^{4}) + c [∵ t = x^{4}]$

Indefinite Integrals exercise 18.9 question 34

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

\int \frac{2 x}{\sqrt[3]{x^{2} - 1}} d x

Solution:
$\begin{aligned} Let I = \int \frac{2 x}{\sqrt[3]{x^{2} - 1}} d x \\ Put x^{2} - 1 = t \Rightarrow 2 x d x = d t then \end{aligned}$
$\Rightarrow I = \int \frac{1}{\sqrt[3]{t}} d t = \int \frac{1}{t^{\frac{1}{3}}} d t = \int t^{- \frac{1}{3}} d t$
$= \frac{t^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1} + c = \frac{t^{\frac{2}{3}}}{\frac{2}{3}} + c$ $[∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \frac{3}{2} t^{\frac{2}{3}} \\ = \frac{3}{2} {(x^{2} - 1)}^{\frac{2}{3}} + c [∵ t = x^{2} - 1] \end{aligned}$

Indefinite Integrals exercise 18.9 question 36

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

\int x^{3} \sin (x^{4} + 1) d x

Solution:

\begin{aligned} Let I = \int x^{3} \sin (x^{4} + 1) d x \\ Put x^{4} + 1 = t \Rightarrow 4 x^{3} d x = d t \\ \Rightarrow d x = \frac{d t}{4 x^{3}} 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}

[∵ \int \sin x d x = - \cos x + c]

= - \frac{\cos (x^{4} + 1)}{4} + c [∵ t = x^{4} + 1]

Indefinite Integrals exercise 18.9 question 37

Answer: $\tan (x e^{x}) + c$
Hint: Use substitution method to solve this integral.
Given:

\int \frac{(x + 1) e^{x}}{\cos^{2} (x e^{x})} d x

Solution:
$\begin{aligned} Let I = \int \frac{(x + 1) e^{x}}{\cos^{2} (x e^{x})} d x \\ Put x e^{x} = t \Rightarrow (x e^{x} + 1 . e^{x}) d x = d t \\ \Rightarrow (x + 1) e^{x} d x = d t then \end{aligned}$
$\Rightarrow I = \int \frac{1}{\cos^{2} t} d t$
$= \int \sec^{2} t d t [∵ \frac{1}{\cos x} = \sec x]$
$\begin{aligned} = \tan t + c \\ = \tan (x e^{x}) + c [∵ t = x e^{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 38

Answer: $\frac{1}{3} \sin (e^{x^{3}}) + c$
Hint: Use substitution method to solve this integral.
Given:

\int x^{2} e^{x^{3}} \cos (e^{x^{3}}) d x

Solution:
$\begin{aligned} Let I = \int x^{2} e^{x^{3}} \cos (e^{x^{3}}) d x \\ Put e^{x^{3}} = t \Rightarrow e^{x^{3}} \cdot 3 x^{2} d x = d t \\ \Rightarrow d x = \frac{d t}{e^{x^{3}} \cdot 3 x^{2}} then \end{aligned}$
$\Rightarrow I = \int e^{x^{3}} \cdot x^{2} \cdot \cos t \frac{d t}{e^{x^{3}} .3 x^{2}} = \frac{1}{3} \int \cos t d t$
$= \frac{1}{3} \sin t d t [∵ \int \cos x d x = \sin x + c]$
$= \frac{1}{3} \sin (e^{x^{3}}) + c [∵ t = e^{x^{3}}]$

Indefinite Integrals exercise 18.9 question 39

Answer: $\frac{1}{3} \sec^{3} (x^{2} + 3) + c$
Hint: Use substitution method to solve this integral.
Given:

\int 2 x \cdot \sec^{3} (x^{2} + 3) \tan (x^{2} + 3) d x

Solution:
$\begin{aligned} Let I = \int 2 x \cdot \sec^{3} (x^{2} + 3) \tan (x^{2} + 3) d x \\ Put x^{2} + 3 = t \Rightarrow 2 x d x = d t \\ \Rightarrow d x = \frac{d t}{2 x} then \end{aligned}$
$\begin{aligned} \Rightarrow I & = \int 2 x \sec^{3} t \cdot \tan t \frac{d t}{2 x} = \int \sec^{3} (t) \cdot \tan (t) d t \\ = \int \sec^{2} (t) \sec (t) \tan (t) d t \end{aligned}$
$Again Put \sec t = u \Rightarrow \sec t \tan t d t = d u then$
$I = \int u^{2} d u = [\frac{u^{2 + 1}}{2 + 1}] + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{array}{ll} = \frac{u^{3}}{3} + c = \frac{\sec^{3} t}{3} + c & [∵ \sec t = u] \end{array}$
$= \frac{1}{3} \sec^{3} (x^{2} + 3) + c [∵ t = x^{2} + 3]$

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 (\frac{x + 1}{x}) (\log x + x)^{2} d x

Solution:

Let I = \int (\frac{x + 1}{x}) (\log x + x)^{2} d x

\begin{aligned} = \int (\frac{x}{x} + \frac{1}{x}) (\log x + x)^{2} d x \\ = \int (1 + \frac{1}{x}) (\log x + x)^{2} d x \end{aligned}

Put \log x + x = t \Rightarrow (\frac{1}{x} + 1) d x = d t

\Rightarrow d x = \frac{1}{(1 + \frac{1}{x})} dt then

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

I = [\frac{t^{2 + 1}}{2 + 1}] + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]

= \frac{t^{3}}{3} + c = \frac{1}{3} (\log x + x)^{3} + c [∵ t = \log x + x]

Indefinite Integrals exercise 18.9 question 41

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

Indefinite Integrals exercise 18 .9 question 42

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

\int \log x \cdot \frac{\sin {1 + (\log x)^{2}}}{x} d x

Solution:
$\begin{aligned} Let I = \int \log x \cdot \frac{\sin {1 + (\log x)^{2}}}{x} d x \\ Put 1 + (\log x)^{2} = t \Rightarrow 2 \log x \cdot \frac{1}{x} \cdot d x = d t \\ \Rightarrow d x = \frac{x \cdot d t}{2 \log x} then \end{aligned}$
$\begin{aligned} \Rightarrow I & = \int \log x \cdot \frac{\sin t}{x} \cdot \frac{x d t}{2 \log x} d t \\ = \frac{1}{2} \int \sin t d t \end{aligned}$
$\begin{aligned} = \frac{1}{2} (- \cos t) + c \\ = - \frac{1}{2} \cos {1 + (\log x)^{2}} + c [∵ t = 1 + (\log x)^{2}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 43

Answer: $- \frac{1}{2 x} - \frac{1}{4} \sin (\frac{2}{x}) + c$
Hint: Use substitution method to solve this integral.
Given:

\int \frac{1}{x^{2}} \cos^{2} (\frac{1}{x}) d x

Solution:
$\begin{aligned} Let I = \int \frac{1}{x^{2}} \cos^{2} (\frac{1}{x}) d x \\ Put \frac{1}{x} = t \Rightarrow - \frac{1}{x^{2}} d x = d t \\ \Rightarrow d x = - x^{2} dt then \end{aligned}$
$I = \int \frac{1}{x^{2}} \cdot \cos^{2} t \cdot (- x^{2}) d t = - \int \cos^{2} t d t$
$= - \int {\frac{1 + \cos 2 t}{2}} d t$ $[\begin{array}{l} ∵ 2 \cos^{2} A - 1 = \cos 2 A \\ \Rightarrow 2 \cos^{2} A = 1 + \cos 2 A \\ \Rightarrow \cos^{2} A = \frac{1 + \cos 2 A}{2} \end{array}]$

\begin{aligned} = - \int {\frac{1}{2} + \frac{\cos 2 t}{2}} 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

[\begin{matrix} ∵ \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{matrix}]

= - \frac{1}{2 x} - \frac{1}{4} \sin (\frac{2}{x}) + 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:
$\begin{aligned} Let I = \int \sec^{4} x \tan x d x \\ Put \tan x = t \Rightarrow \sec^{2} x d x = d t \\ \Rightarrow d x = \frac{d t}{\sec^{2} x} then \end{aligned}$
$I = \int \sec^{4} x \cdot t \cdot \frac{d t}{\sec^{2} x} = \int \sec^{2} x \cdot t d t$
$= \int {1 + \tan^{2} x} . t d t [\begin{array}{l} ∵ \sec^{2} x - \tan^{2} x = 1 \\ \Rightarrow \sec^{2} x = 1 + \tan^{2} x \end{array}]$
$\begin{aligned} = \int (1 + t^{2}) . t d t [∵ \tan x = t] \\ = \int (t + t^{2} t) d t = \int (t + t^{3}) d t \\ = \int t d t + \int t^{3} d t \end{aligned}$
$= \frac{t^{1 + 1}}{1 + 1} + \frac{t^{3 + 1}}{3 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \frac{t^{2}}{2} + \frac{t^{4}}{4} + c \\ = \frac{1}{2} \tan^{2} x + \frac{1}{4} \tan^{4} x + c [∵ t = \tan x] \end{aligned}$

Indefinite Integrals exercise 18.9 question 45

Answer: $2 \sin (e^{\sqrt{x}}) + c$
Hint: Use substitution method to solve this integral.
Given:

\int \frac{e^{\sqrt{x}} \cos (e^{\sqrt{x}})}{\sqrt{x}} d x

Solution:
$\begin{aligned} Let I = \int \frac{e^{\sqrt{x}} \cos (e^{\sqrt{x}})}{\sqrt{x}} d x \\ Put e^{\sqrt{x}} = t \Rightarrow e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} d x = d t \\ \Rightarrow d x = \frac{2 \sqrt{x}}{e^{\sqrt{x}}} dt then \end{aligned}$
$\begin{aligned} I = \int \frac{e^{\sqrt{x}} \cos t}{\sqrt{x}} \frac{2 \sqrt{x}}{e^{\sqrt{x}}} d t \\ = 2 \int \cos t d t \end{aligned}$
$= 2 \sin t + c [∵ \int \cos x d x = \sin x + c]$
$= 2 \sin (e^{\sqrt{x}}) + c [∵ t = e^{\sqrt{x}}]$

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:
$Let I = \int \frac{\cos^{5} x}{\sin x} d x$
$\begin{aligned} Put \sin x = t \Rightarrow \cos x d x = d t \\ \Rightarrow d x = \frac{d t}{\cos x} then \end{aligned}$
$I = \int \frac{\cos^{5} x}{t} \frac{d t}{\cos x} = \int \frac{\cos^{4} x}{t} d t$
$= \int \frac{{(\cos^{2} x)}^{2}}{t} d t = \int \frac{{(1 - \sin^{2} x)}^{2}}{t} d t$ $[\begin{array}{l} ∵ \cos^{2} x + \sin^{2} x = 1 \\ \Rightarrow \cos^{2} x = 1 - \sin^{2} x \end{array}]$
$= \int \frac{{(1 - t^{2})}^{2}}{t} d t$ $[∵ \sin x = t]$
$= \int {\frac{1 + {(t^{2})}^{2} - 2 t^{2}}{t}} d t$ $[∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b]$
$\begin{aligned} = \int {\frac{1 + (t^{4}) - 2 t^{2}}{t}} d t \\ = \int {\frac{1}{t} + \frac{t^{4}}{t} - \frac{2 t^{2}}{t}} d t \\ = \int {\frac{1}{t} + t^{3} - 2 t} d t \end{aligned}$
$= \int \frac{1}{t} d t + \int t^{3} d t - 2 \int t d t$
$= \log | t | + \frac{t^{3 + 1}}{3 + 1} - 2 \frac{t^{1 + 1}}{1 + 1} + c$ $[∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \log | t | + \frac{t^{4}}{4} - 2 \frac{t^{2}}{2} + c \\ = \frac{1}{4} \sin^{4} x - \sin^{2} x + \log | \sin x | + c [∵ t = \sin x] \end{aligned}$

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:
$\begin{aligned} Let I = \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x \\ Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \\ \Rightarrow d x = 2 \sqrt{x} d t then \end{aligned}$
$I = \int \frac{\sin t}{\sqrt{x}} 2 \sqrt{x} d t = 2 \int \sin t d t$
$= 2 [- \cos t] + c [∵ \int \sin x d x = - \cos x + c]$
$\begin{aligned} = - 2 \cos t + c \\ = - 2 \cos \sqrt{x} + c [∵ t = \sqrt{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 48

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

= - \cot (x e^{x}) + c [∵ t = x e^{x}]

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

Solution:
$\begin{aligned} Let I = \int 5^{x + \tan^{- 1} x} \cdot (\frac{x^{2} + 2}{1 + x^{2}}) d x \\ Put x + \tan^{- 1} x = t \\ \Rightarrow (1 + \frac{1}{1 + x^{2}}) d x = d t \Rightarrow (\frac{(1 + x^{2}) + 1}{1 + x^{2}}) d x = d t \\ \Rightarrow (\frac{1 + x^{2} + 1}{1 + x^{2}}) d x = d t \Rightarrow (\frac{2 + x^{2}}{1 + x^{2}}) d x = d t then \end{aligned}$
$I = \int 5^{t} d t = \frac{5^{t}}{\log 5} + c [∵ \int a^{x} d x = \frac{a^{x}}{\log a} + c]$
$= \frac{5^{x + \tan^{- 1} x}}{\log 5} + c [∵ t = x + \tan^{- 1} x]$

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:
$\begin{aligned} Let I = \int \frac{e^{m \sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x \\ Put m \sin^{- 1} x = t \Rightarrow m \frac{1}{\sqrt{1 - x^{2}}} d x = d t \\ \Rightarrow d x = \frac{\sqrt{1 - x^{2}}}{m} dt then \end{aligned}$
$I = \int \frac{e^{t}}{\sqrt{1 - x^{2}}} \cdot \frac{\sqrt{1 - x^{2}}}{m} d t \Rightarrow \frac{1}{m} \int e^{t} d t$
$= \frac{1}{m} e^{t} + c [∵ \int e^{x} d x = e^{x} + c]$
$= \frac{1}{m} e^{m \sin^{- 1} x} + c [∵ t = m \sin^{- 1} x]$

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:
$\begin{aligned} Let I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x \\ Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \\ \Rightarrow d x = 2 \sqrt{x} d t then \end{aligned}$
$\begin{aligned} I & = \int \frac{\cos t}{\sqrt{x}} \cdot 2 \sqrt{x} d t \Rightarrow 2 \int \cos t d t \\ = 2 \sin t + c [∵ \int \cos x d x = \sin x + c] \\ = 2 \sin \sqrt{x} + c [∵ t = \sqrt{x}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 52

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

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:
$Let I = \int \frac{\sin (\log x)}{x} d x$
$\begin{aligned} Put \log x = t \Rightarrow \frac{1}{x} d x = d t \Rightarrow d x = x d t \\ Then I = \int \frac{\sin t}{x} \cdot x d t = \int \sin t d t \end{aligned}$
$\begin{array}{ll} = - \cos t + c & ∵ \sin x d x = - \cos x + c] \\ = - \cos (\log x) + c & [∵ t = \log x] \end{array}$

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:
$Let I = \int \frac{e^{m \tan^{- 1} x}}{1 + x^{2}} d x$
$\begin{aligned} Put m \tan^{- 1} x = t \Rightarrow m \frac{1}{1 + x^{2}} d x = d t \\ \Rightarrow d x = \frac{1 + x^{2}}{m} d t \end{aligned}$
$\begin{aligned} I = \int \frac{e^{t}}{1 + x^{2}} \cdot \frac{1}{m} \cdot (1 + x^{2}) d t \\ = \int \frac{e^{t}}{m} d t \end{aligned}$
$= \frac{1}{m} \int e^{t} d t = \frac{1}{m} (e^{t}) + c [∵ \int e^{x} d x = e^{x} + c]$
$= \frac{1}{m} e^{m \tan^{- 1} x} + c [∵ t = m \tan^{- 1} x]$

Indefinite Integrals exercise 18.9 question 55

Answer: $\frac{1}{6 a^{2}} [{(x^{2} + a^{2})}^{\frac{3}{2}} - {(x^{2} - a^{2})}^{\frac{3}{2}}] + c$
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:
$Let I = \int \frac{x}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}} d x$
On Rationalising we get

I = \int (\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}}}) d x

= \int (\frac{x (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}{(\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}) (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}) d x

= \int \frac{x (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}{{(\sqrt{x^{2} + a^{2}})}^{2} - {(\sqrt{x^{2} - a^{2}})}^{2}} d x

= \int \frac{x (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}{(x^{2} + a^{2}) - (x^{2} - a^{2})} d x

= \int \frac{x (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}{x^{2} + a^{2} - x^{2} + a^{2}} d x

= \int \frac{x (\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} - a^{2}})}{2 a^{2}} d x

= \frac{1}{2 a^{2}} \int (x \sqrt{x^{2} + a^{2}} - x \sqrt{x^{2} - a^{2}}) 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)

Now \frac{1}{2 a^{2}} \int x \sqrt{x^{2} + a^{2}} d x

\begin{aligned} Put x^{2} + a^{2} = t \Rightarrow 2 x d x = d t \Rightarrow d x = \frac{a ı}{2 x} \\ 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} [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]

= \frac{1}{6 a^{2}} {(x^{2} + a^{2})}^{\frac{3}{2}} + c_{1}

......

(i i)

(∵ t = x^{2} + a^{2})

and, \frac{1}{2 a^{2}} \int x \sqrt{x^{2} - a^{2}} d x

\begin{aligned} 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 then, \end{aligned}

= \frac{1}{4 a^{2}} [\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + c_{2} [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]

\begin{aligned} = \frac{1}{4 a^{2}} [\frac{u^{\frac{3}{2}}}{\frac{3}{2}}] + c_{2} \\ = \frac{1}{4 a^{2}} \cdot \frac{2}{3} u_{}^{\frac{3}{2}} + c_{2} \end{aligned}

= \frac{1}{6 a^{2}} {(x^{2} - a^{2})}^{\frac{3}{2}} + c_{2}

....

(i i i)

(∵ u = x^{2} - a^{2})

Putting the values of eqn

(i i)

and eqn

(i i i)

(i)

then,

\begin{aligned} I = \frac{1}{6 a^{2}} {(x^{2} + a^{2})}^{\frac{3}{2}} + c_{1} - \frac{1}{6 a^{2}} {(x^{2} - a^{2})}^{\frac{3}{2}} - c_{2} \\ = \frac{1}{6 a^{2}} [{(x^{2} + a^{2})}^{\frac{3}{2}} - {(x^{2} - a^{2})}^{\frac{3}{2}}] + c (∵ c = c_{1} - c_{2}) \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} {(\tan^{- 1} x^{2})}^{2} + c

Solution:
$I = \int \frac{x \tan^{- 1} x^{2}}{1 + x^{4}} d x$
$\begin{aligned} Put \tan^{- 1} x^{2} = t \Rightarrow \frac{1}{1 + {(x^{2})}^{2}} 2 x d x = d t \\ \Rightarrow \frac{2 x}{1 + x^{4}} d x = d t \Rightarrow d x = \frac{1 + x^{4}}{2 x} d t \\ Then, I = \int \frac{x t}{1 + x^{4}} \cdot \frac{1 + x^{4}}{2 x} d t = \frac{1}{2} \int t d t \end{aligned}$
$= \frac{1}{2} \frac{t^{1 + 1}}{1 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = \frac{1}{2} \cdot \frac{t^{2}}{2} + c = \frac{1}{4} t^{2} + c \\ = \frac{1}{4} {(\tan^{- 1} x^{2})}^{2} + c [∵ t = \tan^{- 1} x^{2}] \end{aligned}$

Indefinite Integrals exercise 18.9 question 57

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

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

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:
$Let I = \int \frac{\sin (2 + 3 \log x)}{x} d x$
$\begin{aligned} Put 2 + 3 \log x = t \Rightarrow 3 \cdot \frac{1}{x} d x = d t \Rightarrow d x = \frac{x}{3} d t \\ Then, I = \int \frac{\sin t}{x} \cdot \frac{x}{3} d t = \frac{1}{3} \int \sin t d t \end{aligned}$
$= \frac{1}{3} (- \cos t) + c [∵ \int \sin x d x = - \cos x + c]$
$\begin{aligned} = \frac{- 1}{3} \cos t + c \\ = \frac{- 1}{3} \cos (2 + 3 \log x) + c \end{aligned}$ $[∵ t = 2 + 3 \log x]$

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:

Let I = \int x e^{x^{2}} d x

\begin{aligned} Put x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow d x = \frac{1}{2 x} d t \\ 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}

[∵ \int e^{x} d x = e^{x} + c]

= \frac{1}{2} e^{x^{2}} + c

[∵ t = x^{2}]

Indefinite Integrals exercise 18.9 question 60

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

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

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:
$Let I = \int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x$
$\begin{aligned} Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \\ Then, I = \int \frac{\sec^{2} t}{\sqrt{x}} \cdot 2 \sqrt{x} d t = 2 \int \sec^{2} t d t \end{aligned}$
$= 2 \tan t + c [∵ \int \sec^{2} x d x = \tan x + c]$
$= 2 \tan \sqrt{x} + c [∵ t = \sqrt{x}]$

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:
$Let I = \int \frac{x + \sqrt{x + 1}}{x + 2}$
$Put x + 1 = t^{2} \Rightarrow d x = 2 t d t then$
$I = \int \frac{(t^{2} - 1) + \sqrt{t^{2}}}{t^{2} + 1} 2 t d t = 2 \int \frac{(t^{2} - 1) + t}{t^{2} + 1} t d t [\begin{array}{l} ∵ x + 1 = t^{2} \\ \Rightarrow x = t^{2} - 1 \end{array}]$
$\Rightarrow I = 2 \int (\frac{t^{2} + t - 1}{t^{2} + 1}) t d t = 2 \int (\frac{t^{2} \cdot t + t . t - t}{t^{2} + 1}) d t$
$\begin{aligned} \Rightarrow I = 2 \int (\frac{t^{3} + t^{2} - t}{t^{2} + 1}) d t = 2 \int {\frac{t^{3}}{t^{2} + 1} + \frac{t^{2}}{t^{2} + 1} - \frac{t}{t^{2} + 1}} d t \\ \Rightarrow I = 2 [\int \frac{t^{3}}{t^{2} + 1} d t + \int \frac{t^{2}}{t^{2} + 1} d t - \int \frac{t}{t^{2} + 1} d t] \end{aligned}$
We can write

I = 2 (I_{1} + I_{2} - I_{3})

. . . . . . . . . (i)

\begin{array}{r} where I_{1} = \int \frac{t^{3}}{t^{2} + 1} d t \end{array}

I_{2} = \int \frac{t^{2}}{t^{2} + 1} d t

and I_{3} = \int \frac{t}{t^{2} + 1} d t

Now I_{1} = \int \frac{t^{3}}{t^{2} + 1} d t = \int (\frac{t^{3} + t - t}{t^{2} + 1}) d t

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

= \int (\frac{t (t^{2} + 1)}{t^{2} + 1} - \frac{t}{t^{2} + 1}) d t = \int (t - \frac{t}{t^{2} + 1}) d t

= \int t d t - \int \frac{t}{t^{2} + 1} d t

\begin{aligned} put t^{2} + 1 = u \Rightarrow 2 t d t = d u \Rightarrow t d t = \frac{d u}{2} 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} [\begin{array}{l} ∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c \\ \int \frac{1}{x} d x = \log | x | + c \end{array}]

= \frac{t^{2}}{2} - \frac{1}{2} \log | 1 + t^{2} | + c_{1}

. . . . . . . (i i)

[∵ u = t^{2} + 1]

And I_{2} = \int \frac{t^{2}}{t^{2} + 1} d t = \int (\frac{t^{2} + t - t}{t^{2} + 1}) d t

\begin{aligned} = \int (\frac{(t^{2} + 1) - 1}{t^{2} + 1}) d t = \int (\frac{t^{2} + 1}{t^{2} + 1} - \frac{1}{t^{2} + 1}) d t \\ = \int (1 - \frac{1}{t^{2} + 1}) 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} [\begin{array}{l} ∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c \\ \int \frac{1}{x} d x = \log | x | + c \end{array}]

= t - \tan^{- 1} (t) + c_{2}

. . . . . . . (i i i)

\begin{aligned} Also I_{3} = \int \frac{t}{t^{2} + 1} d t \\ put t^{2} + 1 = p \Rightarrow 2 t d t = d p \Rightarrow t d t = \frac{d p}{2} 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 | 1 + t^{2} | + c_{3}

. . . . . . (i v)

\begin{aligned} Substituting the values of I_{1}, I_{2}, I_{3} from eqn(ii), (iii) and (iv) in (i) then \\ I = 2 [\frac{t^{2}}{2} - \frac{1}{2} \log | 1 + t^{2} | + c_{1} + t - \tan^{- 1} (t) + c_{2} - \frac{1}{2} \log | 1 + t^{2} | - c_{3}] \end{aligned}

= 2 [\frac{t^{2}}{2} + t - \tan^{- 1} (t) - (\frac{1}{2} + \frac{1}{2}) \log | 1 + t^{2} | + c_{1} + c_{2} - c_{3}]

= 2 [\frac{t^{2}}{2} + t - \tan^{- 1} (t) - \log | 1 + t^{2} | + c_{4}] [∵ c_{4} = c_{1} + c_{2} - c_{3}]

\begin{aligned} = 2 \cdot \frac{t^{2}}{2} + 2 t - 2 \tan^{- 1} (t) - 2 \log | 1 + t^{2} | + 2 c_{4} \\ = t^{2} + 2 t - 2 \tan^{- 1} (t) - 2 \log | 1 + t^{2} | + c \end{aligned}

\begin{aligned} [ since x + 1 = t^{2}] \\ = (x + 1) + 2 \sqrt{x + 1} - 2 \tan^{- 1} \sqrt{x + 1} - 2 \log | 1 + x + 1 | + c \\ 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:
$\begin{aligned} let I = \int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x \\ Putting 5^{5^{5^{x}}} = t \end{aligned}$
$\Rightarrow (5^{5^{5^{x}}} \cdot \log {5.5}^{5^{x}} \cdot \log {5.5}^{x} \log 5) d x = d t$
$\begin{aligned} \Rightarrow 5^{5^{5^{5}}} 5^{5^{x}} 5^{x} (\log 5)^{3} d x = d t \\ \Rightarrow (5^{5^{5^{x}}} 5^{5^{x}} 5^{x}) d x = \frac{d t}{(\log 5)^{3}} then \end{aligned}$
$I = \int \frac{d t}{(\log 5)^{3}} = \frac{1}{(\log 5)^{3}} \int 1 d t$
$= \frac{1}{(\log 5)^{3}} \int t^{0} d t = \frac{1}{(\log 5)^{3}} \frac{t^{0 + 1}}{0 + 1} [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$= \frac{1}{(\log 5)^{3}} \cdot t + c = \frac{1}{(\log 5)^{3}} \cdot 5^{5^{5^{x}}} + c [∵ 5^{5^{5^{x}}} = t]$

Indefinite Integrals exercise 18.9 question 65

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

Given: $\int \frac{1}{x \sqrt{x^{4} - 1}} d x$
Solution:
$\begin{aligned} let I = \int \frac{1}{x \sqrt{x^{4} - 1}} d x = \int \frac{1}{x \sqrt{{(x^{2})}^{2} - 1}} d x \\ Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow d x = \frac{d t}{2 x} then \end{aligned}$
$I = \int \frac{1}{x \sqrt{t^{2} - 1}} \frac{d t}{2 x} = \frac{1}{2} \int \frac{1}{x^{2} \sqrt{t^{2} - 1}} d t$
$= \frac{1}{2} \int \frac{1}{t \sqrt{t^{2} - 1}} d t [∵ x^{2} = t]$
$= \frac{1}{2} \sec^{- 1} t + c [∵ \int \frac{1}{x \sqrt{x^{2} - 1}} d x = \sec^{- 1} x + c]$
$= \frac{1}{2} \sec^{- 1} (x^{2}) + c [∵ t = x^{2}]$

Indefinite Integrals exercise 18.9 question 66

Answer:

2 \sqrt{e^{x} - 1} - 2 \tan^{- 1} (\sqrt{e^{x} - 1}) + c

Hint: Use substitution method to solve this integral

Given: $\int \sqrt{e^{x} - 1} d x$
Solution:
$\begin{aligned} let I = \int \sqrt{e^{x} - 1} d x \\ Putting e^{x} - 1 = t^{2} \Rightarrow e^{x} d x = 2 t d t \Rightarrow d x = \frac{2 t \cdot d t}{e^{x}} then \end{aligned}$
$I = \int \sqrt{t^{2}} \frac{2 t d t}{e^{x}}$
$= 2 \int \frac{t \cdot t}{t^{2} + 1} d t [∵ e^{x} - 1 = t^{2} \Rightarrow t^{2} + 1 = e^{x}]$
$\begin{aligned} = 2 \int \frac{t^{2}}{t^{2} + 1} d t \\ = 2 \int (\frac{t^{2} + 1 - 1}{t^{2} + 1}) d t \end{aligned}$
$\begin{aligned} = 2 \int (\frac{(t^{2} + 1) - 1}{t^{2} + 1}) d t \\ = 2 \int (\frac{(t^{2} + 1)}{t^{2} + 1} - \frac{1}{t^{2} + 1}) d t = 2 \int (1 - \frac{1}{t^{2} + 1}) d t \end{aligned}$
$= 2 \int 1 d t - 2 \int \frac{1}{t^{2} + 1} d t = 2 \int t^{0} d t - 2 \int \frac{1}{t^{2} + 1} d t$
$= 2 \frac{t^{0 + 1}}{0 + 1} - 2 \tan^{- 1} (t) + c$ $[\begin{array}{l} ∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c \\ \int \frac{1}{1 + x^{2}} d x = \tan^{- 1} x + c \end{array}]$
$\begin{aligned} = 2 t - 2 \tan^{- 1} (t) + c \\ = 2 \sqrt{e^{x} - 1} - 2 \tan^{- 1} (\sqrt{e^{x} - 1}) + c \end{aligned}$ $[∵ t^{2} = e^{x} - 1 \Rightarrow t = \sqrt{e^{x} - 1}]$

Indefinite Integrals exercise 18.9 question 67

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

Given: $\int \frac{1}{(x + 1) (x^{2} + 2 x + 2)} d x$
Solution:
$let I = \int \frac{1}{(x + 1) (x^{2} + 2 x + 2)} d x$
$\begin{aligned} = \int \frac{1}{(x + 1) (x^{2} + 2 x + 1 + 1)} d x \\ = \int \frac{1}{(x + 1) {(x^{2} + 2 x + 1) + 1}} d x \end{aligned}$
$= \int \frac{1}{(x + 1) {(x + 1)^{2} + 1}} d x$ $[∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b]$
$Putting x + 1 = \tan u$

. . . . . . (i)

\Rightarrow d x = \sec^{2} u d u then

I = \int \frac{1}{\tan u {\tan^{2} u + 1}} \sec^{2} u d u

= \int \frac{1}{\tan u {\sec^{2} u}} \sec^{2} u d u [∵ 1 + \tan^{2} u = \sec^{2} u]

= \int \frac{1}{\tan u} d u = \int \cot u d u

= \log | \sin u | + c

. . . . . . . . (i i)

[∵ \int \cot x d x = \log | \sin x | + c]

Also, from (i) x + 1 = \tan u = \frac{\sin u}{\cos u} [∵ \tan x = \frac{\sin x}{\cos x}]

\begin{aligned} \Rightarrow (x + 1) \cos u = \sin u \\ \Rightarrow (x + 1)^{2} \cos^{2} u = \sin^{2} u [Squaring on both sides] \\ \Rightarrow (x + 1)^{2} (1 - \sin^{2} u) = \sin^{2} u [∵ \cos^{2} x = 1 - \sin^{2} x] \\ \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} = [1 + (x + 1)^{2}] \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} [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b]

\Rightarrow \sin u = \sqrt{\frac{(x + 1)^{2}}{x^{2} + 2 x + 2}} = \frac{(x + 1)}{\sqrt{x^{2} + 2 x + 2}}

. . . . . . . . (i i i)

\begin{aligned} From (ii) and (iii) we get \\ I = \log | \frac{x + 1}{\sqrt{x^{2} + 2 x + 2}} | + c \end{aligned}

Indefinite Integrals exercise 18.9 question 68

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

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

Indefinite Integrals exercise 18.9 question 69

Answer:

- \frac{20}{3} {(5 - x^{2})}^{\frac{3}{2}} + \frac{4}{5} {(5 - x^{2})}^{\frac{5}{2}} + c

Hint: Use substitution method to solve this integral

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

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:
$\begin{aligned} Let I = \int \frac{1}{\sqrt{x} + x} d x \\ Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \\ \Rightarrow d x = 2 \sqrt{x} d t then \end{aligned}$
$\begin{aligned} I = \int \frac{1}{t + t^{2}} 2 \sqrt{x} d t \\ = \int \frac{1}{t (1 + t)} 2 \sqrt{x} d t = 2 \int \frac{1}{1 + t} d t \end{aligned}$ $[\begin{array}{l} ∵ \sqrt{x} = t \\ \Rightarrow x = t^{2} \end{array}]$
$\begin{aligned} Again put 1 + t = u \Rightarrow d t = d u then \\ I = 2 \int \frac{1}{u} d u = \frac{1}{2} \log | u | + c \end{aligned}$ $[∵ \int \frac{1}{x} d x = \log | x | + c]$
$= \frac{1}{2} \log | t + 1 | + c [∵ u = 1 + t]$
$= \frac{1}{2} \log | 1 + \sqrt{x} | + c [∵ t = \sqrt{x}]$

Indefinite Integrals exercise 18.9 question 71

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

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

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:
$\begin{aligned} Let I = \int \frac{\sin^{5} x}{\cos^{4} x} d x \\ Put \cos x = t \Rightarrow - \sin x d x = d t \\ \Rightarrow d x = \frac{d t}{- \sin x} then \end{aligned}$
$I = \int \frac{\sin^{5} x}{t^{4}} \frac{d t}{- \sin x} = - \int \frac{\sin^{4} x}{t^{4}} d t$
$= - \int \frac{{(\sin^{2} x)}^{2}}{t^{4}} d t = - \int \frac{{(1 - \cos^{2} x)}^{2}}{t^{4}} d t$
$I = - \int \frac{(1 + \cos^{4} x - 2 \cos^{2} x)}{t^{4}} d t [∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b]$
$= - \int [\frac{1 + t^{4} - 2 t^{2}}{t^{4}}] d t [∵ \cos x = t]$
$\begin{aligned} = - \int [\frac{1}{t^{4}} + \frac{t^{4}}{t^{4}} - \frac{2 t^{2}}{t^{4}}] d t \\ = - \int [\frac{1}{t^{4}} + 1 - \frac{2}{t^{2}}] d t \end{aligned}$
$\begin{aligned} = - \int (t^{- 4} + 1 - 2 t^{- 2}) d t \\ = - \int t^{- 4} d t - \int t^{0} d t + 2 \int t^{- 2} d t \end{aligned}$
$= - \frac{t^{- 4 + 1}}{- 4 + 1} - \frac{t^{0 + 1}}{0 + 1} + 2 \frac{t^{- 2 + 1}}{- 2 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$
$\begin{aligned} = - \frac{t^{- 3}}{- 3} - t + 2 \frac{t^{- 1}}{- 1} + c \\ = \frac{1}{3 t^{3}} - t - 2 \cdot \frac{t^{- 1}}{1} + c \end{aligned}$
$= \frac{1}{3 \cos^{2} x} - \cos x - \frac{2}{\cos x} + c [∵ t = \cos x]$

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