NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals 2021

NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:03 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 7 Integrals is one of the most important subjects for those who want to prepare for the advanced levels of Mathematics and also want to score higher in their exams, studying this chapter is crucial. The NCERT Exemplar solutions for Class 12 Maths chapter 7 holds major importance in both board exams of CBSE and also in competitive exams. Students can utilize the NCERT exemplar Class 12 Maths solutions chapter 7 PDF download function to download the solutions and make learning even more convenient and easy to follow and understand

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Question:1

Verify the following:
$\int \frac{2 x-1}{2 x+3} d x=x-\log \left|(2 x+3)^{2}\right|+C$

Answer:

$\begin{aligned} &\text { To Verify; }\\ &\int \frac{2 x-1}{2 x+3} d x=x-\log \left|(2 x+3)^{2}\right|+C\\ &\text { LHS: } \int \frac{2 x-1}{2 x+3} d x\\ &=\int \frac{2 x+3-4}{2 x+3} d x\\ &=\int \mathrm{dx}-\int \frac{4}{2 \mathrm{x}+3} \mathrm{dx} \end{aligned}$

Let t = 2x + 3
$\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2} \\ =\int \mathrm{dx}-4 \int \frac{ \mathrm{dt}}{\mathrm{2t}} \\ =\mathrm{x}-2 \log |\mathrm{t}|+\mathrm{C} \\ \because \int \mathrm{dx}=\mathrm{x} \text { and } \left.\int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}|\right] \\ =\mathrm{x}-\log \left|(2 \mathrm{x}+3)^{2}\right|+\mathrm{C}=\mathrm{RHS} \\ \left[\because 2 \log |\mathrm{x}|=\log \left|\mathrm{x}^{2}\right|\right]$

Hence Verified

Question:2

Verify the following:

$\int \frac{2 x+3}{x^{2}+3 x} d x=\log \left|x^{2}+3 x\right|+C$

Answer:

To Verify;
$\\ \int \frac{2 x+3}{x^{2}+3 x} d x=\log \left|x^{2}+3 x\right|+C\\ \mathrm{LHS}=\int \frac{2 \mathrm{x}+3}{\mathrm{x}^{2}+3 \mathrm{x}} \mathrm{dx}Let; t=x^2 + 3x\\ \Rightarrow dt = 2x + 3\\$
$\\ =\int \frac{d t}{t}=\log |t|+C \\ {\left[\because \int \frac{1}{x} d x=\log |x|\right]} \\ \Rightarrow \log \left|x^{2}+3 x\right|+C=R H S \\$

$[\because t = x^2 + 3x]$

Hence Verified

Question:3

Evaluate the following:
$\int \frac{\left(x^{2}+2\right) d x}{x+1}$

Answer:

Given; $\int \frac{\left(x^{2}+2\right) d x}{x+1}$
Let t = x + 1
$\\\Rightarrow dx = dt \\ =\int \frac{\left((\mathrm{t}-1)^{2}+2\right)}{\mathrm{t}} \mathrm{dt} \\ =\int \frac{\mathrm{t}^{2}-2 \mathrm{t}+1+2}{\mathrm{t}} \mathrm{dt} \\ =\int(\mathrm{t}) \mathrm{dt}-\int 2 \mathrm{dt}+\int \frac{3}{\mathrm{t}} \mathrm{dt} \\ \because \int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1} \text { and } \left.\int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}|\right]$

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

Question:4

Evaluate the following:
$\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x$

Answer:

Given; $\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x$
As we know $n log x = log x^n$

$\begin{aligned} &=\int \frac{e^{\log x^{6}}-e^{\log x^{5}}}{e^{\log x^{4}}-e^{\log x^{2}}} d x\\ &=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x\\ &\text { Take } x^{3} \text { common out of numerator and denominator to get, }\\ &=\int \frac{x^{3}\left(x^{3}-x^{2}\right)}{x^{3}(x-1)} d x\\ &=\int \frac{\left(x^{3}-x^{2}\right)}{(x-1)} d x \end{aligned}$

$\begin{aligned} &=\int \frac{x^{2}(x-1)}{(x-1)} d x\\ &=\int x^{2} d x\\ &\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\\ &\text { So, }\\ &=\frac{x^{3}}{3}+c \end{aligned}$

Question:5

Evaluate the following:
$\int \frac{(1+\cos x)}{x+\sin x} d x$

Answer:

Given; $\int \frac{(1+\cos x)}{x+\sin x} d x$
Let t = x + sin x
$\\\Rightarrow dt = (1 + cos x) dx \\ =\int \frac{1}{t} d t \\ =\log |t|+C \\ =\log |x+\sin x|+C$

Question:6

Evaluate the following:
$\int \frac{d x}{1+\cos x}$

Answer:

Given; $\int \frac{d x}{1+\cos x}$
$\\ =\int \frac{(1-\cos x)}{(1+\cos x)(1-\cos x)} d x \\ =\int \frac{1-\cos x}{1-\cos ^{2} x} d x \\ =\int \frac{1-\cos x}{\sin ^{2} x} d x \\ {\left[\frac{1}{\sin ^{2} x}=\operatorname{cosec}^{2} x \text { and } \frac{\cos x}{\sin ^{2} x}=\operatorname{cosec} x \cot x\right]}$
$\\\begin{aligned} &=\int\left(\operatorname{cosec}^{2} x-\operatorname{cosec} x \cot x\right) d x\\ &\text { As we know, }\\ &\int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+c\\ &\int \operatorname{cosec}^{2} x d x=-\cot x+c\\ &=\operatorname{cosec} x-\cot x+C \end{aligned}\\$

Question:7

Evaluate the following:
$\int \tan ^{2} x \sec ^{4} x d x$

Answer:

Given; $$ \int $ tan\textsuperscript{2} x sec\textsuperscript{4} x dx\\$ $=$ \int $ tan\textsuperscript{2} x sec\textsuperscript{2} x (1+ tan\textsuperscript{2} x) dx\\$ Let; tan x = y
$\\$ \Rightarrow $ sec\textsuperscript{2} x dx = dy\\ \\ =$ \int $ (y\textsuperscript{2}+y\textsuperscript{4} )dy\\$

$\\ =\frac{y^{3}}{3}+\frac{y^{5}}{5}+C \\ =\frac{\tan ^{3} x}{3}+\frac{\tan ^{5} x}{5}+C$

Question:8

Evaluate the following:
$\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x$

Answer:

Given; $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x$
$\\ =\int \begin{array}{c} \sin x+\cos x \\ \sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \\ =\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x \end{array} \\ {\left[\because \sin 2 x=2 \sin x \cos x \text { and } \sin ^{2} x+\cos ^{2} x=1\right]} \\ =\int 1 \mathrm{dx} \\ =x+C$

Question:9

Evaluate the following:
$\int \sqrt{1+\sin x} d x$

Answer:

Given; $\int \sqrt{1+\sin x} d x$

$\\ =\int \sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ {\left[\because \sin 2 x=2 \sin x \cos x \text { and } \sin ^{2} x+\cos ^{2} x=1\right]} \\ =\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x \\ =\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ =2 \sin \frac{x}{2}-2 \cos \frac{x}{2}+c$

Question:10

Evaluate the following:
$\int \frac{x}{\sqrt{x}+1} d x \: \: \: (Hint: Put \sqrt{x} = z)$

Answer:

Given;

Let $z = \sqrt{x}$

$\\ \Rightarrow x = z\textsuperscript{2}\\ \\ \Rightarrow dx = 2z dz\\ =\int \frac{z^{2}}{z+1} 2 z d z\\ =2 \int \frac{\mathrm{z}^{3}}{\mathrm{z}+1} \mathrm{~d} z$$ $\\ Let; $t=z+1$\\ i.e. $t=\sqrt{x}+1$ \\ \Rightarrow dt $=\mathrm{dz}$$ $\\ =2 \int \frac{(\mathrm{t}-1)^{3}}{\mathrm{t}} \mathrm{dt} \\ =2 \int \frac{\mathrm{t}^{3}-3 \mathrm{t}^{2}+3 \mathrm{t}-1}{\mathrm{t}} \mathrm{dt} \\ =2 \int\left(\mathrm{t}^{2}-3 \mathrm{t}+3-\frac{1}{\mathrm{t}}\right) \mathrm{dt} \\ \because \int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1} \text { and } \left.\int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}|\right] \\$

$\\ =\frac{2 t^{3}}{3}-3 t^{2}+3 t-\log |t|+C \\ =\frac{2(\sqrt{x}+1)^{3}}{3}-3(\sqrt{x}+1)^{2}+3(\sqrt{x}+1)-\log |\sqrt{x}+1|+C$

Question:11

Evaluate the following:
$\int \sqrt{\frac{a+x}{a-x}}$

Answer:

Given, $\int \sqrt{\frac{a+x}{a-x}}$
$\\ =\int \frac{\sqrt{a+x}}{\sqrt{a-x}} \times \frac{\sqrt{a+x}}{\sqrt{a+x}} d x \\ =\int \frac{a+x}{\sqrt{a^{2}-x^{2}}} d x \\ =a \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x-\frac{1}{2} \int \frac{-2 x}{\sqrt{a^{2}-x^{2}}} d x$

$\\ \begin{aligned} &\text { Let } t=a^{2}-x^{2}\\ & \Rightarrow-2x \mathrm{dx}=\mathrm{dt}\\ &=\mathrm{a} \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)-\frac{1}{2} \int \frac{1}{\sqrt{\mathrm{t}}} \mathrm{dt}\\ &\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]_{\text {and }}\left[\int \frac{1}{\sqrt{x}} d x=2 \sqrt{x}\right]\\ &=a \sin \left(\frac{x}{a}\right)-\frac{1}{2} \times 2 \sqrt{t}\\ &=a \sin \left(\frac{x}{a}\right)-\sqrt{a^{2}-x^{2}}+c \end{aligned}$

Question:12

Evaluate the following:
$\int \frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}} d x$ (Hint : Put x = z⁴)

Answer:

$\begin{aligned} &\text { Given}\\ &\int \frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}} d x\\ &\text { Let } x=z^{4}\\ &\Rightarrow \mathrm{dx}=4 \mathrm{z}^{3} \mathrm{dz}\\ &=\int \frac{\mathrm{z}^{2}}{1+\mathrm{z}^{3}} 4 \mathrm{z}^{3} \mathrm{~d} z\\ &=4 \int \frac{\mathrm{z}^{5}}{1+\mathrm{z}^{3}} \mathrm{~d} \mathrm{z}\\ &=\frac{4}{3} \int \frac{\mathrm{z}^{3}}{1+\mathrm{z}^{3}} 3 \mathrm{z}^{2} \mathrm{~d} \mathrm{z} \end{aligned}$
$\\ \begin{aligned} &\text { Let } t=1+z^{3}\left[\begin{array}{l} \left.1 . e . t=1+x^{3 / 4}\right] \end{array}\right.\\ &\Rightarrow \mathrm{dt}=3 \mathrm{z}^{2} \mathrm{~d} z\\ &=\frac{4}{3} \int \frac{t-1}{t} d t \end{aligned}$
$\\ =\frac{4}{3} \int \mathrm{dt}-\frac{4}{3} \int \frac{1}{\mathrm{t}} \mathrm{dt} \\ =\frac{4}{3}[\mathrm{t}-\log |\mathrm{t}|]+\mathrm{C} \\ =\frac{4}{3}\left[\left(1+\mathrm{x}^{\frac{3}{4}}\right)-\log \left|1+\mathrm{x}^{\frac{3}{4}}\right|\right]+\mathrm{C}$

Question:13

Evaluate the following:
$\\ \begin{aligned} &\text { Given; }\\ &\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x\\ \end{aligned}$

Answer:

$\\ \begin{aligned} &\text { Given; }\\ &\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x\\ &\text { Let } x=\tan y\\ &\Rightarrow \mathrm{dx}=\sec ^{2} \mathrm{y} \mathrm{dx} \end{aligned}$
$\\ =\int \frac{\sqrt{1+\tan ^{2} y}}{\tan ^{4} y} \sec ^{2} y d y \\ =\int \sec y \times \frac{\cos ^{4} y}{\sin ^{4} y} \times \sec ^{2} y d y \\ =\int \frac{\cos y}{\sin ^{4} y} d y \\ \text { Let } t=\sin y$
$\Rightarrow dt = cos y dy$
$\\ =\int \frac{\mathrm{dt}}{\mathrm{t}^{4}}=-\frac{1}{3 \mathrm{t}^{3}}+\mathrm{C} \\ =-\frac{1}{3 \sin ^{3} \mathrm{y}} \\ =-\frac{1}{3 \sin ^{3}\left(\sin ^{-1} \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}}\right)} \\ =-\frac{\left(\mathrm{x}^{2}+1\right)^{\frac{3}{2}}}{3 \mathrm{x}^{3}}$

Question:14

Evaluate the following:
$\int \frac{d x}{\sqrt{16-9 x^{2}}}$

Answer:

$\\ \begin{aligned} &\text { Given; }\\ &\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}\\ &=\int \frac{d x}{\sqrt{4^{2}-(3 x)^{2}}}\\ &\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]\\ &=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C \end{aligned}$

Question:15

Evaluate the following:
$\\ \int \frac{\mathrm{dt}}{\sqrt{3 \mathrm{t}-2 \mathrm{t}^{2}}}$

Answer:

$\\ \begin{aligned} &\text { Given }\\ &\int \frac{\mathrm{dt}}{\sqrt{3 t-2 t^{2}}}\\ &=\int \frac{\mathrm{dt}}{\sqrt{\left(\frac{3}{2 \sqrt{2}}\right)^{2}-\left(\frac{3}{2 \sqrt{2}}\right)^{2}+2 \times \sqrt{2} t \times \frac{3}{2 \sqrt{2}}-(\sqrt{2} t)^{2}}} \end{aligned}$
$\\ =\int \frac{d t}{\sqrt{\left(\frac{3}{2 \sqrt{2}}\right)^{2}-\left(\sqrt{2} t-\frac{3}{2 \sqrt{2}}\right)^{2}}} \\ =2 \sqrt{2} \int \frac{d t}{\sqrt{3^{2}-(4 t-3)^{2}}} \\ \left.\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right] \\ =\frac{1}{\sqrt{2}} \sin ^{-1} \frac{4 t-3}{3}+C$

Question:16

Evaluate the following:
$\int \frac{3 x-1}{\sqrt{x^{2}+9}} d x$

Answer:

$\\ \text { Given; } \frac{3 x-1}{\sqrt{x^{2}+9}} d x \\ =\int \frac{3 x}{\sqrt{x^{2}+3^{2}}} d x-\int \frac{1}{\sqrt{x^{2}+3^{2}}} d x \\ \text { [Let; } x^{2}+9=y \Rightarrow \left.2 x d x=d y\right] \\ =\frac{3}{2} \int \frac{d y}{\sqrt{y}}-\log \left|x+\sqrt{x^{2}+3^{2}}\right|+c \\ =3 \sqrt{x^{2}+9}-\log \left|x+\sqrt{x^{2}+9}\right|+C$

Question:17

Evaluate the following:
$\int \sqrt{5-2 x+x^{2} }d x$

Answer:

$\\ \text { Given; } \int \sqrt{5-2 x+x^{2}} d x \\ =\int \sqrt{(x-1)^{2}+2^{2}} \\ {\left[\because \int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|\right.} \\ =\frac{x-1}{2} \sqrt{5-2 x+x^{2}}+2 \log \left|(x+1)+\sqrt{5-2 x+x^{2}}\right|+C$

Question:18

Evaluate the following:
$\int \frac{x}{x^{4}-1} d x$

Answer:

$\\ \text { Given; } \int \frac{x}{x^{4}-1} d x \\ \text { [Let; } t=x^{2}\ \Rightarrow \left.d t=2 x d x\right] \\ =\int \frac{1}{\left(t^{2}-1\right)} \frac{d t}{2}$
$\\ =\frac{1}{2} \int \frac{1}{(t+1)(t-1)} d t \\ =\frac{1}{2} \int \frac{1}{2}\left[\frac{1}{(t-1)}-\frac{1}{(t+1)}\right] d t \\ {\left[\because \int \frac{1}{x} d x=\log |x|\right.} \\ =\frac{1}{4}(\log |t-1|-\log |t+1|)+C$
$\\ {\left[\because \log a-\log b=\log \frac{a}{b}\right]} \\ =\frac{1}{4} \log \left|\frac{t-1}{t+1}\right|+c \\ =\frac{1}{4} \log \left|\frac{x^{2}-1}{x^{2}+1}\right|+c$

Question:19

Evaluate the following:
$\int \frac{x^{2}}{1-x^{4}} d x \text { put } x^{2}=t$

Answer:

Given: $\int \frac{x^{2}}{1-x^{4}} d x$
$\\ =\int \frac{1}{2}\left[\frac{2 x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ =\int \frac{1}{2}\left[\frac{x^{2}+x^{2}-1+1}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ =\int \frac{1}{2}\left[\frac{\left(1+x^{2}\right)-\left(1-x^{2}\right)}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ =\int \frac{1}{2}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x-\frac{\left(1-x^{2}\right)}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right]$
$=\int \frac{1}{2}\left[\frac{1}{\left(1-x^{2}\right)} d x-\frac{1}{\left(1+x^{2}\right)} d x\right]$
As we know,
$\\ \int \frac{1}{\left(a^{2}-x^{2}\right)} d x=\frac{1}{2 a} \log \frac{a+x}{a-x} \\ \int \frac{1}{\left(a^{2}+x^{2}\right)} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a} \\ =\frac{1}{2} \times \frac{1}{2} \log \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x+c \\ =\frac{1}{4} \log \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x+c$

Question:20

Evaluate the following:
$\int \sqrt{2 a x-x^{2}} d x$

Answer:

$\\ \text { Given; } \int \sqrt{2 a x-x^{2}} d x \\ =\int \sqrt{a^{2}-a^{2}+2 a x-x^{2}} d x \\ =\int \sqrt{a^{2}-(x-a)^{2}} d x \\ =\frac{(x-a)}{2} \sqrt{2 a x-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x-a}{a}+c$

Question:21

Evaluate the following:
$\\ \int \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} d x$

Answer:

$\\ \begin{aligned}&\text { Given; }\\ &\int \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} d x\\ &\begin{array}{l} \begin{array}{c} \text { Let; } t=\sin ^{-1} x \\ x=\sin t \\ \Rightarrow d t=\frac{1}{\sqrt{1-x^{2}}} d x \end{array} \\ \Rightarrow \int \frac{\sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{\left(1-x^{2}\right)}} d x \\ =\int \frac{t}{\left(1-\sin ^{2} t\right)} d t \\ =\int \frac{t}{\cos ^{2} t} d t \end{array} \end{aligned}$
$\\ =\int t \sec ^{2} t d t\\$
Apply integration by parts
$$$ \left[\int f(x) \cdot g(x) d x=f(x) \int g(x) d x-\int \frac{d}{d x} f(x)\left[\int g(x) d x\right] d x\right]$
$\\ =\mathrm{t} \int \sec ^{2} \mathrm{t} \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})\left[\int \sec ^{2} \mathrm{t} \mathrm{dt}\right] \mathrm{dt} \\ =\mathrm{t} \operatorname{tant}-\int \mathrm{tant} \mathrm{dt} \\ =\mathrm{t} \operatorname{tant}+\int \frac{-\sin \mathrm{t}}{\cos t} \mathrm{dt} \\ =\mathrm{t} \operatorname{tant}+\log |\cos t|+\mathrm{C} \\ =\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}+\log \left|\sqrt{1-\mathrm{x}^{2}}\right|+\mathrm{C}$

Question:22

Evaluate the following:
$\int \frac{(\cos 5 x+\cos 4 x)}{1-2 \cos 3 x} d x$

Answer:

$\\ \text { Given; } \int \frac{(\cos 5 x+\cos 4 x)}{1-2 \cos 3 x} \mathrm{dx} \\ {\left[\cos a+\cos b=2 \cos \frac{1}{2}(a+b) \cos \frac{1}{2}(a-b)\right]} \\ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2}}{1-2 \cos 3 x} d x \\ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{(1-2 \cos 3 x) \cos \frac{3 x}{2}} d x \\ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2}-2 \cos 3 x \cos \frac{3 x}{2}} d x$
$\\ {[\because 2 \cos a \cos b=\cos (a+b)+\cos (a-b)]} \\ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2}-\cos \frac{9 x}{2}-\cos \frac{3 x}{2}} d x \\ =\int-2 \cos \frac{3 x}{2} \cos \frac{x}{2} d x \\ =\int-\cos 2 x-\cos x d x \\ =-\frac{\sin 2 x}{2}-\sin x+C$

Question:23

Evaluate the following:
$\int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cos ^{2} x} d x$

Answer:

$\\ \text { Given; } \int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cos ^{2} x} \mathrm{dx} \\ =\int \frac{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)}{\sin ^{2} x \cos ^{2} x} \mathrm{dx} \\ =\int \frac{\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)}{\sin ^{2} x \cos ^{2} x} d x \\ =\int \frac{\sin ^{2} x}{\cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}-1 d x \\ =\int \tan ^{2} x+\cot ^{2} x-1 d x$
$\\ =\int \sec ^{2} x-1+\operatorname{cosec}^{2} x-1-1 \mathrm{dx} \\ =\tan x-\cot x-3 x+C$

Question:24

Evaluate the following:
$\int \frac{\sqrt{x}}{\sqrt{a^{3}-x^{3}}} d x$

Answer:

$\begin{aligned} &\text { Given; }\\ &\int \frac{\sqrt{x}}{\sqrt{a^{3}-x^{3}}} d x\\ &\left[\text { Let; } t=\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}} \Rightarrow d t=\frac{3 \sqrt{x}}{2 a^{\frac{3}{2}}} d x\right]\\ &=\int \frac{2 a^{\frac{3}{2}} \mathrm{dt}}{3 \sqrt{\mathrm{a}^{3}-\mathrm{a}^{3} \mathrm{t}^{2}}}\\ &=\int \frac{2 \mathrm{dt}}{3 \sqrt{1-\mathrm{t}^{2}}}\\ &=\frac{2}{3} \sin ^{-1} t+C=\frac{2}{3} \sin ^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right)+C \end{aligned}$

Question:25

Evaluate the following:
$\int \frac{\cos x-\cos 2 x}{1-\cos x} d x$

Answer:

Given, $\int \frac{\cos x-\cos 2 x}{1-\cos x} d x$
$\\ =\int \frac{\cos 2 x-\cos x}{\cos x-1} d x \\ =\int \frac{2 \cos ^{2} x-1-\cos x}{\cos x-1} d x \\ =\int \frac{(2 \cos x+1)(\cos x-1)}{\cos x-1} d x \\ =\int(2 \cos x+1) d x \\ =2 \sin x+x+c$

Question:26

Evaluate the following:
$\int \frac{d x}{x \sqrt{x^{4}-1}}\left(H \text { int }: \text { Put } x^{2}=\sec \theta\right)$

Answer:

$\\ \begin{aligned} &\text { Given; }\\ &\int \frac{d x}{x \sqrt{x^{4}-1}}\\ &\text { [Let; } \left.\mathrm{x}^{2}=\sec \theta \Rightarrow 2 \mathrm{xdx}=\sec \theta \tan \theta \mathrm{d} \theta\right]\\ &=\int \frac{\sec \theta \tan \theta}{2 \sec \theta \sqrt{\sec ^{2} \theta-1}} d \theta=\int \frac{\mathrm{d} \theta}{2}\\ &=\frac{\theta}{2}+c\\ &=\frac{\sec ^{-1} x^{2}}{2}+c \end{aligned}$

Question:27

Evaluate the following as limit of sums:
$\int_{0}^{2}\left(x^{2}+3\right) d x$

Answer:

$\\ \text { Given; } \int_{0}^{2}\left(\mathrm{x}^{2}+3\right) \mathrm{d} \mathrm{x} \\ \text { We know } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{h} \sum_{\mathrm{r}=0}^{\mathrm{n}-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \\ \text { Here } \mathrm{a}=0 . \mathrm{b}=2 \\ \mathrm{~h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{\mathrm{n}}=\frac{2}{\mathrm{n}} \\ =>\mathrm{nh}=2$
$\\ =\operatorname{limh}_{h \rightarrow 0} \sum_{r=0}^{n-1} f(r h) \\ =\lim _{h \rightarrow 0} h \sum_{r=0}^{n-1}\left(3+r^{2} h^{2}\right) \\ =\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{(n-1)(n-1+1)(2 n-2+1)}{6}\right)\right.$
$\\ =\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{\left(n^{2}-n\right)(2 n-1)}{6}\right)\right. \\ =\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{2 n^{3}-3 n^{2}+n}{6}\right)\right. \\ =\lim _{h \rightarrow 0}\left(3 n h+\left(\frac{2 n^{3} h^{3}-3 n^{2} h^{3}+n h^{3}}{6}\right)\right.$
$\\ =\lim _{h \rightarrow 0}\left(3.2+\left(\frac{2.2^{3}-3.2^{2}.h+2 h^{2}}{6}\right)\right. \\ =6+\frac{16}{3} \\ =\frac{26}{3}$

Question:28

Evaluate the following as limit of su
$\int_{0}^{2} \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$

Answer:

$\\ \text { We know } \int_{a}^{b} f(x) d x=\lim _{h \rightarrow \infty} h \sum_{r=0}^{n-1} f(a+r h) \\ \text { Here } a=0, b=2 \\ \qquad \\ h=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n} \\ =>n h=2$
$\\ =\operatorname{limh}_{h \rightarrow 0} \sum_{r=0}^{n-1} f(r h) \\ =\lim _{h \rightarrow 0} h\left[1+e^{h}+e^{2 h}+\cdots,+e^{(n-1) h}\right] \\ =\lim _{h \rightarrow 0} h\left[\frac{1\left(e^{h}\right)^{n}-1}{e^{h}-1}\right]$
$\\ =\lim _{h \rightarrow 0} h\left[\frac{e^{n h}-1}{e^{h}-1}\right] \\ =\lim _{h \rightarrow 0} h\left[\frac{e^{2}-1}{e^{h}-1}\right] \\ =(e^{2}-1 )\lim _{h \rightarrow 0}\left[\frac{h}{e^{h}-1}\right] \\ =e^{2}-1$

Question:29

Evaluate the following:
$\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}$

Answer:

$\\ \text { Given; } \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}} \\ =\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1} \\ =\int_{1}^{e} \frac{d t}{t^{2}+1} \\ \text { [Let; } \left.t=e^{x}[\text { when } x=0, t=1 \text { and } x=1, t=e] \Rightarrow d t=e^{x} d x\right] \\ =\left[\tan ^{-1} t\right]_{1}^{e}$
$=\tan ^{-1} \mathrm{e}-\tan ^{-1} 1=\tan ^{-1} \mathrm{e}-\frac{\pi}{4}$

Question:30

Evaluate the following:
$\int_{0}^{\frac{\pi}{2}} \frac{\tan x d x}{1+m^{2} \tan ^{2} x}$

Answer:

$\\ \text { Given; } \int_{0}^{\frac{\pi}{2}} \frac{\tan x}{1+m^{2} \tan ^{2} x} \mathrm{dx} \\ \text { [Let; } \left.t=\tan x \Rightarrow \mathrm{dt}=\sec ^{2} \mathrm{x} \mathrm{dx}\right] \\ \qquad \text { [Let; } \left.\mathrm{u}=\mathrm{t}^{2} \Rightarrow \mathrm{du}=2 \mathrm{tdt}\right] \\ \frac{\mathrm{t}}{1+\mathrm{m}^{2} \mathrm{t}^{2}} \frac{\mathrm{dt}}{\sec ^{2} \mathrm{x}}=\int \frac{\mathrm{t}}{\left(1+\mathrm{m}^{2} \mathrm{t}^{2}\right)} \frac{\mathrm{dt}}{\left(1+\mathrm{t}^{2}\right)} \\ =\int \frac{1}{\left(1+\mathrm{m}^{2} \mathrm{u}\right)(1+\mathrm{u})} \frac{\mathrm{du}}{2}$
By applying partial fraction;
$\\ \frac{1}{\left(1+\mathrm{m}^{2} \mathrm{u}\right)(1+\mathrm{u})}=\frac{\mathrm{A}}{\left(1+\mathrm{m}^{2} \mathrm{u}\right)}+\frac{\mathrm{B}}{(1+\mathrm{u})} \\ 1=\mathrm{A}(1+\mathrm{u})+\mathrm{B}\left(1+\mathrm{m}^{2} \mathrm{u}\right) \\ \mathrm{B}=\frac{1}{1-\mathrm{m}^{2}} \\ \text { When } \mathrm{u}=-1 \\ \text { When } \mathrm{u}=-\frac{1}{\mathrm{~m}^{2}}$
$\\ A=\frac{m^{2}}{m^{2}-1} \\ =\frac{1}{2} \int \frac{m^{2}}{\left(m^{2}-1\right)\left(1+m^{2} u\right)}+\frac{1}{\left(1-m^{2}\right)(1+u)} d u \\ =\frac{1}{2} \times \frac{m^{2}}{m^{2}-1} \times \log \left|1+m^{2} u\right|+\frac{1}{2} \times \frac{1}{1-m^{2}} \times \log |1+u|+C$
$\\ =\frac{1}{2} \times \frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1} \times \log \left|1+\mathrm{m}^{2} \tan ^{2} \mathrm{x}\right|+\frac{1}{2} \times \frac{1}{1-\mathrm{m}^{2}} \times \log \left|1+\tan ^{2} \mathrm{x}\right|+\mathrm{C} \\ =\frac{1}{2} \times \frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}\left[\log \left|1+\mathrm{m}^{2} \tan ^{2} \mathrm{x}\right|+\log \left|1+\tan ^{2} \mathrm{x}\right|\right]+\mathrm{C} \\ =\frac{1}{2} \times \frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}\left[\log \mid\left(1+\mathrm{m}^{2} \tan ^{2} \mathrm{x}\right)\left(1+\tan ^{2} \mathrm{x}\right) \|+\mathrm{C}\right. \\ =\frac{1}{2} \times \frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}\left[\log \left|\left(1+\mathrm{m}^{2} \tan ^{2} \mathrm{x}\right) \sec ^{2} \mathrm{x}\right|\right]+\mathrm{C}$
$\\ =\frac{1}{2} \times \frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}\left[\log \mathrm{g}\left(\cos ^{2} \mathrm{x}+\mathrm{m}^{2} \sin ^{2} \mathrm{x}\right) \sec ^{2} \mathrm{x} \|+\mathrm{C}\right. \\ =\frac{\log \left|\left(\mathrm{m}^{2}-1\right) \sin ^{2} \mathrm{x}+1\right|}{2 \mathrm{~m}^{2}-2}+\mathrm{C}$
By applying the given limits 0 to π/2
$\\ =\frac{\log \left|\left(\mathrm{m}^{2}-1\right) \sin ^{2} \frac{\pi}{2}+1\right|}{2 \mathrm{~m}^{2}-2}-\frac{\log \left|\left(\mathrm{m}^{2}-1\right) \sin ^{2} 0+1\right|}{2 \mathrm{~m}^{2}-2} \\ =\frac{\log \left|\left(\mathrm{m}^{2}\right)\right|}{2 \mathrm{~m}^{2}-2}$

Question:31

Evaluate the following:
$\int_{1}^{2} \frac{d x}{\sqrt{(x-1)(2-x)}}$

Answer:

$\\ Given \int_{1}^{2} \frac{d x}{\sqrt{(x-1)(2-x)}}$\\ $=>_{1}^{2} \frac{\mathrm{dx}}{\sqrt{(x-1)(2-x)}}$\\ $=\int_{1}^{2} \frac{d x}{\sqrt{-\left(x^{2}-3 x+2\right)}}$$ Using perfect square method for the denominator $\Rightarrow x^{2}-3 x+2=x^{2}-3 x+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2$$
$\\ \begin{aligned} &=\left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}\\ &=\int_{1}^{2} \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}}\\ &\text { We know }\\ &\int \frac{\mathrm{dx}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}=\sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}\\ &=\left[\sin ^{-1}\left(\frac{\left(x-\frac{3}{2}\right)}{\frac{1}{2}}\right)\right]_{1}^{2}=\left[\sin ^{-1}(2 x-3)\right]_{1}^{2} \end{aligned}$
$\\ =\sin ^{-1}(1)-\sin ^{-1}(-1) \\ \text {We know } \sin ^{-1}(-\theta)=-\sin \theta \\ =\frac{\pi}{2}+\frac{\pi}{2} \\ =\pi$

Question:32

Evaluate the following:
$\int_{0}^{1} \frac{\mathrm{xdx}}{\sqrt{1+\mathrm{x}^{2}}}$

Answer:

$\\ Given \int_{0}^{1} \frac{x d x}{\sqrt{1+x^{2}}}$\\ Now put $1+x^{2}=t$\\ $=>2 \mathrm{x} \mathrm{dx}=\mathrm{dt}$\\ At $x=0, t=1$ and at x=1, t=2$

$\\ \Rightarrow \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{t}} \\ =\frac{1}{2}[2 \sqrt{t}]_{1}^{2} \\ =\sqrt{2}-1 \\ \Rightarrow \int_{0}^{1} \frac{x d x}{\sqrt{1+x^{2}}}=\sqrt{2}-1$

Question:33

Evaluate the following:
$\int_{0}^{\pi} x \sin x \cos ^{2} x d x$

Answer:

Using Property

$\\ \int_{2}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\ \qquad \begin{array}{l} \text { Let } I=\int_{0}^{\pi} x \sin x \cos ^{2} x d x \\ =>\int_{0}^{\pi} x \sin x \cos ^{2} x d x=\int_{0}^{\pi}(\pi-x) \sin (\pi-x) \cos ^{2}(\pi-x) d x \end{array} \\ \operatorname{ses}(\pi-x)=-\cos x$

$\\ I=\int_{0}^{\pi} \pi \sin x \cos ^{2} x d x-\int_{0}^{\pi} x \sin x \cos ^{2} x d x \\ I=\int_{0}^{\pi} \pi \sin x \cos ^{2} x d x-I \\ {2} I=\int_{0}^{\pi} \pi \sin x \cos ^{2} x d x$

$\\ \begin{aligned} &\Rightarrow \int_{0}^{\pi} \pi \sin x \cos ^{2} x d x=\pi \int_{0}^{\pi} \sin x \cos ^{2} x d x\\ &\text { Now let } \cos x=t\\ &=>-\sin x d x-d t\\ &\text { And, at } x=0, t=1\\ &\text { and at } x=\pi, t=-1 \end{aligned}$

$\\ \Rightarrow 2\mathrm{I}=-\pi \int_{1}^{-1} \mathrm{t}^{2} \mathrm{dt}=-\pi\left[\frac{\mathrm{t}^{2}}{3}\right]_{1}^{-1}=\frac{2 \pi}{3} \\ \Rightarrow{2 \mathrm{I}=\frac{2 \pi}{3}} \\ \Rightarrow{\mathrm{I}=\frac{\pi}{3}} \\ \Rightarrow \int _{0}^{\pi} x \sin x \cos ^{2} x \mathrm{dx}=\frac{\pi}{3}$

Question:34

Evaluate the following:
$\int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}}$ (Hint: let x = sin θ)

Answer:

$\\ \text { Given } \int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} \\ =>\text { Let } x=\sin \theta \\ \qquad \text { At } x=0, \theta=0 \\ \text { and } x=\frac{1}{2}, \theta=\frac{\pi}{6}$
$\\ =\int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}}=\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \sqrt{1-\sin ^{2} \theta}} \\ \text { As } 1-\sin ^{2} \theta=\cos ^{2} \theta \\ =\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \sqrt{1-\sin ^{2} \theta}}=\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \sqrt{\cos ^{2} \theta}} \\ =\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \cos \theta}$
$\\ \begin{aligned} &=>\int_{0}^{\frac{\pi}{6}} \frac{d \theta}{\left(1+\sin ^{2} \theta\right)}\\ &=>_{0}^{\frac{\pi}{6}} \frac{\frac{1}{\cos ^{2} \theta}}{\frac{\left(1+\sin ^{2} \theta\right)}{\cos ^{2} \theta}} \mathrm{d} \theta=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} \theta \mathrm{d} \theta}{\left(\sec ^{2} \theta+\tan ^{2} \theta\right)}\\ &=>\mathrm{As} \sec ^{2} \theta-\tan ^{2} \theta=1\\ &=>\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} \theta \mathrm{d} \theta}{\left(1+2 \tan ^{2} \theta\right)} \end{aligned}$
$\\ \begin{aligned} &\text { Now put } \tan \theta=t\\ &=>\sec ^{2} \theta \mathrm{d} \theta=\mathrm{dt}\\ &\text { At } \theta=0, \mathrm{t}=0\\ &\text { at } \theta=\frac{\pi}{6}, \mathrm{t}=\frac{1}{\sqrt{3}}\\ &=\int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{\left(1+2 t^{2}\right)}=\frac{1}{2} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{\left(\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}\right)}\\ &\text{As } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{2} \tan ^{-1}\left(\frac{x}{a}\right)+c \end{aligned}$

$\\ =>\int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{\left(\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}\right)}=\frac{1}{2}\left[\frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\right]_{0}^{\frac{1}{\sqrt{2}}} \\ =\frac{\sqrt{2}}{2} \tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right) \\ =>\int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

Question:35

Evaluate the following:
$\int \frac{x^{2} d x}{x^{4}-x^{2}-12}$

Answer:

$\\ \text { Given: } \int \frac{x^{2} d x}{x^{4}-x^{2}-12} \\ \text { Put } x^{2}=t \\ =>\frac{x^{2}}{x^{4}-x^{2}-12}=\frac{t}{t^{2}-t-12}$
$\\ \Rightarrow \mathrm{t}^{2}-\mathrm{t}-12=(\mathrm{t}+3)(\mathrm{t}-4) \\ \Rightarrow \frac{\mathrm{t}}{(\mathrm{t}+3)(\mathrm{t}-4)}=\frac{\mathrm{A}}{(\mathrm{t}+3)}+\frac{\mathrm{B}}{(\mathrm{t}-4)} \text { (Concept of partial fraction) } \\ \Rightarrow \mathrm{t}=\mathrm{t}(\mathrm{A}+\mathrm{B})+3 \mathrm{~B}-4 \mathrm{~A}$
On comparing coefficients of ‘t’ we get
$\\ \begin{aligned} &A=\frac{3}{7} \& B=\frac{4}{7}\\ &=\frac{t}{(t+3)(t-4)}=\frac{3}{7(t+3)}+\frac{4}{7(t-4)}\\ &\Rightarrow\text { Now put } t=x^{2} \text { back in the above eq. }\\ &\Rightarrow\frac{x^{2}}{x^{4}-x^{2}-12}=\frac{3}{7\left(x^{2}+3\right)}+\frac{4}{7\left(x^{2}-4\right)} \end{aligned}$
$\\ \Rightarrow\frac{x^{2} d x}{x^{4}-x^{2}-12}=\int\left(\frac{3}{7\left(x^{2}+3\right)}+\frac{4}{7\left(x^{2}-4\right)}\right) d x=\frac{1}{7}\left(\int \frac{3 d x}{\left(x^{2}+3\right)}+\int \frac{4 d x}{\left(x^{2}-4\right)}\right) \\ \Rightarrow\operatorname{Now} \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \ln \left(\frac{x-a}{x+a}\right)+c \& \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ \Rightarrow7\left(\int \frac{3 d x}{\left(x^{2}+3\right)}+\int \frac{4 d x}{\left(x^{2}-4\right)}\right)=\frac{1}{7}\left(\frac{3}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{4}{4} \ln \left(\frac{x-2}{x+2}\right)+c\right) \\ \Rightarrow\int \frac{x^{2} d x}{x^{4}-x^{2}-12}=\frac{\sqrt{3}}{7} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{1}{7} \ln \left(\frac{x-2}{x+2}\right)+c$

Question:36

Evaluate the following:
$\int \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$

Answer:

$\\ \begin{aligned} &\text { Given } \int \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}\\ &\text { Put } x^{2}=t\\ &\Rightarrow \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{t}{\left(t+a^{2}\right)\left(t+b^{2}\right)}\\ &\Rightarrow \frac{t}{\left(t+a^{2}\right)\left(t+b^{2}\right)}=\frac{A}{\left(t+a^{2}\right)}+\frac{B}{\left(t+b^{2}\right)}(\text { Concept of partial fraction }) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow t=t(A+B)+a^{2} B+b^{2} A\\ &\text { On comparing coefficients of "twe get }\\ &\Rightarrow \mathrm{A}=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-\mathrm{b}^{2}} \& \mathrm{~B}=\frac{-\mathrm{b}^{2}}{\mathrm{a}^{2}-\mathrm{b}^{2}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \frac{t}{\left(t+a^{2}\right)\left(t+b^{2}\right)}=\frac{1}{a^{2}-b^{2}}\left(\frac{a^{2}}{\left(t+a^{2}\right)}-\frac{b^{2}}{\left(t+b^{2}\right)}\right)\\ &\Rightarrow \text { Now put } t=x^{2} \text { back in the above eq. }\\ &\Rightarrow \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{a^{2}-b^{2}}\left(\frac{a^{2}}{\left(x^{2}+a^{2}\right)}-\frac{b^{2}}{\left(x^{2}+b^{2}\right)}\right)\\ &\Rightarrow \int \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{a^{2}-b^{2}}\left(\int \frac{a^{2} d x}{\left(x^{2}+a^{2}\right)}-\int \frac{b^{2} d x}{\left(x^{2}+b^{2}\right)}\right)\\ &\Rightarrow_{\text {Now }} \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \end{aligned}$
$\\ =\frac{1}{a^{2}-b^{2}}\left(\int \frac{a^{2} d x}{\left(x^{2}+a^{2}\right)}-\int \frac{b^{2} d x}{\left(x^{2}+b^{2}\right)}\right) \\ \Rightarrow=\frac{1}{a^{2}-b^{2}}\left(\frac{a^{2}}{a} \tan ^{-1}\left(\frac{x}{a}\right)+\frac{b^{2}}{b} \tan ^{-1}\left(\frac{x}{b}\right)+c\right) \\ \Rightarrow \int \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{a^{2}-b^{2}}\left(a \tan ^{-1}\left(\frac{x}{a}\right)+b \tan ^{-1}\left(\frac{x}{b}\right)\right)+c$

Question:37

Evaluate the following:
$\int_{0}^{\pi} \frac{x}{1+\sin x}$

Answer:

$\\ Given \int_{0}^{\pi} \frac{x d x}{1+\sin x}$\\ Let $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{xdx}}{1+\sin \mathrm{x}}$\\ Now using Property $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$\\ $=\int_{0}^{\pi} \frac{x d x}{1+\sin x}=\int_{0}^{\pi} \frac{(\pi-x) d x}{1+\sin (\pi-x)}$$ $\\ \Rightarrow \int_{0}^{\pi} \frac{x d x}{1+\sin x}=\int_{0}^{\pi} \frac{\pi d x}{1+\sin x}-\int_{0}^{\pi} \frac{x d x}{1+\sin x} \\ \Rightarrow 2 \int_{0}^{\pi} \frac{x d x}{1+\sin x}=2 I=\pi \int_{0}^{\pi} \frac{d x}{1+\sin x} \ldots(1) \\ \Rightarrow \int_{0}^{\pi} \frac{d x}{1+\sin x}=\int_{0}^{\pi} \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}=\int_{0}^{\pi} \frac{1-\sin x}{1-\sin ^{2} x} d x \\ \Rightarrow 1-\sin ^{2} x=\cos ^{2} x$
$\\ =\int_{0}^{\pi} \frac{1-\sin x}{1-\sin ^{2} x} d x=\int_{0}^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x=\int_{0}^{\pi} \frac{d x}{\cos ^{2} x}-\int_{0}^{\pi} \frac{\sin x}{\cos ^{2} x} d x \ldots(2) \\ \Rightarrow \int_{0}^{\pi} \frac{d x}{\cos ^{2} x}=\int_{0}^{\pi} \sec ^{2} x d x=[\tan x]_{0}^{\pi}=0 \\ \quad=^{\text {And }} \text { for } \int_{0}^{\pi} \frac{\sin x}{\cos ^{2} x} d x \\ \text { Put } \cos x=t$
$\\ \begin{aligned} &\Rightarrow-\sin x d x=d t\\ &\Rightarrow \mathrm{At}^{\mathrm{x}}=0=>\mathrm{t}=1 \text { and } \mathrm{x}=\pi=>\mathrm{t}=-1\\ &=\int_{1}^{-1}-\frac{d t}{t^{2}}=\left[\frac{1}{t}\right]_{1}^{-1}=-2 \ldots \end{aligned}$
$\\ 2\mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}=\pi\left(\int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{\cos ^{2} \mathrm{x}}-\int_{0}^{\pi} \frac{\sin \mathrm{x}}{\cos ^{2} \mathrm{x}} \mathrm{dx}\right)=\pi(0-(-2))=2 \pi \\ \quad\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{xdx}}{1+\sin \mathrm{x}}=\pi \\$

Question:38

Evaluate the following:
$\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x$

Answer:

Given:
$\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x$
Using the concept of partial fractions,
$\\ \Rightarrow \frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x-3)}$\\ $\Rightarrow 2 x-1=x^{2}(A+B+C)+x(C-4 B-A)+(3 B-2 C-6 A)$\\ Comparing coefficients:\\ $\Rightarrow A+B+C=0 \ldots(1)\\$
$\\ \begin{aligned} &\Rightarrow C-4 B-A=2 \ldots(2)\\ &\Rightarrow 3 B-2 C-6 A=-1 \ldots(3)\\ &\Rightarrow \text { On solving }(1),(2) \text { and }(3) \text { we get }\\ &\Rightarrow A=-\frac{1}{6}, B=-\frac{1}{3} \text { and } C=\frac{1}{2} \end{aligned}$
$\\ =\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{-\frac{1}{6}}{(x-1)}+\frac{-\frac{1}{2}}{(x+2)}+\frac{\frac{1}{2}}{(x-3)} \\ \Rightarrow \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=\int \frac{1}{2(x-3)} d x-\int \frac{1}{3(x+2)} d x-\int \frac{1}{6(x-1)} d x$
$\\ =\frac{1}{2} \ln (x-3)-\frac{1}{3} \ln (x+2)-\frac{1}{6} \ln (x-1)+c \\ \Rightarrow \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=\frac{1}{2} \ln (x-3)-\frac{1}{3} \ln (x+2)-\frac{1}{6} \ln (x-1)+c$

Question:39

Evaluate the following:
$\int \mathrm{e}^{\tan ^{-1} \mathrm{x}}\left(\frac{1+\mathrm{x}+\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$

Answer:

Given: $\int \mathrm{e}^{\tan ^{-1} \mathrm{x}}\left(\frac{1+\mathrm{x}+\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$
$\\ \text { Put } \tan ^{-1} x=t \\ \qquad \begin{array}{l} =\frac{d x}{1+x^{2}}=d t \\ \Rightarrow \int e^{\tan ^{-1} x}\left(\frac{1+x+x^{2}}{1+x^{2}}\right) d x=\int e^{t}\left(1+\tan t+\tan ^{2} t\right) d t \\ \Rightarrow \operatorname{As} \sec ^{2} \theta-\tan ^{2} \theta=1 \end{array} \\ \Rightarrow \int e^{t}\left(1+\tan t+\tan ^{2} t\right) d t=\int e^{t}\left(1+\tan t+\sec ^{2} t-1\right) d t$
$\\ \begin{aligned} &\Rightarrow \int e^{t}\left(\tan t+\sec ^{2} t\right) d t\\ &\text { Now using the property. } \int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})\\ &\Rightarrow \text { Now in } \int e^{t}\left(\tan t+\sec ^{2} t\right) d t \end{aligned}$
$\\=f(t)=\tan t \\ \Rightarrow f^{\prime}(t)=\sec ^{2} x \\ \Rightarrow \int e^{t}\left(\tan t+\sec ^{2} t\right) d t=e^{t} \tan t+C \\ \Rightarrow \int e^{\tan ^{-1} x}\left(\frac{1+x+x^{2}}{1+x^{2}}\right) d x=e^{\tan ^{-1} x} x+C$

Question:40

Evaluate the following:
$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$ (Hint: Put x = a tan2 θ)

Answer:

Given: $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
$\\ \begin{aligned} &\Rightarrow \text { Put } x=a \tan ^{2} \theta\\ &\Rightarrow \mathrm{dx}=2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta\\ &\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=\int \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}} 2 a \tan \theta \sec ^{2} \theta d \theta \end{aligned}$
$\\ \Rightarrow \mathrm{As} \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \quad \int \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a\left(1+\tan ^{2} \theta\right)}} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta \\ \Rightarrow \int \sin ^{-1} \sqrt{\frac{\tan ^{2} \theta}{\sec ^{2} \theta}} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta=\int \sin ^{-1} \sqrt{\sin ^{2} \theta} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta \\ \Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta$
$\\ $\Rightarrow$ \text{Now put }$\tan \theta=t$ $$\\ \Rightarrow \sec ^{2} \theta \mathrm{d} \theta=\mathrm{d} t $$\\ $\Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta=2 \mathrm{a} \int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt} \ldots$\\ $\Rightarrow$ \text{Now apply integration by part on} $\int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt}$\\$ $\\ =\int t \tan ^{-1} t d t=\tan ^{-1} t \int t d t-\int\left(\frac{d}{d t} \tan ^{-1} t\right)\left(\int t d t\right) d t \\ \Rightarrow \frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{t^{2}}{1+t^{2}} d t \ldots(2) \\ \Rightarrow \operatorname{Now} \int \frac{t^{2}}{1+t^{2}} d t=\int \frac{t^{2}+1-1}{1+t^{2}} d t=\int \frac{t^{2}+1}{1+t^{2}} d t-\int \frac{1}{1+t^{2}} d t \\ \Rightarrow \int \frac{t^{2}+1}{1+t^{2}} d t-\int \frac{1}{1+t^{2}} d t=\int d t-\int \frac{1}{1+t^{2}} d t=t-\tan ^{-1} t \ldots(3)$
$\\ \begin{aligned} &\Rightarrow \text { Put }(3) \text { in }(2) \text { and the resulting equation in }(1)\\ &\Rightarrow \int t \tan ^{-1} t d t=\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2}\left(t-\tan ^{-1} t\right)\\ &\Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta=2 \mathrm{a}\left(\frac{\tan ^{2} \theta}{2} \tan ^{-1}(\tan \theta)-\frac{1}{2}\left(\tan \theta-\tan ^{-1} \tan \theta\right)\right)\\ &\Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta=\mathrm{a}\left(\theta \tan ^{2} \theta-\tan \theta+\theta\right) \end{aligned}$
$\\ a\left(\theta \tan ^{2} \theta-\tan \theta+\theta\right)=a\left(\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{2}}\right. \\ \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=(x+a) \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}$

Question:41

Evaluate the following:
$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}}$

Answer:

Given: $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x )^{\frac{5}{2}}}$
Using Trigonometric identities:
$\\ \Rightarrow \cos 2 \mathrm{x}=2 \cos ^{2} \mathrm{x}-1=1-2 \sin ^{2} \mathrm{x} \\ \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}}=\frac{\sqrt{1+\left(2 \cos ^{2} \frac{x}{2}-1\right)}}{\left(1-\left(1-2 \sin ^{2} \frac{x}{2}\right)\right)^{\frac{5}{2}}} \\ =\frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{5}{2}}}$
$\\ =\frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{5}{2}}} \\ =\left(\sqrt{2} \cos \frac{x}{2}\right) /\left(2^{\frac{5}{2}} \sin ^{5} \frac{x}{2}\right) \\ =\frac{\sqrt{2 }\cos \frac{x}{2}}{2^{\frac{5}{2}} \sin ^{5} \frac{x}{2}} \\ =\frac{1}{4} \frac{\cos \frac{x}{2}}{\sin ^{5} \frac{x}{2}}$
$\\ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin ^{5} \frac{x}{2}} d x \\ \text { Put } \sin \left(\frac{x}{2}\right)=t \\ \cos \left(\frac{x}{2}\right) d x=2 d t \\ \text { At } x=\frac{\pi}{3}=>t=\frac{1}{2} \text { and at } x=\frac{\pi}{2}=>t=\frac{1}{\sqrt{2}}$
$\\\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin ^{5} \frac{x}{2}} \mathrm{dx}=\frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}}$
$\\ \frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}}=\frac{1}{2}\left[\frac{t^{-4}}{-4}\right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}=\frac{3}{2} \\ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=\frac{3}{2}$

Question:42

Evaluate the following:
$\int \mathrm{e}^{-3 \mathrm{x}} \cos ^{3} \mathrm{x} \mathrm{d} \mathrm{x}$

Answer:

$\\ \begin{aligned} &\text { Given: } \int e^{-3 x} \cos ^{3} x d x\\ &\text { Using trigonometric identity }\\ &\cos 3 x=4 \cos ^{3} x-3 \cos x\\ &\int e^{-3 x} \cos ^{3} x d x=\frac{1}{4} \int e^{-3 x}(\cos 3 x+3 \cos x) d x\\ &\frac{1}{4} \int e^{-3 x}(\cos 3 x+3 \cos x) d x=\frac{1}{4} \int e^{-3 x} \cos 3 x d x+\frac{3}{4} \int e^{-3 x} \cos x d x ...(1) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Using a generalised formula i.e }\\ &\int e^{a x} \cos b x d x=\frac{e^{2 x}}{a^{2}+b^{2}}(a \cos b x+b \sin b x)\\ &\int e^{-3 x} \cos 3 x d x=\frac{e^{-2 x}}{(-3)^{2}+3^{2}}((-3) \cos 3 x+3 \sin 3 x) \end{aligned}$
$\\ \begin{aligned} &\frac{e^{-2 x}}{(-3)^{2}+3^{2}}((-3) \cos 3 x+3 \sin 3 x)=\frac{e^{-2 x}}{6}(\sin 3 x-\cos 3 x) \ldots(2)\\ &\int e^{-3 x} \cos x d x=\frac{e^{-2 x}}{(-3)^{2}+1^{2}}((-3) \cos x+\sin x)=\frac{e^{-2 x}}{10}(\sin x-3 \cos x)\\ &=\frac{e^{-2 x}}{10}(\sin x-3 \cos x) \ldots(3)\\ &\Rightarrow \text { On putting }(2) \text { and }(3) \text { in }(1) \end{aligned}$
$\frac{1}{4} \int e^{-3 x} \cos 3 x d x+\frac{3}{4} \int e^{-3 x} \cos x d x=\frac{e^{-2 x}}{4 \times 6}(\sin 3 x-\cos 3 x)+\frac{3 e^{-2 x}}{4 \times 10}(\sin x- 3 \cos x) \\ \Rightarrow \int e^{-3 x} \cos ^{3} x d x=e^{-3 x}\left\{\frac{(\sin 3 x-\cos 3 x)}{24}+\frac{3(\sin x-3 \cos x)}{40}\right\}+c$

Question:43

Evaluate the following:
$\int \sqrt{\tan x} d x$ (Hint: Put $tan x = t^2$ )

Answer:

Given:
$\int \sqrt{\tan x} d x$
Put $tan x = t^2$
$\begin{aligned} &\Rightarrow \sec ^{2} x d x=2 t d t\\ &\Rightarrow \mathrm{dx}=\frac{2 t \mathrm{dt}}{\sec ^{2} x}=\frac{2 \mathrm{tdt}}{1+\tan ^{2} \mathrm{x}}=\frac{2 \mathrm{tdt}}{1+\mathrm{t}^{4}}\\ &\Rightarrow \int \sqrt{\tan \mathrm{x}} \mathrm{d} \mathrm{x}=\int \sqrt{\mathrm{t}^{2}} \frac{2 \mathrm{t} \mathrm{dt}}{1+\mathrm{t}^{4}}=\int \frac{2 \mathrm{t}^{2} \mathrm{dt}}{1+\mathrm{t}^{4}}\\ &\Rightarrow \int \frac{2 t^{2} d t}{1+t^{4}}=\int \frac{\left(2 t^{2}+1-1\right) d t}{1+t^{4}}=\int \frac{\left(t^{2}+1\right)+\left(t^{2}-1\right)}{1+t^{4}} d t\\ &\Rightarrow \int \frac{\left(t^{2}+1\right)+\left(t^{2}-1\right)}{1+t^{4}} d t=\int \frac{\left(t^{2}+1\right)}{1+t^{4}} d t+\int \frac{\left(t^{2}-1\right)}{1+t^{4}} d t \end{aligned}$
Taking out $t^2$ common in both the numerators
$\\ \Rightarrow \int \frac{\left(t^{2}+1\right)}{1+t^{4}} d t+\int \frac{\left(t^{2}-1\right)}{1+t^{4}} d t=\int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} d t+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt} \\ \int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}=\int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}+\int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt} \ldots . \text { (1) }$
$\\ \begin{aligned} &\Rightarrow \operatorname{Now} t^{2}+\frac{1}{t^{2}}=\left(t \pm \frac{1}{t}\right)^{2} \mp 2 \ldots(3)\\ &=\operatorname{for}(a) \int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt} \text { taking } t-\frac{1}{t}=z\\ &\Rightarrow\left(1+\frac{1}{t^{2}}\right) d t=d z \end{aligned}$
$\\ =\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \mathrm{dt} \\ =\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \mathrm{dt}=\int \frac{\mathrm{d} z}{z^{2}+2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{z}{\sqrt{2}}\right)+\mathrm{c} \ldots(2)$
$\\ \begin{aligned} &\text { for (b) } \int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \text { dt taking } t+\frac{1}{t}=z\\ &\Rightarrow\\ &\Rightarrow\left(1-\frac{1}{t^{2}}\right) \mathrm{dt}=\mathrm{dz}\\ &=\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\int \frac{\left(1-\frac{1}{\mathrm{t}^{2}}\right)}{\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{2}-2} \mathrm{dt}\\ &=\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} \mathrm{dt}=\int \frac{\mathrm{d} z}{z^{2}-2}=\frac{1}{2 \sqrt{2}} \ln \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|+\mathrm{c} \ldots(3) \end{aligned}$

Put (2) and (3) in (1)
$\\ =\int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}+\int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\left(t+\frac{1}{\mathrm{t}}\right)-\sqrt{2}}{\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)+\sqrt{2}}\right|+\mathrm{c} \\ \Rightarrow \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{t}^{2}-1}{\mathrm{t} \sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\left(\mathrm{t}^{2}+1-\mathrm{t} \sqrt{2}\right.}{\left(\mathrm{t}^{2}+1+\mathrm{t} \sqrt{2}\right.}\right|+\mathrm{C}$
$\\ \Rightarrow \text{Now again putting} t=\sqrt{\tan x}\text{ to obtain the final result} \\ =\int \sqrt{\tan x} d x=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{(\tan x+1-\sqrt{2 \tan x}}{\tan x+1+\sqrt{2 \tan x}}\right| \\$

Question:44

Evaluate the following:
$\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}$
(Hint: Divide Numerator and Denominator by $cos^4x$ )

Answer:

Given: $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}$
Dividing Numerator and Denominator by $cos^4x$
$\\ =\int_{0}^{\frac{\pi}{2}} \frac{\left(1 / \cos ^{4} x\right) d x}{\left(\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right) / \cos ^{2} x\right)^{2}} \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}} \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}} \\ \Rightarrow \text { Put } \tan x=t$
$\\ \Rightarrow \sec ^{2} x d x=d t \\ \Rightarrow A t x=0=>t=0 \text { and at } x=\frac{\pi}{2}=>t=\infty \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x d x}{\left(a^{2}+b^{2} \tan ^{2} x\right)^{2}}=\int_{0}^{\infty} \frac{\left(1+t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}} \\ \Rightarrow \int_{0}^{\infty} \frac{\left(1+t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}+b^{2} t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}$
$\\ =\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}+b^{2} t^{2}\right) d t}{\left(a^{2}+b^{2} t^{2}\right)^{2}}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(a^{2}+b^{2} t^{2}\right)+\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(a^{2}+b^{2} t^{2}\right)+\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t+\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \text { Let } I=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t+\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\left(b^{2}-a^{2}\right)}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \ldots(1)$
$\\ =\operatorname{Let} I_{1}=\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t \\ =\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)} d t=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(\left(a^{2} / b^{2}\right)+t^{2}\right)} d t \\ =1_{1}=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\left(\left(a^{2} / b^{2}\right)+t^{2}\right)} d t=\frac{1}{b^{2}}\left(\frac{b}{a}\right)\left[\tan ^{-1}\left(\frac{b t}{a}\right)\right]_{0}^{\infty}=\frac{\pi}{2 a b} \\ \Rightarrow I_{1}=\frac{\pi}{2 a b} \ldots .(2)$

$\\ \text { Let } I_{2}=\int_{0}^{\infty} \frac{1}{\left(a^{2}+b^{2} t^{2}\right)^{2}} d t \\ \Rightarrow \text { let } b t=a \tan \theta \\ \Rightarrow b d t=\operatorname{asec}^{2} \theta d \theta \\ =I_{2}=\frac{1}{b} \int_{0}^{\frac{\pi}{2}} \frac{\operatorname{asec}^{2} \theta d \theta}{\left(a^{2}+a^{2} \tan ^{2} \theta\right)^{2}}=\frac{1}{b} \int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{a^{4}\left(1+\tan ^{2} \theta\right)^{2}}=\frac{1}{a^{3} b} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{4} \theta}$
$\\ =\frac{1}{a^{3} b} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta \mathrm{d} \theta}{\sec ^{4} \theta}=\frac{1}{a^{3} \mathrm{~b}} \int_{0}^{\frac{\pi}{2}} \cos ^{2} \theta \mathrm{d} \theta=\frac{1}{2 a^{3} \mathrm{~b}} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) \mathrm{d} \theta \\ \Rightarrow \frac{1}{2 a^{3} b} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) \mathrm{d} \theta=\frac{1}{2 a^{3} b}\left[\theta+\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{4 a^{3} b} \ldots(3) \\ \Rightarrow I=\frac{1}{b^{2}}\left(I_{1}+\left(b^{2}-a^{2}\right) \mathrm{I}_{2}\right)=\frac{1}{b^{2}}\left(\frac{\pi}{2 a b}+\left(b^{2}-a^{2}\right) \frac{\pi}{4 a^{3} b}\right)$
$\\\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi}{2 a b^{3}}\left(1+\left(\frac{b^{2}}{a^{2}}-1\right) \pi\right)$

Question:45

Evaluate the following:
$\int_{0}^{1} x \log (1+2 x) d x$

Answer:

Given: $\int_{0}^{1} x \log (1+2 x) d x$
$\\ \text { Let } 1+2 \mathrm{x}=\mathrm{t} \\ \Rightarrow 2 \mathrm{dx}=\mathrm{dt} \\ \Rightarrow \mathrm{At} \mathrm{x}=0=>\mathrm{t}=1 \text { and at } \mathrm{x}=1=>\mathrm{t}=3 \\ \Rightarrow \int_{0}^{1} \mathrm{x} \ln (1+2 \mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{4}\int_{1}^{3}(\mathrm{t}-1) \ln \mathrm{t} \mathrm{dt}....(i)$
$\\ \begin{aligned} &\Rightarrow \int_{1}^{3}(t-1) \ln t d t=\int_{1}^{3} t \ln t d t-\int_{1}^{3} \ln t d t\\ &\Rightarrow \text { Apply Integration by parts }\\ &=\int t \operatorname{ln} t d t=\ln t \int t d t-\int \frac{d}{d t}(\ln t)\left(\int t d t\right) d t=\frac{t^{2}}{2} \ln t-\frac{t^{2}}{4} \ldots(2)\\ &=\int \ln t d t=\ln t \int d t-\int \frac{d}{d t}(\ln t)\left(\int d t\right) d t=t \ln t-t \ldots(3)\\ \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Put }(2) \text { and }(3) \text { in }(1)\\ &=\frac{1}{4}\int_{1}^{3} t \ln t d t-\frac{1}{4}\int_{1}^{3} \ln t d t=\frac{1}{4}\left[\left(\frac{t^{2}}{2} \ln t-\frac{t^{2}}{4}\right)-(t \ln t-t)\right]_{1}^{3}=\frac{3}{8} \ln 3\\ &\Rightarrow \int_{0}^{1} x \ln (1+2 x) d x=\frac{3}{8} \ln 3 \end{aligned}$

Question:46

Evaluate the following:
$\int_{0}^{\pi} x \log \sin x d x$

Answer:

Given: $\int_{0}^{\pi} x \log \sin x d x$
$\\ \text { Using the property: } \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\ \qquad \begin{array}{l} \text { Let } 1=\int_{0}^{\pi} x \ln (\sin x) d x \\ =\int_{0}^{\pi}(\pi-x) \ln (\sin (\pi-x)) d x=\int_{0}^{\pi} \pi \ln (\sin x) d x \int_{0}^{\pi} x \ln (\sin x) d x \end{array} \\ \Rightarrow \text { As } \sin (\pi-x)=\sin x$
$\\ \begin{aligned} &\Rightarrow 2 \mathrm{I}=\int_{0}^{\pi} \pi \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\pi \int_{0}^{\pi} \ln (\sin \mathrm{x}) \mathrm{dx} \ldots(1)\\ &\Rightarrow \text { Now in } \int_{0}^{\pi} \ln (\sin x) d x\\ &\text { Using the property}\\ &\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x(\text { for } f(2 a-x)=f(x))\\ &=\int_{0}^{\pi} \ln (\sin x) d x=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \ldots(2) \end{aligned}$
$\\ \Rightarrow_{\text {Let }} Z=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \\ \Rightarrow \text { Using the property: } \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\ \Rightarrow \quad z=\int_{0}^{\frac{\pi}{2}} \ln \left(\sin \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x \ldots(4)\right. \\ \Rightarrow {2 Z}=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x=\int_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) d x \ldots(5)$
$\\ =\int_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) d x=\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{2 \sin x \cos x}{2}\right) d x \\ \quad=\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{2 \sin x \cos x}{2}\right) d x=\int_{0}^{\frac{\pi}{2}}(\ln (\sin 2 x)-\ln 2) d x \\ \Rightarrow \int_{0}^{\frac{\pi}{2}}(\ln (\sin 2 x)) d x-\int_{0}^{\frac{\pi}{2}}(\ln 2) d x=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x-\frac{\pi \ln 2}{2} \ldots(6) \\ \Rightarrow \operatorname{Now} \operatorname{in} \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x \text { put } 2 x=t$
$\\ \begin{aligned} &\Rightarrow \text { Now in } \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) \mathrm{d} x \text { put } 2 x=t\\ &\Rightarrow 2 \mathrm{dx}=\text { dt and limits changes from } 0 \text { to } \pi\\ &\Rightarrow{2 Z}=\frac{1}{2} \int_{0}^{\pi} \ln (\sin t) d t-\frac{\pi \ln 2}{2} \end{aligned}$
$\\ \Rightarrow \text{from equation (2)} \frac{1}{2} \int_{0}^{\pi} \ln (\operatorname{sint}) \mathrm{dt} \text{again becomes} \\\Rightarrow 2 Z=\frac{2}{2} \int_{0}^{\frac{\pi}{2}} \ln (\sin t) \mathrm{dt}-\frac{\pi \ln 2}{2}\\$
$\\\Rightarrow From eq. (3) \\\Rightarrow 2 \mathrm{Z}=\mathrm{Z}-\frac{\pi \ln 2}{2} \\ \Rightarrow \mathrm{Z}=\int_{0}^{\frac{\pi}{2}} \ln (\sin \mathrm{x}) \mathrm{dx}=-\frac{\pi \ln 2}{2} ......(7)$
$\\ \Rightarrow \text{On putting (7) in (2) and the obtainedresult in(1)} \\$

$\\\Rightarrow 2 \mathrm{I}=-\pi^{2} \ln 2 \\ \quad I=\int_{0}^{\pi} x \ln (\sin x) d x=-\frac{\pi^{2}}{2} \ln 2$

Question:47

Evaluate the following:
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log (\sin x+\cos x) d x$

Answer:

Given: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log (\sin x+\cos x) d x$
$\\ \begin{aligned} &\frac{1}{\text { Let }}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sin x+\cos x) d x \ldots(1)\\ &\text { Using the property: }\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\\ &\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sin x+\cos x) d x=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sin (-x)+\cos (-x)) d x\\ &\Rightarrow \text { As } \sin (-x)=\sin x \text { and } \cos (-x)=\cos x \end{aligned}$
$\\ \begin{aligned} &I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\cos x-\sin x) d x \ldots(2)\\ &\text { Adding equation(1) and(2) }\\ &2 \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sin \mathrm{x}+\cos \mathrm{x}) \mathrm{dx}+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{dx}\\ &2 \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln \left(\cos ^{2} x-\sin ^{2} x\right) \mathrm{d} x=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\cos 2 x) \mathrm{d} x \end{aligned}$
$\\ \Rightarrow \text { Put } 2 \mathrm{x}=\mathrm{t} \\ \Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ 2 \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos t) \mathrm{dt}$

$\\ \begin{aligned} &\Rightarrow \text { As } \cos (-x)=\cos x\\ &\text { Using property: }\\&\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x(f o r f(-x)=f(x))\\ &\Rightarrow{2 \mathrm{I}}=2\int_{0}^{\frac{\pi}{2}} \ln (\cos t) \mathrm{dt}\\ &\text { Using the property: } \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}$
$\\ \begin{aligned} &2I=\int_{0}^{\frac{\pi}{2}} \ln \left(\cos \left(\frac{\pi}{2}-t\right)\right) \mathrm{dt}=\int_{0}^{\frac{\pi}{2}} \ln (\operatorname{sint}) \mathrm{dt}\\ &\Rightarrow \text { Now From previous question eq }(7) \text { we obtained }\\ &\int_{0}^{\frac{\pi}{2}} \ln (\operatorname{sint}) \mathrm{dt}=-\frac{\pi \ln 2}{2}=2 \mathrm{I}\\ &I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sin x+\cos x) d x=-\frac{\pi \ln 2}{4} \end{aligned}$

Question:48

$\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$ is equal to

$\\A. 2(sinx + xcos \theta) + C\\ B. 2(sinx - xcos \theta) + C\\ C. 2(sinx + 2xcos \theta) + C\\ D. 2(sinx -2x cos \theta) + C\\$

Answer:

A)
$\\ \begin{aligned} &\text { Using Trigonometric identity } \cos 2 x=2 \cos ^{2} x-1\\ &\Rightarrow \int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x=\int \frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \theta-1\right)}{\cos x-\cos \theta} d x \end{aligned}$
$\\ =2 \int \frac{\cos ^{2} x-\cos ^{2} \theta}{\cos x-\cos \theta} d x \\ =2 \int\left(\frac{(\cos x+\cos \theta)(\cos x-\cos \theta)}{\cos x-\cos \theta}\right) \\ =2\{(\cos x+\cos \theta) d x \\ =2 \int \cos x d x+2 \int \cos \theta d x$
$\\=2 \int \cos x d x+2 \int \cos \theta d x=2(\sin x+x \cos \theta)+c$

Question:49

$\\ \int \frac{d x}{\sin (x-a) \sin (x-b)}$ is equal to

$\\ A. \sin (b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$\\\\ B. $\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$\\\\ C. $\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$\\\\ D. $\sin (b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$\\$

Answer:

C)
$\\ \begin{aligned} &\text { Given: } \int \frac{d x}{\sin (x-a) \sin (x-b)}\\ &\text { Multiply } \mathrm{Nr} \text { and } \mathrm{Dr} \text { by } \sin (\mathrm{b}-\mathrm{a})\\ &\Rightarrow \frac{1}{\sin (b-a)} \int \frac{\sin (b-a) d x}{\sin (x-a) \sin (x-b)} \end{aligned}$
$\\ \Rightarrow \sin (b-a)=\sin ((x-a)-(x-b)) \\ \Rightarrow \text { Also } \sin (A-B)=\sin A \cos B-\cos A \sin B \\ \Rightarrow \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)}=\frac{\sin ((x-a)-(x-b))}{\sin (x-a) \sin (x-b)}=\frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)} \\ \Rightarrow \quad \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)}=\frac{\cos (x-b)}{\sin (x-b)}-\frac{\cos (x-a)}{\sin (x-a)}$
$\\ =\frac{\cos (x-b)}{\sin (x-b)}-\frac{\cos (x-a)}{\sin (x-a)}=\cot (x-b)-\cot (x-a) \\ \quad \int \frac{d x}{\sin (x-a) \sin (x-b)}=\frac{1}{\sin (b-a)}\left(\int \cot (x-b) d x-\int \cot (x-a) d x\right) \\ \Rightarrow \operatorname{Now} \int \cot x d x=\ln |\sin x|+c \\ \Rightarrow \frac{1}{\sin (b-a)}\left(\int \cot (x-b) d x-\int \cot (x-a) d x\right)$
$\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}(\ln |\sin (\mathrm{x}-\mathrm{b})|-\ln |\sin (\mathrm{x}-\mathrm{a})|) \\ \Rightarrow \int \frac{\mathrm{dx}}{\sin (\mathrm{x}-\mathrm{a}) \sin (\mathrm{x}-\mathrm{b})}=\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \ln \left(\frac{\sin (\mathrm{x}-\mathrm{b})}{\sin (\mathrm{x}-\mathrm{a})}\right)=\operatorname{cosec}(\mathrm{b}-\mathrm{a}) \ln \left(\frac{\sin (\mathrm{x}-\mathrm{b})}{\sin (\mathrm{x}-\mathrm{a})}\right)$

Question:50

$\int \tan ^{-1} \sqrt{x} d x$ is equal to
$\\A.(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C$\\ B. $x \tan ^{-1} \sqrt{x}-\sqrt{x}+C$\\ C. $\sqrt{x}-x \tan ^{-1} \sqrt{x}+C$\\ D. $\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C$$

Answer:

A)
Given: $\int \tan ^{-1} \sqrt{x} d x$
$\\ \text { Put } x=t^{2} \\ \Rightarrow d x=2 d t \\ \Rightarrow \int \tan ^{-1} \sqrt{x} d x=\int 2 \tan ^{-1} \sqrt{t^{2}} d t \\ \Rightarrow 2 \int \tan ^{-1} \operatorname{tdt} \ldots(1)$
$\\ \begin{aligned} &\Rightarrow \text { Now apply integration by part on } \int t \tan ^{-1} t d t\\ &\Rightarrow \int t \tan ^{-1} t d t=\tan ^{-1} t \int t d t-\int\left(\frac{d}{d t} \tan ^{-1} t\right)\left(\int t d t\right) d t\\ &\Rightarrow \frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{t^{2}}{1+t^{2}} d t\\ &\Rightarrow \text { Now } \int \frac{t^{2}}{1+t^{2}} d t=\int \frac{t^{2}+1-1}{1+t^{2}} d t=\int \frac{t^{2}+1}{1+t^{2}} d t-\int \frac{1}{1+t^{2}} d t \end{aligned}$
$\\\begin{aligned} &\Rightarrow \int \frac{t^{2}+1}{1+t^{2}} \mathrm{dt}-\int \frac{1}{1+t^{2}} \mathrm{dt}=\int \mathrm{dt}-\int \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{t}-\tan ^{-1} \mathrm{t} \ldots\\ &\Rightarrow \text { Put }(3) \text { in }(2) \text { and the resulting equation in (1) }\\ &\Rightarrow 2 \int t \tan ^{-1} t d t=2\left(\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2}\left(t-\tan ^{-1} t\right)\right)\\ &\Rightarrow 2 \int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt}=\mathrm{t}^{2} \tan ^{-1} \mathrm{t}-\mathrm{t}+\tan ^{-1} \mathrm{t} \end{aligned}$
$\\ \Rightarrow t^{2} \tan ^{-1} t-t+\tan ^{-1} t=\tan ^{-1} t\left(t^{2}+1\right)-t \\ \Rightarrow \tan ^{-1} t\left(t^{2}+1\right)-t=\tan ^{-1} \sqrt{x}(x+1)-\sqrt{x} \\ \Rightarrow \int \tan ^{-1} \sqrt{x} d x=\tan ^{-1} \sqrt{x}(x+1)-\sqrt{x}+C$

Question:51

$\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{~d} \mathrm{x}$ is equal to
$\\A. \frac{e^{x}}{1+x^{2}}+C \\ B. \frac{-\mathrm{e}^{\mathrm{x}}}{1+\mathrm{x}^{2}}+\mathrm{C}$\\ C.$\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{x}^{2}\right)^{2}}+\mathrm{C}$\\ D. $\frac{-\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{x}^{2}\right)^{2}}+\mathrm{C}$$

Answer:

A)
Given: $\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{~d} \mathrm{x}$
$\\ =\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x=\int e^{x}\left(\frac{1+x^{2}-2 x}{\left(1+x^{2}\right)^{2}}\right) \mathrm{d} x \\ \quad=\int e^{x}\left(\frac{1+x^{2}-2 x}{\left(1+x^{2}\right)^{2}}\right) d x=\int e^{x}\left\{\left(\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}\right)+\left(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\right)\right\} d x \\ =\int e^{x}\left\{\left(\frac{1}{\left(1+x^{2}\right)}\right)+\left(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\right)\right\} d x$
$\\ \begin{aligned} &\text { Now using the property: }\\ &\int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{d} \mathrm{x}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})\\ &\Rightarrow \text { Now in } \int e^{x}\left\{\left(\frac{1}{\left(1+x^{2}\right)}\right)+\left(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\right)\right\} d x\\ &f(x)=\frac{1}{\left(1+x^{2}\right)}\\ &\Rightarrow f^{\prime}(x)=\frac{-2 x}{\left(1+x^{2}\right)^{2}} \end{aligned}$
$\\ =\int e^{x}\left\{\left(\frac{1}{\left(1+x^{2}\right)}\right)+\left(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\right)\right\} d x=\frac{e^{x}}{1+x^{2}}+c \\ \quad \int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x=\frac{e^{x}}{1+x^{2}}+c$

Question:52

$\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ is equal to
$\\\begin{aligned} A.&\frac{1}{5 x}\left(4+\frac{1}{x^{2}}\right)^{-5}+C\\ B.&\frac{1}{5}\left(4+\frac{1}{x^{2}}\right)^{-5}+C\\ C.&\frac{1}{10 x}(1+4)^{-5}+C\\ D.&\frac{1}{10}\left(\frac{1}{x^{2}}+4\right)^{-5}+C \end{aligned}$

Answer:

D)
Given: $\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$
Taking $x^2$ out from the denominator
$\\ \begin{aligned} &\int\left(\frac{x^{9}}{x^{12}\left(4+\frac{1}{x^{2}}\right)^{6}}\right) d x\\ &\int\left(\frac{1}{x^{3}\left(4+\frac{1}{x^{2}}\right)^{6}}\right) \mathrm{dx}=\int\left(\frac{\frac{1}{x^{3}}}{\left(4+\frac{1}{x^{2}}\right)^{6}}\right) \mathrm{dx}\\ &\Rightarrow \text { Now put } 4+\frac{1}{x^{2}}=t \end{aligned}$
$\\ \Rightarrow-\frac{2}{x^{3}} d x=d t \\ \Rightarrow \quad \int\left(\frac{\frac{1}{x^{3}}}{\left(4+\frac{1}{x^{2}}\right)^{6}}\right) d x=\int-\frac{1}{2t^{6}} d t \\ \Rightarrow \quad-\frac{1}{2}t^{-6} d t=\frac{-t^{-5}}{-5\times 2}+C=\frac{1}{10}\left(4+\frac{1}{x^{2}}\right)^{-5}+C$

Question:53

If $\int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$ then
$\\ A.a=\frac{-1}{10}, b=\frac{-2}{5}$\\ B. $a=\frac{1}{10}, b=-\frac{2}{5}$\\ C. $\mathrm{a}=\frac{-1}{10}, \mathrm{~b}=\frac{2}{5}$\\ D. $a=\frac{1}{10}, b=\frac{2}{5}$\\$

Answer:

C)
Given: $\int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$
Using concept of partial fractions
$\\ =\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{(x+2)}+\frac{B x+C}{\left(x^{2}+1\right)} \\ \Rightarrow A\left(x^{2}+1\right)+(B x+C)(x+2)=1$
$\\ \Rightarrow x^{2}(A+B)+x(C+2 B)+(A+2 C)=1 \\ \Rightarrow A+B=0 \quad \ldots(1) \\ \Rightarrow C+2 B=0 \quad \ldots(2) \\ \Rightarrow A+2 C=1...(3)$
On solving the above three equations we get
$\\ \Rightarrow A=\frac{1}{5}, B=-\frac{1}{5} \text { and } C=\frac{2}{5} \\ \Rightarrow \frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{\frac{1}{5}}{(x+2)}+\frac{-\frac{1}{5} x+\frac{2}{5}}{\left(x^{2}+1\right)} \\ \Rightarrow \frac{\frac{1}{5}}{(x+2)}+\frac{-\frac{1}{5} x+\frac{2}{5}}{\left(x^{2}+1\right)}=\frac{1}{5(x+2)}-\frac{x}{5\left(x^{2}+1\right)}+\frac{2}{5\left(x^{2}+1\right)}$
$\\ \int \frac{d x}{(x+2)\left(x^{2}+1\right)}=\int\left(\frac{1}{5(x+2)}-\frac{x}{5\left(x^{2}+1\right)}+\frac{2}{5\left(x^{2}+1\right)}\right) d x \\ =\frac{1}{5} \ln |x+2|-\frac{1}{10} \ln \left|x^{2}+1\right|+\frac{2}{5} \tan ^{-1} x +c \ldots (2)$
On comparing (1) and (2) we get,
$\Rightarrow a=-\frac{1}{10} \text { and } b=\frac{2}{5}$

Question:54

$\int \frac{x^{3}}{x+1}$ is equal to

$\\A. x+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1-x|+C$\\ B. $x+\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1-x|+C$\\ C. $x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1+x|+C$\\ D. $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1+x|+C$\\$

Answer:

D)
Given: $\int \frac{x^{3}}{x+1}$
$\\ \frac{x^{3}}{x+1}=\frac{x^{3}+1-1}{x+1} \\ \Rightarrow \frac{x^{2}+1}{x+1}-\frac{1}{x+1}=\frac{(x+1)\left(x^{2}-x+1\right)}{x+1}-\frac{1}{x+1}$
$\\ \int \frac{x^{3}}{x+1} d x=\int\left(\left(x^{2}-x+1\right)-\frac{1}{x+1}\right) d x \\ \Rightarrow \int\left(x^{2}-x+1\right) d x-\int \frac{1}{x+1} d x=\frac{x^{3}}{3}-\frac{x^{2}}{2}+x-\ln |1+x|+c$

Question:55

$\int \frac{x+\sin x}{1+\cos x} d x$ is equal to

$\\A. \log |1+\cos x|+c$\\ B. $\log |\mathrm{x}+\sin \mathrm{x}|+\mathrm{C}$\\ C. $x-\tan \frac{x}{2}+C$\\ D. $x \cdot \tan \frac{x}{2}+C$\\$

Answer:

$D. x \cdot \tan \frac{x}{2}+C$\\$

Given: $\int \frac{x+\sin x}{1+\cos x} d x$
As we know
$\sin 2 \mathrm{x}=\frac{2 \tan \mathrm{x}}{1+\tan ^{2} \mathrm{x}}, 1+\tan ^{2} \mathrm{x}=\sec ^{2} \mathrm{x} \text { and } \cos 2 \mathrm{x}=\frac{1+\tan ^{2} \mathrm{x}}{1-\tan ^{2} \mathrm{x}}$
$\Rightarrow \frac{x+\sin x}{1+\cos x}=\frac{x+\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}}{1+\frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{1-\tan ^{2}\left(\frac{x}{2}\right)}}$
$\\ \Rightarrow=\frac{x+x \tan ^{2}\left(\frac{x}{2}\right)+2 \tan \left(\frac{x}{2}\right)}{2} \\ \Rightarrow \int \frac{x+\sin x}{1+\cos x} d x=\int \frac{x+x \tan ^{2}\left(\frac{x}{2}\right)+2 \tan \left(\frac{x}{2}\right)}{2} d x \\ \Rightarrow \quad \operatorname{let} \frac{x}{2}=t \\ \Rightarrow \frac{d x}{2}=d t$
$\Rightarrow \int \left(2 t+2 t \tan ^{2} t+2 \tan t\right) d t=2 \int \left(t+t \tan ^{2 }t+\tan t \right) d t \\ \Rightarrow 2 \int \mathrm{tdt}+2 \int \mathrm{t} \left( \sec ^{2} \mathrm{t}-1 \right) \mathrm{dt}+2 \int \left(\tan \mathrm{t} \mathrm{dt}\right) \\ \Rightarrow 2 \int operatorname{tdt}+2 \int t \sec ^{2} t d t-2 \int operatorname{tdt}+2 \int \tan t d t\\ \Rightarrow 2 \int \mathrm{t} \sec ^{2} \mathrm{t} \mathrm{dt}+2 \int \text{ tant } \mathrm{dt} \ldots .(1)$ $\\\text{Applying Integration by parts on } \int \mathrm{t} \sec ^{2} \mathrm{t} dt \\ \Rightarrow \int t \sec ^{2} t d t=t \int \sec ^{2} t d t-\int\left(\frac{d}{d t} t\right)\left(\int \sec ^{2} t d t\right) d t $$$ $\\ \begin{aligned} &\Rightarrow \int \mathrm{t} \sec ^{2} \mathrm{t} \mathrm{dt}=\mathrm{t} \tan \mathrm{t}-\int \mathrm{tan} \mathrm{t} \mathrm{dt} \ldots(2)\\ \end{aligned}$ $\\ \begin{aligned} &\Rightarrow \text { Put }(2) \text { in }(1)\\ &\Rightarrow 2\left(\mathrm{t} \text { tant }-\int \tan t \mathrm{dt}\right)+2 \int \operatorname{tant} \mathrm{dt}=2 \mathrm{t} \operatorname{tant}+\mathrm{c}=\operatorname{xtan}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}\\ &\Rightarrow \int \frac{x+\sin x}{1+\cos x} d x=\operatorname{xtan}\left(\frac{x}{2}\right)+c \end{aligned}$

Question:56

If $\frac{x^{3} d x}{\sqrt{1+x^{2}}}=a\left(1+x^{2}\right)^{\frac{3}{2}}+b \sqrt{1+x^{2}}+C$ , then
$\\ A.a=\frac{1}{3}, b=1$\\ B. $a=\frac{-1}{3}, b=1$\\ C.$a=\frac{-1}{3}, b=-1$\\ D. $a=\frac{1}{3}, b=-1$\\$

Answer:

D)
Given: $\frac{x^{3} d x}{\sqrt{1+x^{2}}}=a\left(1+x^{2}\right)^{\frac{3}{2}}+b \sqrt{1+x^{2}}+C$ ...........(1)
$\\ \begin{aligned} &\text { Put }\\ &1+x^{2}=t\\ &\Rightarrow 2 x d x=d t\\ &\Rightarrow \int \frac{x^{3} d x}{\sqrt{1+x^{2}}}=\int \frac{x^{2} \cdot x d x}{\sqrt{1+x^{2}}} \end{aligned}$
$\\ \Rightarrow=\frac{1}{2} \int \frac{(t-1) \mathrm{dt}}{\sqrt{t}} \\ \Rightarrow \frac{1}{2} \int \frac{(t-1) \mathrm{d} t}{\sqrt{t}} \\ \Rightarrow=\frac{1}{2} \int \frac{t \mathrm{dt}}{\sqrt{t}}-\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{t}} \\ \Rightarrow=\frac{1}{2}\left(\int \sqrt{\mathrm{t}} \mathrm{dt}-\int\left(\frac{1}{\sqrt{\mathrm{t}}}\right) \mathrm{dt}\right)$
$\\ \begin{aligned} &\Rightarrow \frac{1}{2}\left(\int \sqrt{\mathrm{t}} \mathrm{dt}-\int\left(\frac{1}{\sqrt{\mathrm{t}}}\right) \mathrm{dt}\right)=\frac{1}{2}\left(\frac{2}{3} \mathrm{t}^{\frac{3}{2}}-2 \sqrt{\mathrm{t}}+\mathrm{c}\right)\\ &\Rightarrow\left(\frac{1}{3} t^{\frac{3}{2}}-\sqrt{t}+c\right)=\frac{1}{3}\left(1+x^{2}\right)^{\frac{3}{2}}-1 \sqrt{1+x^{2}}+c\\ &\text { Comparing (1) and (3) }\\ &\Rightarrow a=\frac{1}{3} \text { and } b=-1 \end{aligned}$

Question:57

$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x}$ is equal to
A. 1
B. 2
C. 3
D. 4

Answer:

A)
Given: $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x}$
Using trigonometric identities:
$\\ \cos ^{2} x+\sin ^{2} x=1 \text { and } \cos 2 x=\cos ^{2} x-\sin ^{2} x \\ \frac{1}{1+\cos 2 x}=\frac{1}{\left(\cos ^{2} x+\sin ^{2} x+\cos ^{2} x-\sin ^{2} x\right)} \\ =\frac{1}{2 \cos ^{2} x}$
$\\ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{d x}{1+\cos 2 x}\right)=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{d x}{2 \cos ^{2} x}\right)=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \\ \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x=\frac{1}{2}[\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\ =\frac{1+1}{2}=1$

Question:58

$\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x$ is equal to
A. $2\sqrt2$
B. $2(\sqrt2 + 1)$
C. $2$
D. $2 (\sqrt2 -1)$

Answer:

D)
As
$\\ \sin 2 x=2 \sin x \cos x \text { and } \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x} d x \\ \Rightarrow \int_{0}^{\frac{\pi}{2}} \sqrt{(\cos x-\sin x)^{2}} d x=\int_{0}^{\frac{\pi}{2}}|(\cos x-\sin x)| d x$
$\\ \text { From } 0<x<\frac{\pi}{4}, \cos x>\sin x \text { and } \\ \text { from } \frac{\pi}{4}<x<\frac{\pi}{2}, \cos x<\sin x \\ \Rightarrow \int_{0}^{\frac{\pi}{2}}|(\cos x-\sin x)| d x=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x$
$=2[\sin x+\cos x]_{0}^{\pi / 4}$
On solving the Above Integral we get $2 (\sqrt2 -1)$

Question:59

Fill in the blanks in each of the following
$\int_{0}^{\frac{\pi}{2}} \cos x e^{\sin x} d x$ is equal to ___________.

Answer:

$e{-1}$
Given: $\int_{0}^{\frac{\pi}{2}} \cos x e^{\sin x} d x$
$\\ \text { Put } \sin x=t \\ \Rightarrow \cos x d x=d t \\ \Rightarrow \operatorname{At} x=0=>t=0 \text { and } x=\frac{\pi}{2}=>t=1 \\ \Rightarrow \int_{0}^{1} e^{t} d t=\left[e^{t}\right]_{0}^{1}=e-1$

Question:60

Fill in the blanks in each of the following
$\int \frac{x+3}{(x+4)^{2}} e^{x} d x=$ ___________.

Answer:

$\\ \begin{aligned} &\frac{e^{x}}{x+4}+c\\ &\text { Given }\\ &\int \frac{x+3}{(x+4)^{2}} e^{x} d x\\ &=\int \frac{x+3}{(x+4)^{2}} e^{x} d x=\int \frac{(x+3)+1-1}{(x+4)^{2}} e^{x} d x=\int \frac{(x+4)-1}{(x+4)^{2}} e^{x} d x\\ &=\int \frac{(x+4)-1}{(x+4)^{2}} e^{x} d x=\int e^{x}\left(\frac{(x+4)}{(x+4)^{2}}-\frac{1}{(x+4)^{2}}\right) \mathrm{d} x \end{aligned}$
$\\\begin{aligned} &\int e^{x}\left(\frac{(x+4)}{(x+4)^{2}}-\frac{1}{(x+4)^{2}}\right) d x=\int e^{x}\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^{2}}\right) d x\\ &\text { Now using the property: }\\ &\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)\\ &\Rightarrow \text { Now in } \int e^{x}\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^{2}}\right) \mathrm{dx} \end{aligned}$
$\\ \Rightarrow f(x)=\frac{1}{x+4} \\ \quad f^{\prime}(x)=-\frac{1}{(x+4)^{2}} \\ \Rightarrow \int e^{x}\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^{2}}\right) d x=\frac{e^{x}}{x+4}+c \\ \Rightarrow \int \frac{x+3}{(x+4)^{2}} e^{x} d x=\frac{e^{x}}{x+4}+c$

Question:61

Fill in the blanks in each of the following
If $\int_{0}^{a} \frac{1}{1+4 x^{2}} d x=\frac{\pi}{8}$ , then a = ____________.

Answer:

$\\ \mathrm{a}=\frac{1}{2} \\ \text { Given: } \int_{0}^{\mathrm{a}} \frac{1}{1+4 \mathrm{x}^{2}} \mathrm{dx}=\frac{\pi}{8} \\ \frac{1}{1+4 \mathrm{x}^{2}}=\frac{\frac{1}{4}}{\frac{1}{4}+\mathrm{x}^{2}}=\frac{\frac{1}{4}}{\left(\frac{1}{2}\right)^{2}+\mathrm{x}^{2}} \\ \int_{0}^{\mathrm{a}} \frac{1}{1+4 \mathrm{x}^{2}} \mathrm{dx}=\int_{0}^{a} \frac{\frac{1}{4}}{\left(\frac{1}{2}\right)^{2}+\mathrm{x}^{2}} \mathrm{dx}$
$\\ \text { Now } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ \int_{0}^{a} \frac{\frac{1}{4}}{\left(\frac{1}{2}\right)^{2}+x^{2}} d x=\frac{\frac{1}{4}}{\frac{1}{2}} \tan ^{-1}\left(\frac{x}{\frac{1}{2}}\right) \\ =\left[\frac{1}{2} \tan ^{-1}(2 x)\right]_{0}^{a}$
$\\ =\frac{\pi}{8} \\ \frac{1}{2} \tan ^{-1}(2 a)=\frac{\pi}{8} \\ 2 a=\tan \left(\frac{\pi}{4}\right)=1 \\ a=\frac{1}{2}$

Question:62

Fill in the blanks in each of the following
$\int \frac{\sin x}{3+4 \cos ^{2} x} d x=$ __________.

Answer:

$\\ \begin{aligned} &-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+\mathrm{c}\\ &\text { Given }\\ &\int \frac{\sin x}{3+4 \cos ^{2} x} d x\\ &\Rightarrow \text { Now let } \cos \mathrm{x}=\mathrm{t} \end{aligned}$
$\\ \Rightarrow-\sin \mathrm{xdx}=\mathrm{dt} \\ \Rightarrow \quad \int \frac{\sin \mathrm{x}}{3+4 \cos ^{2} \mathrm{x}} \mathrm{dx}=\int \frac{-\mathrm{dt}}{3+4 \mathrm{t}^{2}}=\frac{1}{4} \int \frac{-\mathrm{dt}}{\frac{3}{4}+\mathrm{t}^{2}} \\ \Rightarrow \quad \frac{1}{4} \int \frac{-\mathrm{dt}}{\left(\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{t}^{2}}$
$\\ \Rightarrow \text { Now } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ \quad=\frac{1}{4} \int \frac{-d t}{\left(\frac{2}{2}\right)^{2}+t^{2}}=\frac{-\frac{1}{4}}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 t}{\sqrt{3}}\right)+c \\ \Rightarrow \int \frac{\sin x}{3+4 \cos ^{2} x} d x=-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$

Question:63

Fill in the blanks in each of the following
The value of $\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x$ is ____________.

Answer:

0
Using the property: $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$\\ \text { Let } I=\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x \\ \Rightarrow \int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x=\int_{-\pi}^{\pi} \sin ^{3}(\pi+(-\pi)-x) \cos ^{2}(\pi+(-\pi)-x) d x \\ \Rightarrow \text { As } \sin (-x)=-\sin x \text { and } \cos (-x)=\cos x$
$\\ =\int_{-\pi}^{\pi} \sin ^{3}(\pi+(-\pi)-\mathrm{x}) \cos ^{2}(\pi+(-\pi)-\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{-\pi}^{\pi} \sin ^{3}(-\mathrm{x}) \cos ^{2}(-\mathrm{x}) \mathrm{dx} \\ \Rightarrow \int_{-\pi}^{\pi} \sin ^{3}(-\mathrm{x}) \cos ^{2}(-\mathrm{x}) \mathrm{dx}=-\int_{-\pi}^{\pi} \sin ^{3} \mathrm{x} \cos ^{2} \mathrm{x} \mathrm{dx}=-\mathrm{I} \\ \Rightarrow \mathrm{I}=-\mathrm{I} \\ \Rightarrow 2 \mathrm{I}=0$
$\Rightarrow I=\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x=0$

More About NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Calculus can pose to be a difficult part of mathematics for many, but if one knows what to study and how to study then it can be simplified easily. NCERT exemplar Class 12 Maths solutions chapter 7 are not only crucial for mathematics and higher education point of view. It is also one of the highly scoring chapters of NCERT Class 12 Maths Solutions.

But it is also needed to study physics. In physics integrals play a major role, as it is used in topics like gravitation, work, probability, kinetic energy etc. therefore, the students must pay attention to what integrals are and how it is applied.

NCERT Exemplar Solutions For Class 12 Maths Chapter 7 Integrals Sub-Topics Covered

The sub topics covered in NCERT exemplar Class 12 Maths chapter 7 solutions are:

Introduction

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Geometric interpretation of indefinite integral
Some properties of indefinite integral
Comparison between integration and differentiation

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Methods of integration

Integration by substitution
Integration by trigonometric identities

Integrals of some particular functions

Integration of partial functions

Integration by parts

Integral of the typefxf'xdx
Integral of some other types

Definite integrals

Definite integral as a limit of a sum

The fundamental theorem of calculus

Area function
First fundamental theorem of integral calculus
Second fundamental theorem of integral calculus

Evaluation of definite integrals by substitution

Some properties of definite integrals

What can you learn in Class 12 Maths NCERT Exemplar Solutions Chapter 7?

In NCERT exemplar Class 12 Mathematics chapter 7 Integrals, students will get a better picture of what are integrals and how it is used in advanced calculus. Here, they will learn about integration and geometric representation of the same using graphs. One will learn in detail about various methods of integration, the difference between differentiation and integration, and indefinite integral properties.

Many students tend to find integrals as a chapter that is difficult and hassling. But they should remember that if one wants to score well in exams and also pursue a career in engineering, then understanding and studying integrals is crucial. Therefore, one should ask as many questions as possible to get a hang of the topic.

NCERT exemplar solutions for Class 12 Maths chapter 7 will help in tackling the questions asked in this topic. One should practice the questions as much as possible and should use the Class 12 Maths NCERT exemplar solutions chapter 7 as reference. This will not only make the student confident about integrals but also for tackling any numerical they get in the exam.

NCERT Exemplar Class 12 Maths Solutions

Chapter 1 Relations and Functions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 5 Continuity and Differentiability

Chapter 6 Applications of Derivatives

Chapter 8 Applications of Integrals

Chapter 9 Differential Equations

Chapter 10 Vector Algebra

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 7

In NCERT exemplar Class 12 Maths solutions chapter 7, there are a variety of concepts and topics covered. The main aim is to make the students well aware of integrals and its operations, so much so that any question can be solved easily. Students will learn how to find integrals for various types of functions.
Integrals are the opposite of derivatives of a function and are denoted by the area under the function curve. When it comes to integrals, it can be stated as the opposite of differential calculus. In simple words, integrals describe the numerical value of any type of change like distance like displacement, volume, area and work.
NCERT exemplar Class 12 Maths chapter 7 solutions deals with the important topics coming for the exams which are the methods, functions and operations.

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NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Must Read NCERT Solution subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Are these solutions helpful for competitive examinations?

Indeed, the Class 12 Maths NCERT exemplar solutions chapter 7 covers the syllabus of the competitive exams like NEET and JEE Main to help you ace them.

2. What are the important topics of this chapter?

Methods of Integration, Integration by Parts and Fundamental Theorem of Calculus are some of the important topics of this chapter. However, rest should not be neglected either.

3. How many questions are there in this chapter?

The NCERT exemplar solutions for Class 12 Maths chapter 7 consists of 1 exercise with 63 distinct questions for practice.

4. How many times should one need to read the NCERT books?

Students should read the books enough times months before your exam for better remembering. They can also take help of NCERT exemplar Class 12 Maths solutions chapter 7 pdf download using an online webpage to pdf tool.

Also Read

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
You have to appear for the 2025 12th board exams.
Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
Aim to register before late October to avoid extra fees.
Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer

kumardurgesh1802 10 July,2024

I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Read Complete Answer

satyamercedes0626 23 July,2024

I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.

Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

Hope this resolves your query.

Read Complete Answer

Abhishek 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

More About NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

NCERT Exemplar Solutions For Class 12 Maths Chapter 7 Integrals Sub-Topics Covered

Applications for Admissions are open.

Tallentex 2025 - ALLEN's Talent Encouragement Exam

Aakash iACST Scholarship Test 2024

What can you learn in Class 12 Maths NCERT Exemplar Solutions Chapter 7?

NCERT Exemplar Class 12 Maths Solutions

Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 7

JEE Main Important Mathematics Formulas

NCERT Exemplar Class 12 Solutions

Also, check NCERT Solutions for questions given in the book:

Must Read NCERT Solution subject wise

Read more NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

Also Read

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