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

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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

Integration as an inverse process of differentiation

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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.

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 Questions (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 Class 12 Maths Notes - Download Chapter wise Free PDFs

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 Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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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 ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

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.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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

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

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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

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

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