NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Edited By Ravindra Pindel | Updated on Sep 15, 2022 - 5:03 p.m. 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

Integration as an inverse process of differentiation

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

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

Exemplar

Frequently Asked Question (FAQs) - NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Question: How many times should one need to read the NCERT books?

Answer:

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.

Question: How many questions are there in this chapter?

Answer:

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

Question: What are the important topics of this chapter?

Answer:

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.

Question: Are these solutions helpful for competitive examinations?

Answer:

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.

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Questions related to CBSE Class 12th

Showing 147 out of 147 Questions

251 Views

i couldnt clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

Upasana Barman 24th Aug, 2022

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

307 Views

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Mayankjha.student2007 24th Aug, 2022

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

99 Views

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Nuthalapati Vyshnavi 21st Aug, 2022

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

65 Views

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Muskan Dhalwal 18th Aug, 2022

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

747 Views

I will have my boards in Feb 2023 (Class 12) and I still havent completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

khushi.thakurbbk 7th Sep, 2022

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

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NCERT Solutions

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

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

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs) - NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Question: How many times should one need to read the NCERT books?

Question: How many questions are there in this chapter?

Question: What are the important topics of this chapter?

Question: Are these solutions helpful for competitive examinations?

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