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NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

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NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

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

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

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: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: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: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: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 xis 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)

More About NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

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

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

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

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

  • Introduction

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  • 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

Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 7

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

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

Must Read NCERT Solution subject wise

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Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Are these solutions helpful for competitive examinations?

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

2. What are the important topics of this chapter?

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

3. How many questions are there in this chapter?

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

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

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

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Hello aspirant,

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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

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.

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.

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.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive. 

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Stock Analyst

Individuals who opt for a career as a stock analyst examine the company's investments makes decisions and keep track of financial securities. The nature of such investments will differ from one business to the next. Individuals in the stock analyst career use data mining to forecast a company's profits and revenues, advise clients on whether to buy or sell, participate in seminars, and discussing financial matters with executives and evaluate annual reports.

2 Jobs Available
Researcher

A Researcher is a professional who is responsible for collecting data and information by reviewing the literature and conducting experiments and surveys. He or she uses various methodological processes to provide accurate data and information that is utilised by academicians and other industry professionals. Here, we will discuss what is a researcher, the researcher's salary, types of researchers.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Veterinary Doctor
5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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