NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Komal MiglaniUpdated on 31 Mar 2025, 03:53 AM IST

Integrals play a significant role in mathematics. Have you ever thought about how to find the area under a curve, the distance travelled by an object accelerating throughout the distance, or even just the total amount of a quantity accumulated over time? Integrals can be used to find the solution to those types of continuous problems. Within the scope of Class 12 Mathematics, integration is introduced as the anti-derivative, or reverse differentiation, ultimately giving us the ability to retrieve a function from its derivative. In addition, this chapter regarding integration has indefinite integrals, which implies that such a family is represented by an arbitrary constant (∫f(x)dx = F(x) + C).

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  1. NCERT Exemplar Class 12 Maths Solutions Chapter 7
  2. NCERT Exemplar Class 12 Maths Solutions
  3. Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 7
  4. NCERT Solutions for Class 12 Maths: Chapter Wise
  5. NCERT solutions of class 12 - Subject-wise
  6. NCERT Notes of class 12 - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. NCERT Exemplar Class 12 Solutions - Subject Wise

To become proficient in this subject, students should work through the NCERT Solutions for Class 12 Maths, which include clear explanations and very easy-to-follow methods for solving integral problems. The NCERT Solutions encompass a breadth of exercises, such as properties of integrals, applications of integration in physics and engineering, as well as the Fundamental Theorem of Calculus, which connects differentiation and integration. The conceptual understanding and work as preparation for the board exam and competitive examinations, like JEE and NEET.

NCERT Exemplar Class 12 Maths Solutions Chapter 7

Class 12 Maths Chapter 7 exemplar solutions Exercise: 7.3
Page number: 163-169
Total questions: 63

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;

$
\begin{aligned}
& \int \frac{2 x+3}{x^2+3 x} d x=\log \left|x^2+3 x\right|+C \\
& \text { LHS }=\int \frac{2 \mathrm{x}+3}{\mathrm{x}^2+3 \mathrm{x}} \mathrm{dx} \text { Let; } t=x^2+3 x \\
& \Rightarrow d t=2 x+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=\text { RHS } \\
& {\left[\because t=x^2+3 x\right]}
\end{aligned}
$

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; $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x$
$\\ =\int{c} \sin x+\cos x \\ \sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \\$

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

Question:9

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

Answer:

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

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

$=\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x \\ =\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ =2 \sin \frac{x}{2}-2 \cos \frac{x}{2}+c$

Question:10

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

Answer:

Given;

Let $z = \sqrt{x}$

\\ \Rightarrow x = z\textsuperscript{2}\\ \\ \Rightarrow dx = 2z dz\\ =\int \frac{z^{2}}{z+1} 2 z d z\\ =2 \int \frac{\mathrm{z}^{3}}{\mathrm{z}+1} \mathrm{~d} z$

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

$\\ =\frac{2 t^{3}}{3}-3 t^{2}+3 t-\log |t|+C \\$

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

Question:11

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

Answer:

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

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

Question:12

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

Answer:

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

Question:13

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

Answer:

$\\ \begin{aligned} &\text { Given; }\\ &\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x\\ &\text { Let } x=\tan y\\ &\Rightarrow \mathrm{dx}=\sec ^{2} \mathrm{y} \mathrm{dx} \end{aligned}$
$\\ =\int \frac{\sqrt{1+\tan ^{2} y}}{\tan ^{4} y} \sec ^{2} y d y \\$

$ =\int \sec y \times \frac{\cos ^{4} y}{\sin ^{4} y} \times \sec ^{2} y d y \\ $

$=\int \frac{\cos y}{\sin ^{4} y} d y \\$

$ \text { Let } t=\sin y$
$\Rightarrow dt = cos y dy$
$\\ =\int \frac{\mathrm{dt}}{\mathrm{t}^{4}}=-\frac{1}{3 \mathrm{t}^{3}}+\mathrm{C} \\ =-\frac{1}{3 \sin ^{3} \mathrm{y}} \\ =-\frac{1}{3 \sin ^{3}\left(\sin ^{-1} \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}}\right)} \\ =-\frac{\left(\mathrm{x}^{2}+1\right)^{\frac{3}{2}}}{3 \mathrm{x}^{3}}$

Question:14

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

Answer:

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

Question:15

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

Answer:

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

$ =\frac{1}{\sqrt{2}} \sin ^{-1} \frac{4 t-3}{3}+C$

Question:16

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

Answer:

$\\ \text { Given; } \frac{3 x-1}{\sqrt{x^{2}+9}} d x \\$

$ =\int \frac{3 x}{\sqrt{x^{2}+3^{2}}} d x-\int \frac{1}{\sqrt{x^{2}+3^{2}}} d x \\$

$ \text { [Let; } x^{2}+9=y \Rightarrow \left.2 x d x=d y\right] \\$

$ =\frac{3}{2} \int \frac{d y}{\sqrt{y}}-\log \left|x+\sqrt{x^{2}+3^{2}}\right|+c \\ $

$=3 \sqrt{x^{2}+9}-\log \left|x+\sqrt{x^{2}+9}\right|+C$

Question:17

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

Answer:

$\\ \text { Given; } \int \sqrt{5-2 x+x^{2}} d x \\$

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

$ {\left[\because \int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|\right.} \\$

$ =\frac{x-1}{2} \sqrt{5-2 x+x^{2}}+2 \log \left|(x+1)+\sqrt{5-2 x+x^{2}}\right|+C$

Question:18

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

Answer:

$\\ \text { Given; } \int \frac{x}{x^{4}-1} d x \\$

$ \text { [Let; } t=x^{2}\ \Rightarrow \left.d t=2 x d x\right] \\$

$ =\int \frac{1}{\left(t^{2}-1\right)} \frac{d t}{2}$
$\\ =\frac{1}{2} \int \frac{1}{(t+1)(t-1)} d t \\ =\frac{1}{2} \int \frac{1}{2}\left[\frac{1}{(t-1)}-\frac{1}{(t+1)}\right] d t \\ $${[\because \int \frac{1}{x} d x}={\log |x|}]$

$=\frac{1}{4}(\log |t-1|-\log |t+1|)+C$$\\ {\left[\because \log a-\log b=\log \frac{a}{b}\right]} \\ $

$=\frac{1}{4} \log \left|\frac{t-1}{t+1}\right|+c \\ =\frac{1}{4} \log \left|\frac{x^{2}-1}{x^{2}+1}\right|+c$

Question:19

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

Answer:

Given: $\int \frac{x^{2}}{1-x^{4}} d x$
$\\ =\int \frac{1}{2}\left[\frac{2 x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ =\int \frac{1}{2}\left[\frac{x^{2}+x^{2}-1+1}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ $

$=\int \frac{1}{2}\left[\frac{\left(1+x^{2}\right)-\left(1-x^{2}\right)}{\left(1+x^{2}\right)\left(1-x^{2}\right)} d x\right] \\ $

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

$ \int \frac{1}{\left(a^{2}+x^{2}\right)} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a} \\$

$ =\frac{1}{2} \times \frac{1}{2} \log \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x+c \\ =\frac{1}{4} \log \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x+c$

Question:20

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

Answer:

$\\ \text { Given; } \int \sqrt{2 a x-x^{2}} d x \\ =\int \sqrt{a^{2}-a^{2}+2 a x-x^{2}} d x \\$

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

$ =\frac{(x-a)}{2} \sqrt{2 a x-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x-a}{a}+c$

Question:21

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

Answer:

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

$ =\mathrm{t} \operatorname{tant}-\int \mathrm{tant} \mathrm{dt} \\ =\mathrm{t} \operatorname{tant}+\int \frac{-\sin \mathrm{t}}{\cos t} \mathrm{dt} \\ $

$=\mathrm{t} \operatorname{tant}+\log |\cos t|+\mathrm{C} \\ =\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}+\log \left|\sqrt{1-\mathrm{x}^{2}}\right|+\mathrm{C}$

Question:22

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

Answer:

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

$ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2}}{1-2 \cos 3 x} d x \\ =\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{(1-2 \cos 3 x) \cos \frac{3 x}{2}} d x \\ $

$=\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2}-2 \cos 3 x \cos \frac{3 x}{2}} d x$

$\\ {[\because 2 \cos a \cos b=\cos (a+b)+\cos (a-b)]} \\ $

$=\int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2}-\cos \frac{9 x}{2}-\cos \frac{3 x}{2}} d x \\ =\int-2 \cos \frac{3 x}{2} \cos \frac{x}{2} d x \\$

$ =\int-\cos 2 x-\cos x d x \\ =-\frac{\sin 2 x}{2}-\sin x+C$

Question:23

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

Answer:

$\\ \text { Given; } \int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cos ^{2} x} \mathrm{dx} \\ $

$=\int \frac{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)}{\sin ^{2} x \cos ^{2} x} \mathrm{dx} \\$

$ =\int \frac{\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)}{\sin ^{2} x \cos ^{2} x} d x \\$

$ =\int \frac{\sin ^{2} x}{\cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}-1 d x \\$

$ =\int \tan ^{2} x+\cot ^{2} x-1 d x$
$\\$

$ =\int \sec ^{2} x-1+\operatorname{cosec}^{2} x-1-1 \mathrm{dx} \\ $

$=\tan x-\cot x-3 x+C$

Question:24

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

Answer:

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

Question:25

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

Answer:

Given, $\int \frac{\cos x-\cos 2 x}{1-\cos x} d x$
$\\ =\int \frac{\cos 2 x-\cos x}{\cos x-1} d x \\ $

$=\int \frac{2 \cos ^{2} x-1-\cos x}{\cos x-1} d x \\ $

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

Question:26

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

Answer:

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

Question:27

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

Answer:

$\\ \text { Given; } \int_{0}^{2}\left(\mathrm{x}^{2}+3\right) \mathrm{d} \mathrm{x} \\$

$ \text { We know } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{h} \sum_{\mathrm{r}=0}^{\mathrm{n}-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \\$

$ \text { Here } \mathrm{a}=0 . \mathrm{b}=2 \\ \mathrm{~h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{\mathrm{n}}=\frac{2}{\mathrm{n}} \\$

$\Rightarrow\mathrm{nh}=2$
$\\ =\operatorname{limh}_{h \rightarrow 0} \sum_{r=0}^{n-1} f(r h) \\ =\lim _{h \rightarrow 0} h \sum_{r=0}^{n-1}\left(3+r^{2} h^{2}\right) \\$

$ =\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{(n-1)(n-1+1)(2 n-2+1)}{6}\right)\right.$
$\\ =\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{\left(n^{2}-n\right)(2 n-1)}{6}\right)\right. \\ $

$=\operatorname{limh}_{h \rightarrow 0} h\left(3 n+h^{2}\left(\frac{2 n^{3}-3 n^{2}+n}{6}\right)\right. \\ =\lim _{h \rightarrow 0}\left(3 n h+\left(\frac{2 n^{3} h^{3}-3 n^{2} h^{3}+n h^{3}}{6}\right)\right.$
$\\ =\lim _{h \rightarrow 0}\left(3.2+\left(\frac{2.2^{3}-3.2^{2}.h+2 h^{2}}{6}\right)\right. \\ =6+\frac{16}{3} \\ =\frac{26}{3}$

Question:28

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

Answer:

$\\ \text { We know } \int_{a}^{b} f(x) d x=\lim _{h \rightarrow \infty} h \sum_{r=0}^{n-1} f(a+r h) \\$

$ \text { Here } a=0, b=2 \\$

$\\ h=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n} \\$

$\Rightarrow h=2$
$\\ =\operatorname{limh}_{h \rightarrow 0} \sum_{r=0}^{n-1} f(r h) \\ =\lim _{h \rightarrow 0} h\left[1+e^{h}+e^{2 h}+\cdots,+e^{(n-1) h}\right] \\ =\lim _{h \rightarrow 0} h\left[\frac{1\left(e^{h}\right)^{n}-1}{e^{h}-1}\right]$
$\\ =\lim _{h \rightarrow 0} h\left[\frac{e^{n h}-1}{e^{h}-1}\right] \\ =\lim _{h \rightarrow 0} h\left[\frac{e^{2}-1}{e^{h}-1}\right] \\ =(e^{2}-1 )\lim _{h \rightarrow 0}\left[\frac{h}{e^{h}-1}\right] \\ =e^{2}-1$

Question:29

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

Answer:

$\\ \text { Given; } \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}} \\$

$ =\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1} \\ =\int_{1}^{e} \frac{d t}{t^{2}+1} \\ \text { [Let; }.t=e^{x}[\text { when } x=0, t=1 \text { and } x=1, t=e] ]$

$\Rightarrow d t=[e^{x} d x] \\ =\left[\tan ^{-1} t\right]_{1}^{e}$
$=\tan ^{-1} \mathrm{e}-\tan ^{-1} 1=\tan ^{-1} \mathrm{e}-\frac{\pi}{4}$

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 a 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

$\\ \int_{2}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\$

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

$\Rightarrow\int_{0}^{\pi} x \sin x \cos ^{2} x d x=\int_{0}^{\pi}(\pi-x) \sin (\pi-x) \cos ^{2}(\pi-x) d x \\$

$ \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$

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

$\\ \Rightarrow 2\mathrm{I}=-\pi \int_{1}^{-1} \mathrm{t}^{2} \mathrm{dt}=-\pi\left[\frac{\mathrm{t}^{2}}{3}\right]_{1}^{-1}=\frac{2 \pi}{3} \\$

$ \Rightarrow{2 \mathrm{I}=\frac{2 \pi}{3}} \\ \Rightarrow{\mathrm{I}=\frac{\pi}{3}} \\$

$ \Rightarrow \int _{0}^{\pi} x \sin x \cos ^{2} x \mathrm{dx}=\frac{\pi}{3}$

Question:34

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

Answer:

$\\ \text { Given } \int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} \\ $

$\Rightarrow\text { Let } x=\sin \theta \\$

$\text { At } x=0, \theta=0 \\ \text { and } x=\frac{1}{2}, \theta=\frac{\pi}{6}$
$\\ =\int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}}=\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \sqrt{1-\sin ^{2} \theta}} \\$

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

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

$\Rightarrow\int_{0}^{\frac{1}{2}} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-x^{2}}}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

Question:35

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

Answer:

$\\ \text { Given: } \int \frac{x^{2} d x}{x^{4}-x^{2}-12} \\ \text { Put } x^{2}=t \\ $

$=>\frac{x^{2}}{x^{4}-x^{2}-12}=\frac{t}{t^{2}-t-12}$
$\\ \Rightarrow \mathrm{t}^{2}-\mathrm{t}-12=(\mathrm{t}+3)(\mathrm{t}-4) \\$

$ \Rightarrow \frac{\mathrm{t}}{(\mathrm{t}+3)(\mathrm{t}-4)}=$

$\frac{\mathrm{A}}{(\mathrm{t}+3)}+\frac{\mathrm{B}}{(\mathrm{t}-4)} \text { (Concept of partial fraction) } \\$

$ \Rightarrow \mathrm{t}=\mathrm{t}(\mathrm{A}+\mathrm{B})+3 \mathrm{~B}-4 \mathrm{~A}$
On comparing coefficients of ‘t’ we get
$\\ 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)} $

$ \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:

$\\ \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)}\\ $

$\Rightarrow \frac{t}{\left(t+a^{2}\right)\left(t+b^{2}\right)}=\frac{A}{\left(t+a^{2}\right)}+\frac{B}{\left(t+b^{2}\right)}(\text { Concept of partial fraction })$

$\Rightarrow t=t(A+B)+a^{2} B+b^{2} A\\ \text { On comparing coefficients of "twe get }\\ $

$\Rightarrow \mathrm{A}=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-\mathrm{b}^{2}} \ \mathrm{~B}=\frac{-\mathrm{b}^{2}}{\mathrm{a}^{2}-\mathrm{b}^{2}}$

$\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 $
$\\ =\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 \\$

And $\text { for } \int_{0}^{\pi} \frac{\sin x}{\cos ^{2} x} d x \\ \text { Put } \cos x=t$
$\\ \Rightarrow-\sin x d x=d t\\ $

$\Rightarrow \mathrm{At}^{\mathrm{x}}=0=>\mathrm{t}=1 \text { and } \mathrm{x}=\pi=>\mathrm{t}=-1\\ =\int_{1}^{-1}-\frac{d t}{t^{2}}=\left[\frac{1}{t}\right]_{1}^{-1}=-2 \ldots$

$\\ 2\mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}=$

$\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 \\ $

${l} =\frac{d x}{1+x^{2}}=d t \\ \Rightarrow \int e^{\tan ^{-1} x}\left(\frac{1+x+x^{2}}{1+x^{2}}\right) d x=\int e^{t}\left(1+\tan t+\tan ^{2} t\right) d t \\$

$ \Rightarrow \operatorname{As} \sec ^{2} \theta-\tan ^{2} \theta=1 \\ $

$\Rightarrow \int e^{t}\left(1+\tan t+\tan ^{2} t\right) d t=\int e^{t}\left(1+\tan t+\sec ^{2} t-1\right) d t$
$\\ \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 $
$\\=f(t)=\tan t \\ \Rightarrow f^{\prime}(t)=\sec ^{2} x \\ $

$\Rightarrow \int e^{t}\left(\tan t+\sec ^{2} t\right) d t=e^{t} \tan t+C \\$

$ \Rightarrow \int e^{\tan ^{-1} x}\left(\frac{1+x+x^{2}}{1+x^{2}}\right) d x=e^{\tan ^{-1} x} x+C$

Question:40

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

Answer:

Given: $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
$\\ \Rightarrow \text { Put } x=a \tan ^{2} \theta\\ \Rightarrow \mathrm{dx}=2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta\\ \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$$\int \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}} 2 a \tan \theta \sec ^{2} \theta d \theta $
$\\ \Rightarrow \mathrm{As} \sec ^{2} \theta-\tan ^{2} \theta=1 \\$

$ =\int \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a\left(1+\tan ^{2} \theta\right)}} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta \\ \Rightarrow \int \sin ^{-1} \sqrt{\frac{\tan ^{2} \theta}{\sec ^{2} \theta}} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta$

$=\int \sin ^{-1} \sqrt{\sin ^{2} \theta} 2 \mathrm{a} \tan \theta \sec ^{2} \theta \mathrm{d} \theta \\ $

$\Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta$

$\Rightarrow$ $\text{Now put }$$\tan \theta=t$

$\\ \Rightarrow \sec ^{2} \theta \mathrm{d} \theta=\mathrm{d} t $

$\Rightarrow 2 \mathrm{a} \int \theta \tan \theta \sec ^{2} \theta \mathrm{d} \theta=2 \mathrm{a} \int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt} \ldots$

$\Rightarrow$ $\text{Now apply integration by part on}$$\int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt}$

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:
$\\ \Rightarrow \cos 2 \mathrm{x}=2 \cos ^{2} \mathrm{x}-1=1-2 \sin ^{2} \mathrm{x} \\ \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}}=\frac{\sqrt{1+\left(2 \cos ^{2} \frac{x}{2}-1\right)}}{\left(1-\left(1-2 \sin ^{2} \frac{x}{2}\right)\right)^{\frac{5}{2}}} \\ =\frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{5}{2}}}$
$\\ =\frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{5}{2}}} \\ =\left(\sqrt{2} \cos \frac{x}{2}\right) /\left(2^{\frac{5}{2}} \sin ^{5} \frac{x}{2}\right) \\ =\frac{\sqrt{2 }\cos \frac{x}{2}}{2^{\frac{5}{2}} \sin ^{5} \frac{x}{2}} \\ =\frac{1}{4} \frac{\cos \frac{x}{2}}{\sin ^{5} \frac{x}{2}}$
$\Rightarrow\\ \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 \\ $

$\text { Put } \sin \left(\frac{x}{2}\right)=t \\ \cos \left(\frac{x}{2}\right) d x=2 d t \\ \text { At } x=\frac{\pi}{3}$

$\Rightarrow t=\frac{1}{2} \text { and at } x=\frac{\pi}{2}$

$\Rightarrow t=\frac{1}{\sqrt{2}}$
$\Rightarrow\\\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin ^{5} \frac{x}{2}} \mathrm{dx}=\frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}}$
$\Rightarrow\\ \frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}}=\frac{1}{2}\left[\frac{t^{-4}}{-4}\right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}=\frac{3}{2} \\ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=\frac{3}{2}$

Question:42

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

Answer:

$\\ \begin{aligned} &\text { Given: } \int e^{-3 x} \cos ^{3} x d x\\ &\text { Using trigonometric identity }\\ &\cos 3 x=4 \cos ^{3} x-3 \cos x\\ &\int e^{-3 x} \cos ^{3} x d x=\frac{1}{4} \int e^{-3 x}(\cos 3 x+3 \cos x) d x\\ &\frac{1}{4} \int e^{-3 x}(\cos 3 x+3 \cos x) d x=\frac{1}{4} \int e^{-3 x} \cos 3 x d x+\frac{3}{4} \int e^{-3 x} \cos x d x ...(1) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Using a generalised formula i.e }\\ &\int e^{a x} \cos b x d x=\frac{e^{2 x}}{a^{2}+b^{2}}(a \cos b x+b \sin b x)\\ &\int e^{-3 x} \cos 3 x d x=\frac{e^{-2 x}}{(-3)^{2}+3^{2}}((-3) \cos 3 x+3 \sin 3 x) \end{aligned}$
$=\\ \frac{e^{-2 x}}{(-3)^{2}+3^{2}}((-3) \cos 3 x+3 \sin 3 x)$

$=\frac{e^{-2 x}}{6}(\sin 3 x-\cos 3 x) \ldots(2)\\ \int e^{-3 x} \cos x d x$

$=\frac{e^{-2 x}}{(-3)^{2}+1^{2}}((-3) \cos x+\sin x)$

$=\frac{e^{-2 x}}{10}(\sin x-3 \cos x)\\ =\frac{e^{-2 x}}{10}(\sin x-3 \cos x) \ldots(3)\\ $

$\Rightarrow \text { On putting }(2) \text { and }(3) \text { in }(1)$
$\frac{1}{4} \int e^{-3 x} \cos 3 x d x+\frac{3}{4} \int e^{-3 x} \cos x d x=\frac{e^{-2 x}}{4 \times 6}(\sin 3 x-\cos 3 x)+\frac{3 e^{-2 x}}{4 \times 10}(\sin x- 3 \cos x) \\$

$ \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
$\\ \Rightarrow \int \frac{\left(t^{2}+1\right)}{1+t^{4}} d t+\int \frac{\left(t^{2}-1\right)}{1+t^{4}} d t=\int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} d t+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt} \\ \int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}$

$=\int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}+\int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt} \ldots . \text { (1) }$
$\\ \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\\ $

$\Rightarrow\left(1+\frac{1}{t^{2}}\right) d t=d z$

$\\ =\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \mathrm{dt} \\$

$ =\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)$
$\\ \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\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) $

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} \\ $

$\Rightarrow \mathrm{At} \mathrm{x}=0$

$\mathrm{t}=1 \text { and at } \mathrm{x}=1$

$\Rightarrow\mathrm{t}=3 \\ $

$\Rightarrow \int_{0}^{1} \mathrm{x} \ln (1+2 \mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{4}\int_{1}^{3}(\mathrm{t}-1) \ln \mathrm{t} \mathrm{dt}....(i)$

$\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)\\ $
$\\ \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 $

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 \\ $

$\text { Let } I=\int_{0}^{\pi} x \ln (\sin x) d x \\ =\int_{0}^{\pi}(\pi-x) \ln (\sin (\pi-x)) d x=\int_{0}^{\pi} \pi \ln (\sin x) d x \int_{0}^{\pi} x \ln (\sin x) d x \\$

$ \Rightarrow \text { As } \sin (\pi-x)=\sin x$
$\\ \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) $
$\\ \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 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. \\ $

$\Rightarrow {2 Z}=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x=\int_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) d x \ldots(5)$
$\\ =\int_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) d x=\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{2 \sin x \cos x}{2}\right) d x \\$

$=\int_{0}^{\frac{\pi}{2}} \ln \left(\frac{2 \sin x \cos x}{2}\right) d x=\int_{0}^{\frac{\pi}{2}}(\ln (\sin 2 x)-\ln 2) d x \\$

$ \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) \\ $

$\Rightarrow \operatorname{Now} \operatorname{in} \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x \text { put } 2 x=t$
$\\ \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\\ $

$\Rightarrow{2 Z}=\frac{1}{2} \int_{0}^{\pi} \ln (\sin t) d t-\frac{\pi \ln 2}{2}$
$\\ \Rightarrow \text{from equation (2)} \frac{1}{2} \int_{0}^{\pi} \ln (\operatorname{sint}) \mathrm{dt} \text{again becomes} \\$

$\Rightarrow 2 Z=\frac{2}{2} \int_{0}^{\frac{\pi}{2}} \ln (\sin t) \mathrm{dt}-\frac{\pi \ln 2}{2}\\$
$\\\Rightarrow From eq. (3) \\$

$\Rightarrow 2 \mathrm{Z}=\mathrm{Z}-\frac{\pi \ln 2}{2} \\ $

$\Rightarrow \mathrm{Z}=\int_{0}^{\frac{\pi}{2}} \ln (\sin \mathrm{x}) \mathrm{dx}=-\frac{\pi \ln 2}{2} ......(7)$
$\\ \Rightarrow \text{On putting (7) in (2) and the obtained result in(1)} \\$

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

Question:47

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

Answer:

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

$\Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ 2 \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos t) \mathrm{dt}$

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

Question:48

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

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

Answer:

A)

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

$ \\ =2\{(\cos x+\cos \theta) d x \\ =2 \int \cos x d x+2 \int \cos \theta d x$
$\\=2 \int \cos x d x+2 \int \cos \theta d x=2(\sin x+x \cos \theta)+c$

Question:49

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

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

Answer:

C)

$\\ \begin{aligned} &\text { Given: } \int \frac{d x}{\sin (x-a) \sin (x-b)}\\ &\text { Multiply } \mathrm{Nr} \text { and } \mathrm{Dr} \text { by } \sin (\mathrm{b}-\mathrm{a})\\ &\Rightarrow \frac{1}{\sin (b-a)} \int \frac{\sin (b-a) d x}{\sin (x-a) \sin (x-b)} \end{aligned}$
$\\ \Rightarrow \sin (b-a)=\sin ((x-a)-(x-b)) \\$

$ \Rightarrow \text { Also } \sin (A-B)=\sin A \cos B-\cos A \sin B \\ $

$\Rightarrow \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)}=\frac{\sin ((x-a)-(x-b))}{\sin (x-a) \sin (x-b)}=\frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)} \\$

$ \Rightarrow \quad \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)}=\frac{\cos (x-b)}{\sin (x-b)}-\frac{\cos (x-a)}{\sin (x-a)}$
$\\ =\frac{\cos (x-b)}{\sin (x-b)}-\frac{\cos (x-a)}{\sin (x-a)}$

$=\cot (x-b)-\cot (x-a) \\ \quad \int \frac{d x}{\sin (x-a) \sin (x-b)}$

$=\frac{1}{\sin (b-a)}\left(\int \cot (x-b) d x-\int \cot (x-a) d x\right) \\$

$ \Rightarrow \operatorname{Now} \int \cot x d x=\ln |\sin x|+c \\$

$ \Rightarrow \frac{1}{\sin (b-a)}\left(\int \cot (x-b) d x-\int \cot (x-a) d x\right)$
$\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}(\ln |\sin (\mathrm{x}-\mathrm{b})|-\ln |\sin (\mathrm{x}-\mathrm{a})|)$ $\\ \Rightarrow \int \frac{\mathrm{dx}}{\sin (\mathrm{x}-\mathrm{a}) \sin (\mathrm{x}-\mathrm{b})}=\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \ln \left(\frac{\sin (\mathrm{x}-\mathrm{b})}{\sin (\mathrm{x}-\mathrm{a})}\right)=\operatorname{cosec}(\mathrm{b}-\mathrm{a}) \ln \left(\frac{\sin (\mathrm{x}-\mathrm{b})}{\sin (\mathrm{x}-\mathrm{a})}\right)$

Question:50

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

Answer:

A)

Given: $\int \tan ^{-1} \sqrt{x} d x$
$\\ \text { Put } x=t^{2} \\ \Rightarrow d x=2 d t \\$

$ \Rightarrow \int \tan ^{-1} \sqrt{x} d x=\int 2 \tan ^{-1} \sqrt{t^{2}} d t \\ $

$\Rightarrow 2 \int \tan ^{-1} \operatorname{tdt} \ldots(1)$
$\\ \Rightarrow \text { Now apply integration by part on } \int t \tan ^{-1} t d t\\$

$ \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\\ $

$\Rightarrow \text { Now } \int \frac{t^{2}}{1+t^{2}} d t=\int \frac{t^{2}+1-1}{1+t^{2}} d t=\int \frac{t^{2}+1}{1+t^{2}} d t-\int \frac{1}{1+t^{2}} d t$
$\\ \Rightarrow \int \frac{t^{2}+1}{1+t^{2}} \mathrm{dt}-\int \frac{1}{1+t^{2}} \mathrm{dt}=\int \mathrm{dt}-\int \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{t}-\tan ^{-1} \mathrm{t} \ldots\\$

$ \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)\\ $

$\Rightarrow 2 \int \mathrm{t} \tan ^{-1} \mathrm{t} \mathrm{dt}=\mathrm{t}^{2} \tan ^{-1} \mathrm{t}-\mathrm{t}+\tan ^{-1} \mathrm{t}$
$\\ \Rightarrow t^{2} \tan ^{-1} t-t+\tan ^{-1} t=\tan ^{-1} t\left(t^{2}+1\right)-t \\$

$ \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: $\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{~d} \mathrm{x}$
$\\ =\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x=\int e^{x}\left(\frac{1+x^{2}-2 x}{\left(1+x^{2}\right)^{2}}\right) \mathrm{d} x \\$

$=\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
$\\ \int\left(\frac{x^{9}}{x^{12}\left(4+\frac{1}{x^{2}}\right)^{6}}\right) d x\\$

$\Rightarrow \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}\\ $

$\Rightarrow \text { Now put } 4+\frac{1}{x^{2}}=t$
$\\ \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} \\ $

$\Rightarrow \frac{x^{2}+1}{x+1}-\frac{1}{x+1}=\frac{(x+1)\left(x^{2}-x+1\right)}{x+1}-\frac{1}{x+1}$
$\\\Rightarrow \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}} \\ $

$\Rightarrow=\frac{1}{2}\left(\int \sqrt{\mathrm{t}} \mathrm{dt}-\int\left(\frac{1}{\sqrt{\mathrm{t}}}\right) \mathrm{dt}\right)$
$\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 $

Question:57

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

Answer:

A)

Given: $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x}$
Using trigonometric identities:
$\\ \cos ^{2} x+\sin ^{2} x=1 \text { and } \cos 2 x=\cos ^{2} x-\sin ^{2} x \\$

$ \frac{1}{1+\cos 2 x}=\frac{1}{\left(\cos ^{2} x+\sin ^{2} x+\cos ^{2} x-\sin ^{2} x\right)} \\ =\frac{1}{2 \cos ^{2} x}$
$\\ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{d x}{1+\cos 2 x}\right)=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{d x}{2 \cos ^{2} x}\right)=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \\$

$=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x=\frac{1}{2}[\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\ =\frac{1+1}{2}=1$

Question:58

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

Answer:

D)
As

$\\ \sin 2 x=2 \sin x \cos x \text { and } \sin ^{2} x+\cos ^{2} x=1 \\ $

$\Rightarrow \int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x} d x \\$

$ \Rightarrow \int_{0}^{\frac{\pi}{2}} \sqrt{(\cos x-\sin x)^{2}} d x=\int_{0}^{\frac{\pi}{2}}|(\cos x-\sin x)| d x$
$\\ \text { From } 0<x<\frac{\pi}{4}, \cos x>\sin x \text { and } \\ \text { from } \frac{\pi}{4}<x<\frac{\pi}{2}, \cos x<\sin x \\ $

$\Rightarrow \int_{0}^{\frac{\pi}{2}}|(\cos x-\sin x)| d x=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x$
$=2[\sin x+\cos x]_{0}^{\pi / 4}$
On solving the Above Integral we get $2 (\sqrt2 -1)$

Question:59

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

Answer:

$e{-1}$
Given: $\int_{0}^{\frac{\pi}{2}} \cos x e^{\sin x} d x$
$\\ \text { Put } \sin x=t \\ \Rightarrow \cos x d x=d t \\$

$ \Rightarrow \operatorname{At} x=0$

$\Rightarrow t=0 \text { and } x=\frac{\pi}{2}$

$\Rightarrow t=1 \\ $

$\Rightarrow \int_{0}^{1} e^{t} d t=\left[e^{t}\right]_{0}^{1}=e-1$

Question:60

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

Answer:

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

$ \Rightarrow f^{\prime}(x)=-\frac{1}{(x+4)^{2}} \\$

$ \Rightarrow \int e^{x}\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^{2}}\right) d x=\frac{e^{x}}{x+4}+c \\ $

$\Rightarrow \int \frac{x+3}{(x+4)^{2}} e^{x} d x=\frac{e^{x}}{x+4}+c$

Question:61

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

Answer:

$\\ \mathrm{a}=\frac{1}{2} \\$

$ \text { Given: } \int_{0}^{\mathrm{a}} \frac{1}{1+4 \mathrm{x}^{2}} \mathrm{dx}=\frac{\pi}{8} \\$

$ \frac{1}{1+4 \mathrm{x}^{2}}=\frac{\frac{1}{4}}{\frac{1}{4}+\mathrm{x}^{2}}=\frac{\frac{1}{4}}{\left(\frac{1}{2}\right)^{2}+\mathrm{x}^{2}} \\ $

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

$ \int_{0}^{a} \frac{\frac{1}{4}}{\left(\frac{1}{2}\right)^{2}+x^{2}} d x=\frac{\frac{1}{4}}{\frac{1}{2}} \tan ^{-1}\left(\frac{x}{\frac{1}{2}}\right) \\$

$ =\left[\frac{1}{2} \tan ^{-1}(2 x)\right]_{0}^{a}\\ =\frac{\pi}{8} \\ $

$\frac{1}{2} \tan ^{-1}(2 a)=\frac{\pi}{8} \\$

$ 2 a=\tan \left(\frac{\pi}{4}\right)=1 \\$

$ a=\frac{1}{2}$

Question:62

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

Answer:

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

$ \Rightarrow \quad \int \frac{\sin \mathrm{x}}{3+4 \cos ^{2} \mathrm{x}} \mathrm{dx}=\int \frac{-\mathrm{dt}}{3+4 \mathrm{t}^{2}}=\frac{1}{4} \int \frac{-\mathrm{dt}}{\frac{3}{4}+\mathrm{t}^{2}} \\ $

$\Rightarrow \quad \frac{1}{4} \int \frac{-\mathrm{dt}}{\left(\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{t}^{2}}$
$\\ \Rightarrow \text { Now } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ \quad=\frac{1}{4} \int \frac{-d t}{\left(\frac{2}{2}\right)^{2}+t^{2}}=\frac{-\frac{1}{4}}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 t}{\sqrt{3}}\right)+c \\$

$ \Rightarrow \int \frac{\sin x}{3+4 \cos ^{2} x} d x=-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$

Question:63

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

Answer:

0

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

$\Rightarrow \int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x=\int_{-\pi}^{\pi} \sin ^{3}(\pi+(-\pi)-x) \cos ^{2}(\pi+(-\pi)-x) d x \\ $

$\Rightarrow \text { As } \sin (-x)=-\sin x \text { and } \cos (-x)=\cos x$
$\\ =\int_{-\pi}^{\pi} \sin ^{3}(\pi+(-\pi)-\mathrm{x}) \cos ^{2}(\pi+(-\pi)-\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{-\pi}^{\pi} \sin ^{3}(-\mathrm{x}) \cos ^{2}(-\mathrm{x}) \mathrm{dx} \\$

$ \Rightarrow \int_{-\pi}^{\pi} \sin ^{3}(-\mathrm{x}) \cos ^{2}(-\mathrm{x}) \mathrm{dx}=-\int_{-\pi}^{\pi} \sin ^{3} \mathrm{x} \cos ^{2} \mathrm{x} \mathrm{dx}=-\mathrm{I} \\ $

$\Rightarrow \mathrm{I}=-\mathrm{I} \\ \Rightarrow 2 \mathrm{I}=0$
$\Rightarrow I=\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x d x=0$

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 7

Class 12 Maths NCERT exemplar solutions Chapter 7 Integrals touches upon an exhaustive explanation of how we do integration by using properties.

  • These solutions provide a wide range of problems that will help students during their exams.
  • These solutions are easy to understand as they are well explained.
  • These also cover some major properties and rules, like King's property and Leibnitz's rule.
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NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:

Frequently Asked Questions (FAQs)

Q: Are these solutions helpful for competitive examinations?
A:

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.

Q: What are the important topics of this chapter?
A:

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.

Q: How many questions are there in this chapter?
A:

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

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

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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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

For the 2025-2026 academic session, the CBSE plans to conduct board exams from 17 February 2026 to 20 May 2026.

You can download it in pdf form from below link

CBSE DATE SHEET 2026

all the best for your exam!!

Hii neeraj!

You can check CBSE class 12th registration number in:

  • Your class 12th board exam admit card. Please do check admit card for registration number, it must be there.
  • You can also check the registration number in your class 12th marksheet in case you have got it.
  • Alternatively you can also visit your school and ask for the same in the administration office they may tell you the registration number.

Hope it helps!