Exercise 7.2 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download

Ex 7.2 Question 1.

$\frac{2 x}{1+x^2}$

Answer.

Putting $1+x^2=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t \\
& \therefore \int \frac{2 x}{1+x^2} d x=\int \frac{d t}{t} \\
& =\int \frac{1}{t} d t=\log |t|+c \\
& =\log \left|1+x^2\right|+c
\end{aligned}
$

Ex 7.2 Question 2.

$\frac{(\log x)^2}{x}$

Answer.

Putting $\log x=t$
$
\Rightarrow \frac{1}{x}=\frac{d t}{d x}
$

$
\begin{aligned}
& \Rightarrow \frac{d x}{x}=d t \\
& \therefore \int \frac{(\log x)^2}{x} d x \\
& =\int t^2 d t \\
& =\frac{t^3}{3}+c \\
& =\frac{1}{3}(\log x)^3+c
\end{aligned}
$

Ex 7.2 Question 3

 $\frac{1}{x+x \log x}$

Answer.

Putting $1+\log x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{x}=\frac{d t}{d x} \\
& \Rightarrow \frac{d x}{x}=d t \\
& \therefore \int \frac{1}{x+x \log x} d x \\
& =\int \frac{1}{1+\log x} \frac{d x}{x} \\
& =\int \frac{1}{t} d t=\log |t|+c \\
& =\log |1+\log x|+c
\end{aligned}
$

Ex 7.2 Question 4.

$\sin x \sin (\cos x)$

Answer.

Putting $\cos x=t$
$
\begin{aligned}
& \Rightarrow-\sin x=\frac{d t}{d x} \\
& \Rightarrow-\sin x d x=d t \\
& \therefore \int \sin x \sin (\cos x) d x \\
& =-\int \sin (\cos x)(-\sin x d x) \\
& =-\int \sin t d t \\
& =-(-\cos t)+c \\
& =\cos t+c=\cos (\cos x)+c
\end{aligned}
$

Ex 7.2 Question 5.

$\sin (c x+b) \cos (c x+b)$
Answer

$\begin{aligned}
& \text {} \int \sin (a x+b) \cos (a x+b) d x \\
& =\frac{1}{2} \int 2 \sin (a x+b) \cos (a x+b) d x \\
& =\frac{1}{2} \int \sin 2(a x+b) d x \\
& =\frac{1}{2} \int \sin (2 a x+2 b) d x
\end{aligned}$

$\begin{array}{ll}
=\frac{1}{2}\left[\frac{-\cos (2 a x+2 b)}{2 a}\right]+c & \text { [ because } \left.\int \operatorname{Sin}(a x+b) d x=-\frac{1}{a} \operatorname{Cos}(a x+b)\right] \\
=\frac{-1}{4 a} \cos 2(a x+b)+c &
\end{array}$

Ex 7.2 Question 6.

$\sqrt{a x+b}$

Answer.

$\int \sqrt{a x+b} d x$
$
=\int(a x+b)^{\frac{1}{2}} d x
$

Using $\int(a x+b)^n d x=\frac{(a x+b)^{n+1}}{a(n+1)}+c$ We have
$
\begin{aligned}
& \int(a x+b)^{\frac{1}{2}} d x=\frac{(a x+b)^{\frac{3}{2}}}{\frac{3}{2} a}+c \\
& =\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+c
\end{aligned}
$

Ex 7.2 Question 7.

$x \sqrt{x+2}$

Answer.

$\int x \sqrt{x+2} d x$
$
\begin{aligned}
& =\int\{(x+2)-2\} \sqrt{x+2} d x \\
& =\int\left\{(x+2)(x+2)^{\frac{1}{2}}-2(x+2)^{\frac{1}{2}}\right\} d x \\
& =\int\left\{(x+2)^{\frac{3}{2}}-2(x+2)^{\frac{1}{2}}\right\} d x \\
& =\int(x+2)^{\frac{3}{2}} d x-2 \int(x+2)^{\frac{1}{2}} d x
\end{aligned}
$

Using $\int(a x+b)^n d x=\frac{(a x+b)^{n+1}}{a(n+1)}+c$

$\begin{aligned}
& =\frac{(x+2)^{\frac{5}{2}}}{\frac{5}{2}}-2 \frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}+c \\
& =\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+c
\end{aligned}$

Ex 7.2 Question 8.

$x \sqrt{1+2 x^2}$

Answer.

Let $\mathrm{I}=\int x \sqrt{1+2 x^2} d x$
$
=\frac{1}{4} \int \sqrt{1+2 x^2}(4 x d x)
$

Putting $1+2 x^2=t$
$
\begin{aligned}
& \Rightarrow 4 x=\frac{d t}{d x} \\
& \Rightarrow 4 x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\frac{1}{4} \int \sqrt{t} d t \\
& =\frac{1}{4} \int t^{\frac{1}{2}} d t \\
& =\frac{1}{4} \frac{t^{3 / 2}}{3 / 2}+c \\
& =\frac{1}{4} \cdot \frac{2}{3} t^{3 / 2}+c
\end{aligned}
$

$=\frac{1}{6}\left(1+2 x^2\right)^{3 / 2}+c$

Ex 7.2 Question 9

$(4 x+2) \sqrt{x^2+x+1}$
$
\begin{aligned}
& \text { Answer. Let } \mathrm{I}=\int(4 x+2) \sqrt{x^2+x+1} d x \\
& =\int 2(2 x+1) \sqrt{x^2+x+1} d x \\
& =\int 2 \sqrt{x^2+x+1}(2 x+1) d x \ldots . . \text { (i) }
\end{aligned}
$

Putting $x^2+x+1=t$
$
\begin{aligned}
& \Rightarrow(2 x+1)=\frac{d t}{d x} \\
& \Rightarrow(2 x+1) d x=d t \\
& \therefore \text { From eq. (i), } \mathrm{I}=\int 2 \sqrt{t} d t \\
& =2 \int t^{\frac{1}{2}} d t \\
& =2 \frac{t^{3 / 2}}{3 / 2}+c \\
& =\frac{4}{3} t^{3 / 2}+c \\
& =\frac{4}{3}\left(x^2+x+1\right)^{3 / 2}+c
\end{aligned}
$

Ex 7.2 Question 10.

$\frac{1}{x-\sqrt{x}}$

Answer.

Let $\mathrm{I}=\int \frac{1}{x-\sqrt{x}} d x$.
Putting $\sqrt{x}=t$
$
\begin{aligned}
& \Rightarrow x=t^2 \\
& \Rightarrow \frac{d x}{d t}=2 t \\
& \Rightarrow d x=2 t d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{t^2-t} 2 t d t \\
& =2 \int \frac{t}{t(t-1)} d t \\
& =2 \int \frac{1}{(t-1)} d t \\
& =2 \log |t-1|+c \\
& =2 \log |\sqrt{x}-1|+c
\end{aligned}
$

Ex 7.2 Question 11.

$\frac{x}{\sqrt{x+4}}, x>0$

Answer.

$\int \frac{x}{\sqrt{x+4}} d x$

$\begin{aligned}
& =\int \frac{x+4-4}{\sqrt{x+4}} d x \\
& =\int \frac{x+4}{\sqrt{x+4}}-\frac{4}{\sqrt{x+4}} d x \\
& =\int \sqrt{x+4} d x-4 \int \frac{1}{\sqrt{x+4}} d x \\
& =\int(x+4)^{\frac{1}{2}} d x-4 \int(x+4)^{\frac{-1}{2}} d x \\
& =\frac{(x+4)^{\frac{3}{2}}}{\frac{3}{2}(1)}-\frac{4(x+4)^{\frac{1}{2}}}{\frac{1}{2}(1)}+c \quad \text { Using } \int(a x+b)^n d x=\frac{(a x+b)^{n+1}}{a(n+1)}+c \\
& =\frac{2}{3}(x+4)^{\frac{3}{2}}-8(x+4)^{\frac{1}{2}}+c \\
& =2 \sqrt{x+4}\left(\frac{x+4}{3}-4\right)+c \\
& =2 \sqrt{x+4}\left(\frac{x+4-12}{3}\right)+c \\
& =\frac{2}{3} \sqrt{x+4}(x-8)+c \text { ans. }
\end{aligned}$

Ex 7.2 Question 12.

$\left(x^3-1\right)^{1 / 3} x^5$

Answer.

Let $\mathrm{I}=\int\left(x^3-1\right)^{\frac{1}{3}} x^5 d x$
$
\begin{aligned}
& =\int\left(x^3-1\right)^{\frac{1}{3}} x^3 x^2 d x \\
& =\frac{1}{3} \int\left(x^3-1\right)^{\frac{1}{3}} x^3\left(3 x^2 d x\right) .
\end{aligned}
$

Putting $x^3-1=t$
$
\begin{aligned}
& \Rightarrow x^3=t+1 \\
& \Rightarrow 3 x^2=\frac{d t}{d x} \\
& \Rightarrow 3 x^2 d x=d t \\
& \therefore \text { From eq. (i), } \mathrm{I}=\frac{1}{3} \int t^{\frac{1}{3}}(t+1) d t \\
& =\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t \\
& =\frac{1}{3}\left(\int t^{\frac{4}{3}} d t+\int t^{\frac{1}{3}} d t\right) \\
& =\frac{1}{3}\left(\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right)+c \\
& =\frac{1}{3}\left(\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}\right)+c \\
& =\frac{1}{7} t^{\frac{7}{3}}+\frac{1}{4} t^{\frac{4}{3}}+c \\
& =\frac{1}{7}\left(x^3-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^3-1\right)^{\frac{4}{3}}+c \\
&
\end{aligned}
$

Ex 7.2 Question 13.

$\text {} \frac{x^2}{\left(2+3 x^3\right)^3}$

Answer.

Let $\mathrm{I}=\int \frac{x^2}{\left(2+3 x^3\right)^3} d x$
$
=\frac{1}{9} \int \frac{9 x^2}{\left(2+3 x^3\right)^3} d x .
$

Putting $2+3 x^3=t$
$
\begin{aligned}
& \Rightarrow 9 x^2=\frac{d t}{d x} \\
& \Rightarrow 9 x^2 d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), I $=\frac{1}{9} \int t^{-3} d t$
$
\begin{aligned}
& =\frac{1}{9} \cdot \frac{t^{-2}}{-2}+c \\
& =\frac{-1}{18 t^2}+c \\
& =\frac{-1}{18\left(2+3 x^3\right)^2}+c
\end{aligned}
$

Ex 7.2 Question 14.

$\frac{1}{x(\log x)^m}, x>0$

Answer.

Let $\mathrm{I}=\int \frac{1}{x(\log x)^m} d x$
$
=\int \frac{\frac{1}{x} d x}{(\log x)^m} \ldots \ldots \ldots .(\text { (i) }
$

Putting $\log x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{x}=\frac{d t}{d x} \\
& \Rightarrow \frac{d x}{x}=d t
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=\int \frac{d t}{t^m}=\int t^{-m} d t$
$
\begin{aligned}
& =\frac{t^{-m+1}}{-m+1}+c \\
& =\frac{(\log x)^{1-m}}{1-m}+c
\end{aligned}
$

Ex 7.2 Question 15.

$\frac{x}{9-4 x^2}$

Answer.

Let $\mathrm{I}=\int \frac{x}{9-4 x^2} d x$
$
=\frac{-1}{8} \int \frac{-8 x}{9-4 x^2} d x \text {. }
$

Putting $9-4 x^2=t$
$
\begin{aligned}
& \Rightarrow-8 x=\frac{d t}{d x} \\
& \Rightarrow-8 x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=\frac{-1}{8} \int \frac{d t}{t}=\frac{-1}{8} \int \frac{1}{t} d t$
$
=\frac{-1}{8} \log |t|+c
$

$=\frac{-1}{8} \log \left|9-4 x^2\right|+c$

Ex 7.2 Question 16.

$e^{2 x+3}$

Answer.

$\int e^{2 x+3} d x$
$
=\frac{1}{2} e^{2 x+3}+c \quad \text { Using } \int e^{a x+b} d x=\frac{e^{a x+b}}{a}+c
$

Ex 7.2 Question 17.

$\frac{x}{e^{x^2}}$

Answer.

Let $\mathrm{I}=\int \frac{x}{e^{x^2}} d x$
$
=\frac{1}{2} \int \frac{2 x}{e^{x^2}} d x \ldots \ldots \ldots \text { (i }
$

Putting $x^2=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), I $=\frac{1}{2} \int \frac{d t}{e^t}=\frac{1}{2} \int e^{-t} d t$
using $\int e^{a x+b} d x=\frac{e^{a x+b}}{a}+c$
We have $\frac{1}{2} \int e^{-t} d t=\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+c$
$
=\frac{-1}{2\left(e^t\right)}+c
$

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

Ex 7.2 Question 18.

$\frac{e^{\tan ^{-2} x}}{1+x^2}$

Answer

Let $\mathrm{I}=\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x$
Putting $\tan ^{-1} x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{1+x^2}=\frac{d t}{d x} \\
& \Rightarrow \frac{d x}{1+x^2}=d t \\
& \therefore \text { From eq. (i), I }=\int e^t d t \\
& =e^t+c \\
& =e^{\operatorname{ta}^{-t} x}+c
\end{aligned}
$

Ex 7.2 Question 19.

$\frac{e^{2 x}-1}{e^{2 x}+1}$

Answer.

Let $\mathrm{I}=\int \frac{e^{2 x}-1}{e^{2 x}+1} d x$
$=\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x$ [Multiplying each term by $\left.e^{-x}\right]$
Putting $e^x+e^{-x}=t$

$\begin{aligned}
& \Rightarrow e^x+e^{-x} \frac{d}{d x}(-x)=\frac{d t}{d x} \\
& \Rightarrow\left(e^x-e^{-x}\right) d x=d t \\
& \therefore \text { From eq. (i), I }=\int \frac{d t}{t}=\int \frac{1}{t} d t \\
& =\log |t|+c \\
& =\log \left|e^x+e^{-x}\right|+c \\
& =\log \left(e^x+e^{-x}\right)+c
\end{aligned}$

Ex 7.2 Question 20.

$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$

Answer.

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

Putting $e^{2 x}+e^{-2 x}=t$
$\Rightarrow e^{2 x} \frac{d}{d x} 2 x+e^{-2 x} \frac{d}{d x}(-2 x)=\frac{d t}{d x}$
$\Rightarrow e^{2 x} \cdot 2-2 e^{-2 x}=\frac{d t}{d x}$
$\Rightarrow 2\left(e^{2 x}-e^{-2 x}\right) d x=d t$
$\therefore$ From eq. (i), $\mathrm{I}=\frac{1}{2} \int \frac{d t}{t}$

$\begin{aligned}
& =\frac{1}{2} \log |t|+c \\
& =\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+c \\
& =\frac{1}{2} \log \left(e^{2 x}+e^{-2 x}\right)+c
\end{aligned}$

Ex 7.2 Question 21

$\tan ^2(2 x-3)$

Answer.

$\int \tan ^2(2 x-3) d x$
$
\begin{aligned}
& =\int\left\{\sec ^2(2 x-3)-1\right\} d x \\
& =\int \sec ^2(2 x-3) d x-\int 1 d x
\end{aligned}
$

Using $\int \sec ^2(a x+b) d x=\frac{\tan (a x+b)}{a}+c$
$
=\frac{\tan (2 x-3)}{2}-x+c
$
Ex 7.2 Question 22.

$\sec ^2(7-4 x)$

Answer.

$\int \sec ^2(7-4 x) d x$
Using $\int \sec ^2(a x+b) d x=\frac{\tan (a x+b)}{a}+c$
$
=\frac{-1}{4} \tan (7-4 x)+c
$

Ex 7.2 Question 23. 

$\text {} \frac{\sin ^{-1} x}{\sqrt{1-x^2}}$

Answer.

Let $\mathrm{I}=\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$
Putting $\sin ^{-1} x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{d t}{d x} \\
& \Rightarrow \frac{d x}{\sqrt{1-x^2}}=d t \\
& \therefore \text { From eq. (i), I }=\int t d t \\
& =\frac{t^2}{2}+c \\
& =\frac{1}{2}\left(\sin ^{-1} x\right)^2+c
\end{aligned}
$

Ex 7.2 Question 24.

$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$

Answer.

Let $\mathrm{I}=\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x$
$=\int \frac{2 \cos x-3 \sin x}{2(2 \sin x+3 \cos x)} d x$
$=\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{2 \sin x+3 \cos x} d x$.
Putting $2 \sin x+3 \cos x=t$
$\Rightarrow 2 \cos x-3 \sin x=\frac{d t}{d x}$
$\Rightarrow(2 \cos x-3 \sin x) d x=d t$

$\therefore$ From eq. (i), I $=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t|+c$
$
=\frac{1}{2} \log |2 \sin x+3 \cos x|+c
$

Ex 7.2 Question 25.

$\frac{1}{\cos ^2 x(1-\tan x)^2}$

Answer.

Let $\mathrm{I}=\int \frac{1}{\cos ^2 x(1-\tan x)^2} d x$
$=\int \frac{\sec ^2 x}{(1-\tan x)^2} d x$
$=-\int \frac{-\sec ^2 x}{(1-\tan x)^2} d x$.
Putting $1-\tan x=t$
$
\begin{aligned}
& \Rightarrow-\sec ^2 x=\frac{d t}{d x} \\
& \Rightarrow-\sec ^2 x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), I $=-\int \frac{d t}{t^2}=-\int t^{-2} d t$
$
\begin{aligned}
& =\frac{-t^{-1}}{-1}+c \\
& =\frac{1}{t}+c \\
& =\frac{1}{1-\tan x}+c
\end{aligned}
$

Ex 7.2 Question 26.

$\frac{\cos \sqrt{x}}{\sqrt{x}}$

Answer.

Let $I=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$
Putting $\sqrt{x}=t$
$
\begin{aligned}
& \Rightarrow x=t^2 \\
& \Rightarrow \frac{d x}{d t}=2 t \\
& \Rightarrow d x=2 t d t
\end{aligned}
$
$\therefore$ From eq. (i), I $=\int \frac{\cos t}{t} 2 t d t$
$
\begin{aligned}
& =2 \int \cos t d t \\
& =2 \sin t+c \\
& =2 \sin \sqrt{x}+c
\end{aligned}
$

Ex 7.2 Question 27.

$\sqrt{\sin 2 x} \cos 2 x$

Answer.

Let $\mathrm{I}=\int \sqrt{\sin 2 x} \cos 2 x d x$
$
=\frac{1}{2} \int \sqrt{\sin 2 x}(2 \cos 2 x d x)
$

Putting $\sin 2 x=t$
$
\Rightarrow \cos 2 x \frac{d}{d x}(2 x)=\frac{d t}{d x}
$

$
\Rightarrow 2 \cos 2 x d x=d t
$
$\therefore$ From eq. (i), $\mathrm{I}=\frac{1}{2} \int \sqrt{t} d t$
$
\begin{aligned}
& =\frac{1}{2} \int t^{\frac{1}{2}} d t \\
& =\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \\
& =\frac{1}{2} \cdot \frac{t^{3 / 2}}{3 / 2}+c \\
& =\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+c
\end{aligned}
$

Ex 7.2 Question 28.

$\frac{\cos x}{\sqrt{1+\sin x}}$

Answer.

Let $\mathrm{I}=\int \frac{\cos x}{\sqrt{1+\sin x}} d x$.
Putting $1+\sin x=t$
$
\begin{aligned}
& \Rightarrow \cos x=\frac{d t}{d x} \\
& \Rightarrow \cos x d x=d t \\
& \therefore \text { From eq. (i), I }=\int \frac{d t}{\sqrt{t}} \\
& =\int t^{\frac{-1}{2}} d t
\end{aligned}
$

$\begin{aligned}
& =\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c \\
& =\frac{t^{1 / 2}}{1 / 2}+c \\
& =2 \sqrt{t+c} \\
& =2 \sqrt{1+\sin x}+c
\end{aligned}$

Ex 7.2 Question 29.

$\cot x \log \sin x$

Answer.

Let $\mathrm{I}=\int \cot x \log \sin x d x$ $\qquad$
Putting $\log \sin x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{\sin x} \frac{d}{d x}(\sin x)=\frac{d t}{d x} \\
& \Rightarrow \frac{1}{\sin x} \cos x=\frac{d t}{d x} \\
& \Rightarrow \cot x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=\int t d t$
$
\begin{aligned}
& =\frac{t^2}{2}+c \\
& =\frac{1}{2}(\log \sin x)^2+c
\end{aligned}
$

Ex 7.2 Question 30.

$\text {} \frac{\sin x}{1+\cos x}$

$
\begin{aligned}
& \text { Answer. Let } \mathrm{I}=\int \frac{\sin x}{1+\cos x} d x \\
& =-\int \frac{-\sin x}{1+\cos x} d x \ldots \ldots . . \text { (i) }
\end{aligned}
$

Putting $1+\cos x=t$
$
\begin{aligned}
& \Rightarrow-\sin x=\frac{d t}{d x} \\
& \Rightarrow-\sin x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=-\int \frac{d t}{t}$
$
\begin{aligned}
& =-\log |t|+c \\
& =-\log |1+\cos x|+c
\end{aligned}
$

Ex 7.2 Question 31.

$\frac{\sin x}{(1+\cos x)^2}$

Answer.

Let $\mathrm{I}=\int \frac{\sin x}{(1+\cos x)^2} d x$
$=-\int \frac{-\sin x}{(1+\cos x)^2} d x$
Putting $1+\cos x=t$
$
\begin{aligned}
& \Rightarrow-\sin x=\frac{d t}{d x} \\
& \Rightarrow-\sin x d x=d t
\end{aligned}
$

$\begin{aligned}
& \therefore \text { From eq. (i), } \mathrm{I}=-\int \frac{d t}{t^2} \\
& =-\int t^{-2} d t \\
& =\frac{-t^{-1}}{-1}+c \\
& =\frac{1}{t}+c \\
& =\frac{1}{1+\cos x}+c
\end{aligned}$

Ex 7.2 Question 32.

$\frac{1}{1+\cot x}$

Answer.

Let $I=\int \frac{1}{1+\cot x} d x$
$
\begin{aligned}
& =\int \frac{1}{1+\frac{\cos x}{\sin x}} d x \\
& =\int \frac{1}{\left(\frac{\sin x+\cos x}{\sin x}\right)} d x \\
& =\int \frac{\sin x}{\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{\sin x+\sin x}{\sin x+\cos x} d x
\end{aligned}
$

Adding and subtracting $\cos x$ in the numerator,

$
\begin{aligned}
& =\frac{1}{2} \int \frac{\sin x+\cos x-\cos x+\sin x}{\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{(\sin x+\cos x)-(\cos x-\sin x)}{\sin x+\cos x} d x \\
& =\frac{1}{2} \int\left(\frac{\sin x+\cos x}{\sin x+\cos x}-\frac{\cos x-\sin x}{\sin x+\cos x}\right) d x \\
& =\frac{1}{2} \int\left(1-\frac{\cos x-\sin x}{\sin x+\cos x}\right) d x \\
& =\frac{1}{2}\left[\int 1 d x-\int \frac{\cos x-\sin x}{\sin x+\cos x} d x\right]=\frac{1}{2}\left[x-I_1\right]
\end{aligned}
$
where $I_1=\int \frac{\cos x-\sin x}{\sin x+\cos x} d x$
Putting $\sin x+\cos x=t$
$
\begin{aligned}
& \Rightarrow \cos x-\sin x=\frac{d t}{d x} \\
& \Rightarrow(\cos x-\sin x) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{t}=\log |t| \\
& =\log |\sin x+\cos x|
\end{aligned}
$

Putting this value in eq. (i), we get required integral,
$
=\frac{1}{2}[x-\log |\sin x+\cos x|]+c
$

Ex 7.2 Question 33.

$\text {} \frac{1}{1-\tan x}$

Answer.

Let I $=\int \frac{1}{1-\tan x} d x$
$
\begin{aligned}
& =\int \frac{1}{1-\frac{\sin x}{\cos x}} d x \\
& =\int \frac{1}{\left(\frac{\cos x-\sin x}{\cos x}\right)} d x \\
& =\int \frac{\cos x}{\cos x-\sin x} d x \\
& =\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} d x \\
& =\frac{1}{2} \int \frac{\cos x+\cos x}{\cos x-\sin x} d x
\end{aligned}
$

Adding and subtracting $\sin x$ in the numerator,

$\begin{aligned}
& =\frac{1}{2} \int \frac{\cos x-\sin x+\sin x+\cos x}{\cos x-\sin x} d x \\
& =\frac{1}{2} \int \frac{(\cos x-\sin x)+(\sin x+\cos x)}{\cos x-\sin x} d x \\
& =\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x-\sin x}+\frac{\sin x+\cos x}{\cos x-\sin x} d x \\
& =\frac{1}{2} \int\left(1+\frac{\sin x+\cos x}{\cos x-\sin x}\right) d x \\
& =\frac{1}{2}\left[\int 1 d x-\int \frac{-\sin x-\cos x}{\cos x-\sin x} d x\right] \\
& =\frac{1}{2}[x-\log |\cos x-\sin x|]+c
\end{aligned}$

Ex 7.2 Question 34.

$\frac{\sqrt{\tan x}}{\sin x \cos x}$

Answer.

Let $\mathrm{I}=\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x$
$
\begin{aligned}
& =\int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cos x \cdot \cos x} d x \\
& =\int \frac{\sqrt{\tan x}}{\tan x \cos ^2 x} d x \\
& =\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x \ldots . \text { (i) }
\end{aligned}
$

Putting $\tan x=t$
$
\begin{aligned}
& \Rightarrow \sec ^2 x=\frac{d t}{d x} \\
& \Rightarrow \sec ^2 x d x=d t \\
& \therefore \text { From eq. (i), I }=\int \frac{d t}{\sqrt{t}} \\
& =\int t^{\frac{-1}{2}} d t \\
& =\frac{t^{1 / 2}}{1 / 2}+c \\
& =2 \sqrt{t}+c \\
& =2 \sqrt{\tan x}+c
\end{aligned}
$

Ex 7.2 Question 35.

$\text {} \frac{(1+\log x)^2}{x}$

Answer.

Let I $=\int \frac{(1+\log x)^2}{x} d x \ldots \ldots \ldots .$. (i)
Putting $1+\log x=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{x}=\frac{d t}{d x} \\
& \Rightarrow \frac{d x}{x}=d t \\
& \therefore \text { From eq. (i), I }=\int t^2 d t \\
& =\frac{t^3}{3}+c \\
& =\frac{1}{3}(1+\log x)^3+c
\end{aligned}
$

Ex 7.2 Question 36.

$\frac{(x+1)(x+\log x)^2}{x}$

Answer.

Let $\mathrm{I}=\int \frac{(x+1)(x+\log x)^2}{x} d x$
Putting $x+\log x=t$
$
\begin{aligned}
& \Rightarrow 1+\frac{1}{x}=\frac{d t}{d x} \\
& \Rightarrow \frac{x+1}{x}=\frac{d t}{d x} \\
& \Rightarrow\left(\frac{x+1}{x}\right) d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=\int t^2 d t$

$\begin{aligned}
& =\frac{t^3}{3}+c \\
& =\frac{1}{3}(x+\log x)^3+c
\end{aligned}$

Ex 7.2 Question 37.

$\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}$

Answer.

Let $\mathrm{I}=\int \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8} d x$
$
=\frac{1}{4} \int \sin \left(\tan ^{-1} x^4\right) \cdot \frac{4 x^3}{1+x^3} d x
$

Putting $\tan ^{-1} x^4=t$
$
\begin{aligned}
& \Rightarrow \frac{1}{1+\left(x^4\right)^2} \frac{d}{d x} x^4=\frac{d t}{d x} \\
& \Rightarrow \frac{4 x^3}{1+x^8} d x=d t
\end{aligned}
$
$\therefore$ From eq. (i), I $=\frac{1}{4} \int \sin t d t$
$
\begin{aligned}
& =\frac{-1}{4} \cos t+c \\
& =\frac{-1}{4} \cos \left(\tan ^{-1} x^4\right)+c
\end{aligned}
$

Ex 7.2 Question 38.

$\text {} \int \frac{10 x^9+10^x \log _{\ell} 10 d x}{x^{10}+10^x} \text { equals }$

(A) $10^x-x^{10}+\mathrm{C}$
(B) $10^x+x^{10}+\mathrm{C}$
(C) $\left(10^x-x^{10}\right)^{-1}+\mathrm{C}$
(D) $\log \left(10^x+x^{10}\right)+c$

Answer.

Let $\mathrm{I}=\int \frac{10 x^9+10^* \log _\epsilon 10}{x^{10}+10^x} d x$
Putting $x^{10}+10^x=t$
$\Rightarrow\left(10 x^9+10^x \log _{\varepsilon} 10\right) d x=d t$
$\therefore$ From eq. (i), I $=\int \frac{d t}{t}$
$=\log |t|+c$
$=\log \left|x^{10}+10^x\right|+c$
Therefore, option (D) is correct.

Ex 7.2 Question 39.

$\int \frac{d x}{\sin ^2 x \cos ^2 x}$ equals
(A) $\tan x+\cot x+C$
(B) $\tan x-\cot x+C$
(C) $\tan x \cot x+C$
(D) $\tan x-\cot 2 x+\mathrm{C}$

Answer.

$\int \frac{d x}{\sin ^2 x \cos ^2 x}$

$
\begin{aligned}
& =\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\
& =\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\
& =\int \frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x} d x \\
& =\int\left(\sec ^2 x+\operatorname{cosec} e^2 x\right) d x \\
& =\int \sec ^2 x d x+\int \operatorname{cosec} x d x \\
& =\tan x-\cot x+c
\end{aligned}
$

Therefore, option (B) is correct.