Exercise 7.3 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download
Ex 7.3 Question 1.
$\sin ^2(2 x+5)$
Answer.
$\int \sin ^2(2 x+5) d x$
$
=\int \frac{1}{2}\{1-\cos 2(2 x+5)\} d x
$
Using $\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}$
$
\begin{aligned}
& =\frac{1}{2} \int\{1-\cos (4 x+10)\} d x \\
& =\frac{1}{2}\left[\int 1 d x-\int \cos (4 x+10) d x\right]
\end{aligned}
$
Using $\int \cos (a x+b) d x=\frac{\sin (a x+b)}{a}+c$
$
\begin{aligned}
& =\frac{1}{2}\left[x-\frac{\sin (4 x+10)}{4}\right]+c \\
& =\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+c \text { Ans. }
\end{aligned}
$
Ex 7.3 Question 2.
$\sin 3 x \cos 4 x$
Answer.
$\int \sin 3 x \cos 4 x d x=\frac{1}{2} \int 2 \sin 3 x \cos 4 x d x$ $==\frac{1}{2} \int\{\sin (4 x+3 x)-\sin (4 x-3 x)\} d x \quad$ Using $2 \sin \mathrm{B} \cos \mathrm{A}=\sin (\mathrm{A}+\mathrm{B})-\sin (\mathrm{A}-\mathrm{B})$
$\begin{aligned}
& =\frac{1}{2} \int(\sin 7 x-\sin x) d x \\
& =\frac{1}{2}\left[\int \sin 7 x d x-\int \sin x d x\right] \\
& =\frac{1}{2}\left[\frac{-\cos 7 x}{7}-(-\cos x)\right]+c \\
& =\frac{-1}{14} \cos 7 x+\frac{1}{2} \cos x+c \text { Ans }
\end{aligned}$
Ex 7.3 Question 3.
$\cos 2 x \cos 4 x \cos 6 x$
$
\begin{aligned}
& \text { Answer. } \int \cos 2 x \cos 4 x \cos 6 x d x \\
& =\frac{1}{2} \int 2(\cos 6 x \cos 4 x) \cos 2 x d x \\
& =\frac{1}{2}\left[\int\{\cos 10 x+\cos 2 x\} \cos 2 x d x\right]
\end{aligned}
$
Using $2 \cos A \cos B=\cos (A+B)+\cos (A-B)$
$
\begin{aligned}
& =\frac{1}{2}\left[\int\left(\cos 10 x \cos 2 x+\cos ^2 2 x\right) d x\right] \\
& =\frac{1}{4}\left[\int\left\{(2 \cos 10 x \cos 2 x)+2 \cos ^2 2 x\right\} d x\right] \\
& =\frac{1}{4}\left[\int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x\right]
\end{aligned}
$
Using $2 \cos A \cos B=\cos (A+B)+\cos (A-B)$
and $2 \cos ^2 \theta=1+\cos 2 \theta$
$
=\frac{1}{4}\left[\int \cos 12 x d x+\int \cos 8 x d x+\int \cos 4 x d x+\int 1 d x\right]
$
Ex 7.3 Question 4.
$\sin ^3(2 x+1)$
Answer.
$\int \sin ^3(2 x+1) d x$
$
\begin{aligned}
& =\int\left(\frac{3}{4} \sin (2 x+1)-\frac{1}{4} \sin 3(2 x+1)\right) d x \\
& {\left[\because \sin ^3 \theta=\frac{3}{4} \sin \theta-\frac{1}{4} \sin 3 \theta\right]} \\
& =\int\left(\frac{3}{4} \sin (2 x+1)-\frac{1}{4} \sin (6 x+3)\right) d x \\
& =\frac{3}{4} \int \sin (2 x+1) d x-\frac{1}{4} \int \sin (6 x+3) d x \\
& =\frac{3}{4}\left(\frac{-\cos (2 x+1)}{2}\right)-\frac{1}{4}\left(\frac{-\cos (6 x+3)}{6 \rightarrow \operatorname{coeff} . \text { of } x}\right)+c \\
& =\frac{-3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+c
\end{aligned}
$
Another Method
$\int \sin ^3(2 x+1) d x$
$
\begin{aligned}
& =\int \operatorname{Sin}^2(2 x+1) \operatorname{Sin}(2 x+1) d x \\
& =\int\left(1-\cos ^2(2 x+1)\right) \operatorname{Sin}(2 x+1) d x
\end{aligned}
$
Using $\sin ^2 \theta=1-\cos ^2 \theta$
let $\cos (2 x+1)=t$
therefore $-2 \sin (2 x+1) d x=d t$
$\begin{aligned}
& \text { thus } \sin (2 x+1) d x=-\frac{1}{2} d t \\
= & -\frac{1}{2} \int\left(1-t^2\right) d t \\
= & -\frac{1}{2}\left(\int 1 d t-\int t^2 d t\right) \\
= & -\frac{1}{2}\left(t-\frac{t^3}{3}\right)+c \\
= & -\frac{1}{2} \operatorname{Cos}(2 x+1)+\frac{1}{6} \operatorname{Cos}^3(2 x+1)+c \text { Ans }
\end{aligned}$
Ex 7.3 Question 5.
$\text {} \sin ^3 x \cos ^3 x$
Answer.
$\begin{aligned}
& \text {} \int \sin ^3 x \cos ^3 x d x \\
& =\int(\sin x \cos x)^3 d x \\
& =\frac{1}{8} \int(2 \sin x \cos x)^3 d x \\
& =\frac{1}{8} \int(\sin 2 x)^3 d x \\
& =\frac{1}{8} \int \sin ^3 2 x d x \\
& =\frac{1}{8} \int\left(\frac{3}{4} \sin 2 x-\frac{1}{4} \sin 6 x\right) d x \\
& =\frac{3}{32} \int \sin 2 x d x-\frac{1}{32} \int \sin 6 x d x \\
& =\frac{-3}{32} \frac{\cos 2 x}{2}-\frac{1}{32}\left(\frac{-\cos 6 x}{6}\right)+c \\
& =\frac{-3}{64} \cos 2 x+\frac{1}{192} \cos 6 x+c
\end{aligned}$
Another method
let $\mathrm{I}=\int \sin ^3 x \cos ^3 x d x$
$
\begin{aligned}
& I=\int\left(1-\cos ^2 x\right) \cos ^3 x \sin x d x \\
& =\int\left(\cos ^3 x-\cos ^5 x\right) \sin x d x
\end{aligned}
$
let $\cos x=t$
$
\begin{aligned}
& -\sin \mathrm{dx}=\mathrm{dt} \\
& \sin \mathrm{x} \mathrm{dx}=-\mathrm{dt} \\
& I=-\int\left(t^3-t^5\right) d t \\
& I=\int\left(t^5-t^3\right) d t \\
& =\frac{t^6}{6}-\frac{t^4}{4}+c \\
& =\frac{\cos ^6}{6}-\frac{\cos ^4}{4}+c \text { ans. }
\end{aligned}
$
Ex 7.3 Question 6.
$\sin x \sin 2 x \sin 3 x$
Answer.
$\int \sin x \sin 2 x \sin 3 x d x$
$
\begin{aligned}
& =\frac{1}{2} \int(2 \sin 3 x \sin 2 x) \sin x d x \\
& =\frac{1}{2} \int(\cos x-\cos 5 x) \sin x d x \quad \text { Using } 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \left.=\frac{1}{2} \int(\cos x \sin x-\cos 5 x \sin x) d x\right) \\
& \left.=\frac{1}{4} \int(2 \cos x \sin x-2 \cos 5 x \sin x) d x\right) \\
& ==\frac{1}{4} \int(2 \sin x \cos x-\{\sin (5 x+x)-\sin (5 x-x)\}) d x
\end{aligned}
$
using $2 \cos A \sin B=\sin (A+B)-\sin (A-B)$
$\begin{aligned}
&= \frac{1}{4} \int(\sin 2 x-\sin 6 x+\sin 4 x) d x \\
&= \frac{1}{4}\left[\int \sin 2 x d x+\int \sin 4 x d x-\int \sin 6 x d x\right] \\
& \quad \text { as } \int \sin (a x+b) d x=\frac{-\cos (a x+b)}{a}+c \\
&= \frac{1}{4}\left(\frac{-\cos 2 x}{2}-\frac{\cos 4 x}{4}+\frac{\cos 6 x}{6}\right)+c \text { Ans }
\end{aligned}$
Ex 7.3 Question 7.
$\sin 4 x \sin 8 x$
Answer.
$\int \sin 4 x \sin 8 x d x$
$
\begin{aligned}
& =\frac{1}{2} \int 2 \sin 4 x \sin 8 x d x \quad \text { Using } 2 \sin \mathrm{A} \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B}) \\
& =\frac{1}{2} \int(\cos (4 x-8 x)-\cos (4 x+8 x)) d x \\
& =\frac{1}{2} \int(\cos (-4 x)-\cos 12 x) d x \\
& =\frac{1}{2} \int(\cos 4 x-\cos 12 x) d x
\end{aligned}
$
As $\cos (-x)=\cos x$
$
=\frac{1}{2}\left[\int \cos 4 x d x-\int \cos 12 x d x\right]
$
$=\frac{1}{2}\left(\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right)+c$ Ans.
As $\int \cos (a x+b) d x=\frac{\sin (a x+b)}{a}+c$
Ex 7.3 Question 8.
$\frac{1-\cos x}{1+\cos x}$
Answer.
$\int \frac{1-\cos x}{1+\cos x} d x$
$
\begin{aligned}
& =\int \frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x \\
& 2 \sin ^2 \frac{\theta}{2}=1-\cos \theta
\end{aligned}
$
As
$
\begin{aligned}
& 2 \cos ^2 \frac{\theta}{2}=1+\cos \theta \\
= & \int \tan ^2 \frac{x}{2} d x \\
= & \int\left(\sec ^2 \frac{x}{2}-1\right) d x \\
= & \int \sec ^2 \frac{x}{2} d x-\int 1 d x
\end{aligned}
$
As $\int \sec ^2(a x+b) d x=\frac{\tan (a x+b)}{a}+c$
$
=2 \tan \frac{x}{2}-x+c \text { Ans. }
$
Ex 7.3 Question 9.
$\frac{\cos x}{1+\cos x}$
Answer.
$\int \frac{\cos x}{1+\cos x} d x$
$=\int \frac{1+\cos x-1}{1+\cos x} d x$
$
\begin{aligned}
& =\int \frac{1+\cos x}{1+\cos x}-\frac{1}{1+\cos x} d x \\
& =\int\left(1-\frac{1}{2 \cos ^2 \frac{x}{2}}\right) d x
\end{aligned}
$
As $2 \cos ^2 \frac{\theta}{2}=1+\cos \theta$
$
=\int 1 d x-\frac{1}{2} \int \sec ^2 \frac{x}{2} d x
$
as $\int \sec ^2(a x+b) d x=\frac{\tan (a x+b)}{a}+c$
$
=x-\frac{1}{2} \frac{\tan \frac{x}{2}}{\frac{1}{2}}+c
$
$=x-\tan \frac{x}{2}+c$ Ans.
Ex 7.3 Question 10.
$\sin ^4 x$
Answer.
$\int \sin ^4 x d x$
$
\begin{aligned}
& =\int\left(\sin ^2 x\right)^2 d x \\
& =\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x
\end{aligned}
$
as $2 \sin ^2 \theta=1-\cos 2 \theta$
$\begin{aligned}
& =\int \frac{(1-\cos 2 x)^2}{4} d x \\
& =\frac{1}{4} \int\left(1+\cos ^2 2 x-2 \cos 2 x\right) d x \\
& =\frac{1}{4} \int\left\{1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right\} d x \\
& \text { as } 2 \cos ^2 2 \theta=1+\cos 4 \theta \\
& =\frac{1}{4} \int\left(\frac{2+1+\cos 4 x-4 \cos 2 x}{2}\right) d x \\
& =\frac{1}{8} \int(3+\cos 4 x-4 \cos 2 x) d x \\
& =\frac{1}{8}\left[3 \int 1 d x+\int \cos 4 x d x-4 \int \cos 2 x d x\right] \\
& =\frac{1}{8}\left[3 x+\frac{\sin 4 x}{4}-\frac{4 \sin 2 x}{2}\right]+c \\
& \text { As } \int \cos (a x+b) d x=\frac{\sin (a x+b)}{a}+c \\
& =\frac{3}{8} x+\frac{1}{32} \sin 4 x-\frac{1}{4} \sin 2 x+c \text { Ans }
\end{aligned}$
Ex 7.3 Question 11.
$\cos ^4 2 x$
$
\begin{aligned}
& \text { Answer. } \int \cos ^4 2 x d x \\
& =\int\left(\cos ^2 2 x\right)^2 d x \\
& =\int\left(\frac{1+\cos 4 x}{2}\right)^2 d x
\end{aligned}
$
As $2 \cos ^2 2 \theta=1+\cos 4 \theta$
$
\begin{aligned}
& =\int \frac{(1+\cos 4 x)^2}{4} d x \\
& =\frac{1}{4} \int\left(1+\cos ^2 4 x+2 \cos 4 x\right) d x \\
& =\frac{1}{4} \int\left\{1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right\} d x
\end{aligned}
$
as $2 \cos ^2 4 \theta=1+\cos 8 \theta$
$
\begin{aligned}
& =\frac{1}{4} \int\left(\frac{2+1+\cos 8 x+4 \cos 4 x}{2}\right) d x \\
& =\frac{1}{8} \int(3+\cos 8 x+4 \cos 4 x) d x \\
& =\frac{1}{8}\left[3 \int 1 d x+\int \cos 8 x d x+4 \int \cos 4 x d x\right] \\
& =\frac{1}{8}\left[3 x+\frac{\sin 8 x}{8}+\frac{4 \sin 4 x}{4}\right]+c
\end{aligned}
$
using $\int \cos (a x+b) d x=\frac{\sin (a x+b)}{a}+c$
$
=\frac{3}{8} x+\frac{1}{64} \sin 8 x+\frac{1}{8} \sin 4 x+c \quad \text { Ans. }
$
Ex 7.3 Question 12.
$\frac{\sin ^2 x}{1+\cos x}$
Answer.
$\int \frac{\sin ^2 x}{1+\cos x} d x$
$\begin{aligned}
& =\int \frac{1-\cos ^2 x}{1+\cos x} d x \\
& =\int \frac{(1-\cos x)(1+\cos x)}{1+\cos x} d x \\
& =\int(1-\cos x) d x \\
& =\int 1 d x-\int \cos x d x \\
& =x-\sin x+c \text { Ans. }
\end{aligned}$
Ex 7.3 Question 13.
$\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$
Answer.
$\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x$
$
=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x
$
$
\begin{aligned}
& \text { using } \cos 2 \theta=2 \cos ^2 \theta-1 \\
& =\int \frac{2 \cos ^2 x-1-2 \cos ^2 \alpha+1}{\cos x-\cos \alpha} d x \\
& =\int \frac{2 \cos ^2 x-2 \cos ^2 \alpha}{\cos x-\cos \alpha} d x \\
& =2 \int \frac{\cos ^2 x-\cos ^2 \alpha}{\cos x-\cos \alpha} d x \\
& =2 \int \frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{(\cos x-\cos \alpha)} d x \\
& =2 \int(\cos x+\cos \alpha) d x
\end{aligned}
$
$\begin{aligned}
& =2\left[\int \cos x d x+\int \cos \alpha d x\right] \\
& =2\left[\sin x+\cos \alpha \int 1\right]+c \\
& =2[\sin x+\cos \alpha(x)]+c \\
& =2 \sin x+2 x \cos \alpha+c
\end{aligned}$
Ex 7.3 Question 14.
$\frac{\cos x-\sin x}{1+\sin 2 x}$
Answer.
Let $\mathrm{I}=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x$
$
=\int \frac{\cos x-\sin x}{\cos ^2 x+\sin ^2 x+2 \sin x \cos x} d x
$
using identity $1=\sin ^2 \theta+\cos ^2 \theta$
$
=\int \frac{\cos -\sin x}{(\cos x+\sin x)^2} d x
$
Putting $\cos x+\sin x=t$
$
\begin{aligned}
& \Rightarrow-\sin x+\cos x=\frac{d t}{d x} \\
& \Rightarrow(\cos x-\sin x) d x=d t \\
& \therefore \text { From eq. (i), I }=\int \frac{d t}{t^2} \\
& =\int t^{-2} d t \\
& =\frac{t^{-1}}{-1}+c
\end{aligned}
$
$\therefore$ From eq. (i), $\mathrm{I}=\int \frac{d t}{t^2}$
$\begin{aligned}
& =\frac{-1}{t}+c \\
& =\frac{-1}{\cos x+\sin x}+c
\end{aligned}$
Ex 7.3 Question 15
$
\begin{aligned}
& \text {} \tan ^3 2 x \sec 2 x \\
& \text { Ans. Let } \mathrm{I}=\int \tan ^3 2 x \sec 2 x d x \\
& =\int \tan ^2 2 x \tan 2 x \sec 2 x d x \\
& =\int\left(\sec ^2 2 x-1\right) \tan 2 x \sec 2 x d x \\
& =\frac{1}{2} \int\left(\sec ^2 2 x-1\right)(2 \sec 2 x \tan 2 x) d x
\end{aligned}
$
Putting $\sec 2 x=t$
$
\begin{aligned}
& \Rightarrow \sec 2 x \tan 2 x \frac{d}{d x}(2 x)=\frac{d t}{d x} \\
& \Rightarrow 2 \sec 2 x \tan 2 x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\frac{1}{2} \int\left(t^2-1\right) d t \\
& =\frac{1}{2}\left(\int t^2 d t-\int 1 d t\right) \\
& =\frac{1}{2}\left(\frac{t^3}{3}-t\right)+c \\
& =\frac{1}{6} t^3-\frac{1}{2} t+c
\end{aligned}
$
$=\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c$
Ex 7.3 Question 16
$
\begin{aligned}
& \text {} \tan ^4 x \\
& \text { Answer. } \int \tan ^4 x d x \\
& =\int \tan ^2 x \tan ^2 x d x \\
& =\int \tan ^2 x\left(\sec ^2 x-1\right) d x \\
& =\int \tan ^2 x \sec ^2 x d x-\int\left(\sec ^2 x-1\right) d x \\
& =\int \tan ^2 x \sec ^2 x d x-\int \sec ^2 x d x+\int 1 d x .
\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), } \\
& =\int t^2 d t-\tan x+x+c \\
& =\frac{t^3}{3}-\tan x+x+c \\
& =\frac{1}{3} \tan ^3 x-\tan x+x+c
\end{aligned}
$
Ex 7.3 Question 17.
$\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}$
Answer.
$\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x$
$
\begin{aligned}
& =\int \frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x} d x \\
& =\int \frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x} d x \\
& =\int \frac{\sin x}{\cos x \cos x}+\frac{\cos x}{\sin x \cos x} d x \\
& =\int(\tan x \sec x+\cot x \operatorname{cosec} x) d x \\
& =\int \tan x \sec x d x+\int \operatorname{cosec} x \cot x d x \\
& =\sec x-\operatorname{cosec} x+c
\end{aligned}
$
Ex 7.3 Question 18.
$\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}$
Answer.
$\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x$
$
=\int \frac{\left(1-2 \sin ^2 x\right)+2 \sin ^2 x}{\cos ^2 x} d x
$
using $\cos 2 \theta=1-2 \sin ^2 \theta$
$
\begin{aligned}
& =\int \frac{1}{\cos ^2 x} d x \\
& =\int \sec ^2 x d x \\
& =\tan x+c
\end{aligned}
$
Ex 7.3 Question 19.
$\frac{1}{\sin x \cos ^3 x}$
Answer.
Let $\mathrm{I}=\int \frac{1}{\sin x \cos ^3 x} d x \ldots \ldots \ldots .(\mathrm{i})$
Dividing numerator and denominator by $\cos ^4 \theta$
$
\begin{aligned}
& =\int \frac{1 / \cos ^4 x d x}{\sin x / \cos x} \\
& I=\int \frac{\sec ^4 x d x}{\tan x} \\
& =\int \frac{\sec ^2 x \sec ^2 x}{\tan x} d x \\
& I=\int \frac{\left(1+\tan ^2 x\right) \sec ^2 x d x}{\tan x}
\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. (ii), } \\
& \mathrm{I}=\int \frac{\left(1+t^2\right)}{t} d t \\
& =\int\left(\frac{1}{t}+\frac{t^2}{t}\right) d t \\
& =\int\left(\frac{1}{t}+t\right) d t
\end{aligned}
$
$\begin{aligned}
& =\int \frac{1}{t} d t+\int t d t \\
& =\log |t|+\frac{t^2}{2}+c \\
& =\log |\tan x|+\frac{1}{2} \tan ^2 x+c
\end{aligned}$
Ex 7.3 Question 20.
$\frac{\cos 2 x}{(\cos x+\sin x)^2}$
Answer.
Let $\mathrm{I}=\int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x$
$
\begin{aligned}
& =\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x \\
& =\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)(\cos x+\sin x)} d x \\
& =\int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x \ldots \ldots . .(\mathrm{i})
\end{aligned}
$
Putting $\cos x+\sin x=t$
$
\begin{aligned}
& \Rightarrow-\sin x+\cos x=\frac{d t}{d x} \\
& \Rightarrow(\cos x-\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 |\cos x+\sin x|+c
\end{aligned}
$
Ex 7.3 Question 21.
$\sin ^{-1}(\cos x)$
Answer.
$\int \sin ^{-1}(\cos x) d x$
$
\begin{aligned}
& =\int \sin ^{-1} \sin \left(\frac{\pi}{2}-x\right) d x \\
& =\int\left(\frac{\pi}{2}-x\right) d x \\
& =\int \frac{\pi}{2} d x-\int x d x \\
& =\frac{\pi}{2} \int 1 d x-\int x d x \\
& =\frac{\pi}{2} x-\frac{x^2}{2}+c
\end{aligned}
$
Ex 7.3 Question 22.
$\frac{1}{\cos (x-a) \cos (x-b)}$
Answer.
Let $\mathrm{I}=\int \frac{1}{\cos (x-a) \cos (x-b) d x}$
Multiplying and dividing by
$\sin (b-a)$ as $(x-a)-(x-b)=b-a$,
$\mathrm{I}=\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\cos (x-a) \cos (x-b)} d x$
$
\begin{aligned}
& =\frac{1}{\sin (b-a)} \int \frac{\sin [(x-a)-(x-b)]}{\cos (x-a) \cos (x-b)} d x \\
& =\frac{1}{\sin (b-a)} \int \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\cos (x-a) \cos (x-b)} d x
\end{aligned}
$
using $\sin (A-B)=\sin A \cos B-\cos A \sin B$
$
\begin{aligned}
& =\frac{1}{\sin (b-a)} \int\left[\frac{\sin (x-a) \cos (x-b)}{\cos (x-a) \cos (x-b)}-\frac{\cos (x-a) \sin (x-b)}{\cos (x-a) \cos (x-b)}\right] d x \\
& =\frac{1}{\sin (b-a)} \int[\tan (x-a)-\tan (x-b)] d x \\
& =\frac{1}{\sin (b-a)}[-\log |\cos (x-a)|+\log |\cos (x-b)|]+c \\
& =\frac{1}{\sin (b-a)} \log \left|\frac{\cos (x-b)}{\cos (x-a)}\right|+c
\end{aligned}
$
Ex 7.3 Question 23.
$\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$ is equal to:
(A) $\tan x+\cot x+C$
(B) $\tan x+\operatorname{cosec} x+C$
(C) $-\tan x+\cot x+C$
(D) $\tan x+\sec x+C$
Answer.
$\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$
$
\begin{aligned}
& =\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 \sec ^2 x-\operatorname{cosec}{ }^2 x d x \\
& =\int \sec ^2 x d x-\int \operatorname{cosec}{ }^2 x d x \\
& =\tan x-(-\cot x)+C \\
& =\tan x+\cot x+C
\end{aligned}
$
Therefore, option (A) is correct.
Ex 7.3 Question 24.
$\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x$ is equal to:
(A) $-\cot \left(e x^x\right)+\mathrm{C}$
(B) $\tan \left(x e^x\right)+\mathrm{C}$
(C) $\tan \left(e^x\right)+\mathrm{C}$
(D) $\cot \left(e^x\right)+\mathrm{C}$
Answer.
Let $\mathrm{I}=\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x$ ...(i)
Putting $e^x x=t$
$
\Rightarrow e^x \cdot 1+x e^x=\frac{d t}{d x}
$
$
\begin{aligned}
& \Rightarrow e^x(1+x) d x=d t \\
& \therefore \text { From eq. (i), } \\
& \mathrm{I}=\int \frac{d t}{\cos ^2 t} \\
& =\int \sec ^2 t d t \\
& =\tan t+\mathrm{C} \\
& =\tan \left(x \cdot e^x\right)+\mathrm{C}
\end{aligned}
$
Therefore, option (B) is correct.
