Miscellaneous Exercise (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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Miscellaneous Exercise Question 1.
$\frac{1}{x-x^3}$
Answer.
Let $I=\int \frac{1}{x-x^3} d x$
$
\begin{aligned}
& \text { Here, } \frac{1}{x-x^3} \\
& =\frac{1}{x\left(1-x^2\right)} \\
& =\frac{1}{x(1-x)(1+x)} \\
& =\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-x}+\frac{\mathrm{C}}{1+x} \ldots \ldots \ldots .(\mathrm{i}) \\
& \Rightarrow 1=\mathrm{A}(1-x)(1+x)+\mathrm{B} x(1+x)+\mathrm{C} x(1-x) \\
& \Rightarrow 1=\mathrm{A}\left(1-x^2\right)+\mathrm{B}\left(x+x^2\right)+\mathrm{C}\left(x-x^2\right) \\
& \Rightarrow 1=\mathrm{A}-\mathrm{A} x^2+\mathrm{B} x+\mathrm{B} x^2+\mathrm{C} x-\mathrm{C} x^2
\end{aligned}
$
Comparing the coefficients of $x^2-\mathrm{A}+\mathrm{B}-\mathrm{C}=0$
Comparing the coefficients of $x \mathrm{~B}+\mathrm{C}=0$ ..(iii)
Comparing constants $\mathrm{A}=1$ (iv)
On solving eq. (ii), (iii) and (iv), we get $A=1, B=\frac{1}{2}, C=\frac{-1}{2}$
Putting these values in eq. (i),
$
\begin{aligned}
& \frac{1}{x-x^3} \\
&= \frac{1}{x}+\frac{\frac{1}{2}}{1-x}+\frac{\frac{-1}{2}}{1+x} \\
& \Rightarrow \int \frac{1}{x-x^3} d x=\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{1-x} d x-\frac{1}{2} \int \frac{1}{1+x} d x \\
&= \log |x|+\frac{1}{2} \frac{\log |1-x|}{-1}-\frac{1}{2} \log |1+x|+c \\
&= \frac{1}{2}[2 \log |x|-\log |1-x|-\log |1+x|]+c \\
&= \frac{1}{2}\left[\log |x|^2-(\log |1-x|+\log |1+x|)\right]+c \\
&= \frac{1}{2}\left[\log |x|^2-\log |1-x||1+x|\right]+c \\
&= \frac{1}{2}\left[\log |x|^2-\log \left|1-x^2\right|\right]+c \\
&= \frac{1}{2} \log \left|\frac{x^2}{1-x^2}\right|+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question2.
$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$
Answer.
Let $\mathrm{I}=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$
$\begin{aligned}
& =\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} d x \\
& =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{x+a-x-b} d x \\
& =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} d x \\
& =\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x \\
& =\frac{1}{a-b}\left[\int(x+a)^{\frac{1}{2}} d x-\int(x+b)^{\frac{1}{2}} d x\right] \\
& =\frac{1}{a-b}\left[\frac{(x+a)^{\frac{3}{2}}}{3}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}(1)}\right]+c \\
& =\frac{1}{a-b}\left[\frac{2}{3}(x+a)^{\frac{3}{2}}-\frac{2}{3}(x+b)^{\frac{3}{2}}\right]+c \\
& =\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question3.
$\frac{1}{x \sqrt{a x-x^2}}$
Answer.
Let I $=\int \frac{1}{x \sqrt{a x-x^2}} d x$ $\qquad$
Putting $x=\frac{1}{t}=t^{-1}$
$
\Rightarrow d x=\frac{-1}{t^2} d t
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{\frac{-1}{t^2} d t}{\frac{1}{t} \sqrt{\frac{a}{t}-\frac{1}{t^2}}} \\
& =-\int \frac{d t}{\sqrt{a t-1}} \\
& =-\int(a t-1)^{\frac{-1}{2}} d t \\
& =\frac{-(a t-1)^{\frac{1}{2}}}{\frac{1}{2} \times a}+c \\
& =\frac{-2}{a} \sqrt{\frac{a}{x}-1}+c \\
& =\frac{-2}{a} \sqrt{\frac{a-x}{x}}+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 4.
$\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}$
Answer.
Let $\mathrm{I}=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} d x$
$
\begin{aligned}
& =\int \frac{1}{x^2\left[x^4\left(1+\frac{1}{x^4}\right)\right]^{-\frac{3}{4}}} d x \\
& =\int \frac{1}{x^2 \cdot x^3\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} d x \\
& =\int \frac{1}{x^5}\left(1+\frac{1}{x^4}\right)^{\frac{-3}{4}} d x
\end{aligned}
$
Putting $1+\frac{1}{x^4}=t$
$
\begin{aligned}
& \Rightarrow-4 x^{-5} d x=d t \\
& \Rightarrow \frac{1}{x^5} d x=\frac{-1}{4} d t \\
& \therefore \mathrm{I}=\frac{-1}{4} \int t^{\frac{-3}{4}} d t \\
& =\frac{-1}{4} \cdot \frac{t^{1 / 4}}{1 / 4}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{1 / 4}+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 5.
$\frac{1}{x^{1 / 2}+x^{1 / 3}}$
Answer.
Let $\mathrm{I}=\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x$
$\begin{aligned}
& \text { Putting } x^{\frac{1}{6}}=t \\
& \Rightarrow x=t^6 \\
& \Rightarrow d x=6 t^5 d t \\
& \therefore \mathrm{I}=\int \frac{6 t^5}{t^3+t^2} d t \\
& =6 \int \frac{t^5}{t^2(t+1)} d t \\
& =6 \int \frac{t^3}{t+1} d t \\
& =6 \int \frac{\left(t^3+1\right)-1}{t+1} d t \\
& =6\left[\int \frac{t^3+1}{t+1}-\frac{1}{t+1} d t\right] \\
& =6\left[\int\left(\frac{(t+1)\left(t^2-t+1\right)}{t+1}-\frac{1}{t+1}\right) d t\right]
\end{aligned}$
$\begin{aligned}
& =6\left[\int\left(t^2-t+1-\frac{1}{t+1}\right) d t\right] \\
& =6\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\log |t+1|\right]+c \\
& =2 t^3-3 t^2+6 t-6 \log |t+1|+c \\
& =2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left|x^{\frac{1}{6}}+1\right|+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 6.
$\frac{5 x}{(x+1)\left(x^2+9\right)}$
Answer.
Let $\mathrm{I}=\int \frac{5 x}{(x+1)\left(x^2+9\right)} d x$ .(i) $\qquad$
Let $\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+9}$ $\qquad$
$\Rightarrow 5 x=\mathrm{A}\left(x^2+9\right)+(\mathrm{B} x+\mathrm{C})(x+1)$
$\Rightarrow 5 x=\mathrm{A} x^2+9 \mathrm{~A}+\mathrm{B} x^2+\mathrm{B} x+\mathrm{C} x+\mathrm{C}$
Comparing coefficients of $x^2 \mathrm{~A}+\mathrm{B}=0$ $\qquad$
Comparing coefficients of $x \mathrm{~B}+\mathrm{C}=5$ $\qquad$
Comparing constants $9 \mathrm{~A}+\mathrm{C}=0$ $\qquad$ .(v)
On solving eq. (iii), (iv) and (v), we get $A=\frac{-1}{2}, B=\frac{1}{2}, C=\frac{9}{2}$
Putting these values of A, B and C in eq. (ii),
$
\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{\frac{-1}{2}}{x+1}+\frac{\frac{1}{2} x+\frac{9}{2}}{x^2+9}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{5 x}{(x+1)\left(x^2+9\right)} d x=\frac{-1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{x}{x^2+9} d x+\frac{9}{2} \int \frac{1}{x^2+9} d x \\
& =\frac{-1}{2} \log |x+1|+\frac{1}{4} \int \frac{2 x}{x^2+9} d x+\frac{9}{2} \cdot \frac{1}{3} \tan ^{-1} \frac{x}{3}+c
\end{aligned}
$
$\begin{aligned}
& =\frac{-1}{2} \log |x+1|+\frac{1}{4} \log \left|x^2+9\right|+\frac{3}{2} \tan ^{-1} \frac{x}{3}+c \\
& =\frac{-1}{2} \log |x+1|+\frac{1}{4} \log \left(x^2+9\right)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 7.
$\frac{\sin x}{\sin (x-a)}$
Answer.
Let $\mathrm{I}=\int \frac{\sin x}{\sin (x-a)} d x$
$
\begin{aligned}
& =\int \frac{\sin (x-a+a)}{\sin (x-a)} d x \\
& =\int \frac{\sin (x-a) \cos a+\cos (x-a) \sin a}{\sin (x-a)} d x \\
& =\int\left(\frac{\sin (x-a) \cos a}{\sin (x-a)}+\frac{\cos (x-a) \sin a}{\sin (x-a)}\right) d x \\
& =\int(\cos a+\sin a \cot (x-a)) d x \\
& =\int \cos a d x+\int \sin a \cot (x-a) d x \\
& =\cos a \int 1 d x+\sin a \int \cot (x-a) d x \\
& =(\cos a) x+\sin a \frac{\log |\sin (x-a)|}{1}+c \\
& =x \cos a+\sin a \log |\sin (x-a)|+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 8.
$\text {} \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$
$\begin{aligned}
& \text { Answer. Let } \mathrm{I}=\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x \\
& =\int \frac{e^{\log x^3}-e^{\log x^4}}{e^{\log x^3}-e^{\log x^2}} d x \\
& =\int \frac{x^5-x^4}{x^3-x^2} d x \\
& =\int \frac{x^4(x-1)}{x^2(x-1)} d x \\
& \Rightarrow \mathrm{I}=\int x^2 d x \\
& =\frac{x^3}{3}+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 9.
$\frac{\cos x}{\sqrt{4-\sin ^2 x}}$
Answer.
Let $\mathrm{I}=\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x$ $\qquad$
Putting $\sin x=t$
$\Rightarrow \cos x=\frac{d t}{d x}$
$\Rightarrow \cos x d x=d t$
$\therefore$ From eq. (i),
$
\mathrm{I}=\int \frac{d t}{\sqrt{4-t^2}}
$
$\begin{aligned}
& =\sin ^{-1}\left(\frac{t}{2}\right)+c \\
& =\sin ^{-1}\left[\frac{1}{2} \sin x\right]+c \text { Ans. }
\end{aligned}$
$\begin{aligned}
& =\sin ^{-1}\left(\frac{t}{2}\right)+c \\
& =\sin ^{-1}\left[\frac{1}{2} \sin x\right]+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 10.
$\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}$
Answer.
Let $\mathrm{I}=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int \frac{\left(\sin ^4 x\right)^2-\left(\cos ^4 x\right)^2}{1-2 \sin ^2 x \cos ^2 x} d x \\
& =\int \frac{\left(\sin ^4 x-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right)}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \Rightarrow \mathrm{I}=\int \frac{\left\{\left(\sin ^2 x\right)^2-\left(\cos ^2 x\right)^2\right\}\left\{\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^2\right\}}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \Rightarrow \mathrm{I}=\int \frac{\left(\sin ^2 x-\cos ^2 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left\{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \tan ^2 x\right\}}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \Rightarrow \mathrm{I}=\int \frac{-\left(\cos ^2 x-\sin ^2 x\right)\left\{1-2 \sin ^2 x \tan ^2 x\right\}}{1-2 \sin ^2 x \cos ^2 x} d x \\
& =\int \frac{-\cos 2 x\left\{1-2 \sin ^2 x \tan ^2 x\right\}}{1-2 \sin ^2 x \cos ^2 x} d x \\
& \Rightarrow \mathrm{I}=-\int \cos 2 x d x \\
& =\frac{-\sin 2 x}{2}+c \text { Ans. } \\
&
\end{aligned}
$
Miscellaneous Exercise Question 11.
$\frac{1}{\cos (x+a) \cos (x+b)}$
Answer.
Let $\mathrm{I}=\int \frac{1}{\cos (x+a) \cos (x+b)} d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x+a) \cos (x+b)} d x \\
& =\frac{1}{\sin (a-b)} \int \frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)} d x \\
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \int \frac{\sin (x+a) \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)} d x \\
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \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 \\
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \int(\tan (x+a)-\tan (x+b)) d x \\
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b)|]+c \\
& \Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \log \left|\frac{\cos (x+b)}{\cos (x+a)}\right|+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 12.
$\text {} \frac{x^3}{\sqrt{1-x^8}}$
Answer.
Let $\mathrm{I}=\int \frac{x^3}{\sqrt{1-x^8}} d x$
$
=\frac{1}{4} \int \frac{4 x^3}{\sqrt{1-x^8}} d x \ldots \ldots \ldots . \text { (i) }
$
Putting $x^4=t$
$
\begin{aligned}
& \Rightarrow 4 x^3=\frac{d t}{d x} \\
& \Rightarrow 4 x^3 d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}} \\
& =\frac{1}{4} \sin ^{-1} t+c \\
& =\frac{1}{4} \sin ^{-1}\left(x^4\right)+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 13.
$\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}$
Answer.
Let $\mathrm{I}=\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x$
Putting $e^x=t$
$
\begin{aligned}
& \Rightarrow e^x=\frac{d t}{d x} \\
& \Rightarrow e^x d x=d t
\end{aligned}
$
$\begin{aligned}
& \therefore \text { From eq. (i), } \mathrm{I}=\int \frac{d t}{(1+t)(2+t)} \\
& =\int \frac{1}{(t+1)(t+2)} d t \ldots \ldots \ldots . .(\text { iii) } \\
& =>I=\int \frac{(t+2)-(t+1)}{(t+1)(t+2)} d t \\
& =>I=\int\left(\frac{t+2}{(t+1)(t+2)}-\frac{t+1}{(t+1)(t+2)}\right) d t \\
& \Rightarrow \mathrm{I}=\int\left(\frac{1}{(t+1)}-\frac{1}{(t+2)}\right) d t \\
& =\log |t+1|-\log |t+2|+c \\
& \Rightarrow \mathrm{I}=\log \left|\frac{t+1}{t+2}\right|+c \\
& =\log \left|\frac{e^x+1}{e^x+2}\right|+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 14.
$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}$
Answer.
Let I $=\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} d x$ $\qquad$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\frac{1}{3} \int \frac{3}{\left(x^2+1\right)\left(x^2+4\right)} d x \\
& =\frac{1}{3} \int \frac{\left(x^2+4\right)-\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2+4\right)} d x
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\frac{1}{3} \int\left(\frac{1}{\left(x^2+1\right)}-\frac{1}{\left(x^2+4\right)}\right) d x \\
& =\frac{1}{3}\left[\int \frac{1}{x^2+1} d x-\int \frac{1}{x^2+4} d x\right] \\
& \Rightarrow \mathrm{I}=\frac{1}{3}\left[\tan ^{-1} x-\frac{1}{2} \tan ^{-1} \frac{x}{2}\right]+\mathrm{c} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 15.
$\cos ^3 x e^{\log \sin x}$
Answer.
Let $\mathrm{I}=\int \cos ^3 x e^{\mathrm{log} \sin x} d x$
$
\begin{aligned}
& =\int \cos ^3 x \sin x d x \\
& =-\int \cos ^3 x(-\sin x) d x
\end{aligned}
$
Putting $\cos x=t$
$
\begin{aligned}
& \Rightarrow-\sin x=\frac{d t}{d x} \\
& \Rightarrow-\sin x d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
& \mathrm{I}=-\int t^3 d t \\
& =\frac{-t^4}{4}+c \\
& =\frac{-1}{4} \cos ^4 x+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 16.
$\text {} e^{3 \log x}\left(x^4+1\right)^{-1}$
Answer.
Let $\mathrm{I}=\int e^{3 \log x}\left(x^4+1\right)^{-1} d x$
$
\begin{aligned}
& =\int \frac{e^{3 \log x}}{\left(x^4+1\right)} d x \\
& =\int \frac{e^{\log x^3}}{\left(x^4+1\right)} d x \\
& =\int \frac{x^3}{\left(x^4+1\right)} d x \\
& \Rightarrow \mathrm{I}=\frac{1}{4} \int \frac{4 x^3}{\left(x^4+1\right)} d x
\end{aligned}
$
Putting $x^4+1=t$
$
\begin{aligned}
& \Rightarrow 4 x^3=\frac{d t}{d x} \\
& \Rightarrow 4 x^3 d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\mathrm{I}=\frac{1}{4} \int \frac{d t}{t}
$
$\begin{aligned}
& =\frac{1}{4} \log |t|+c \\
& \Rightarrow \mathrm{I}=\frac{1}{4} \log \left|x^4+1\right|+c \\
& =\frac{1}{4} \log \left(x^4+1\right)+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 17.
$\int f^{\prime}(a x+b)(f(a x+b))^n d x$
Answer.
Let $\mathrm{I}=\int f^{\prime}(a x+b)\{f(a x+b)\}^n d x$
$
=\frac{1}{a} \int f^{\prime}(a x+b)^n a f(a x+b) d x
$
Putting $f(a x+b)=t$
$
\begin{aligned}
& \Rightarrow f^{\prime}(a x+b) \frac{d}{d x}(a x+b)=\frac{d t}{d x} \\
& \Rightarrow a f^{\prime}(a x+b) d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\frac{1}{a} \int t^n d t \\
& =\frac{1}{a} \cdot \frac{t^{n+1}}{n+1}+c \text { if } n \neq-1
\end{aligned}
$
$
\Rightarrow \mathrm{I}=\frac{\{f(a x+b)\}^{n+1}}{a(n+1)}+c \text { Ans. }
$
Miscellaneous Exercise Question 18.
$\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$
Answer.
Let I $=\int \frac{d x}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$
$
=\int \frac{d x}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\int \frac{d x}{\sqrt{\sin ^3 x \cdot \sin x(\cos \alpha+\cot x \sin \alpha)}} \\
& =\int \frac{d x}{\sin ^2 x \sqrt{\cos \alpha+\cot x \sin \alpha}} \\
& \Rightarrow \mathrm{I}=\int \frac{\operatorname{cosec}^2 x d x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} \\
& \text { Putting } \cos \alpha+\cot x \sin \alpha=t \\
& \Rightarrow-\operatorname{cosec}^2 x \sin \alpha d x=d t \\
& \Rightarrow \operatorname{cosec}^2 x d x=-\frac{d t}{\sin \alpha} \\
& \therefore \mathrm{I}=-\int \frac{d t}{\sin \alpha \sqrt{t}} \\
& =\frac{-1}{\sin \alpha} \int t^{\frac{-1}{2}} d t \\
& =\frac{-1}{\sin \alpha} \cdot \frac{t^{1 / 2}}{1 / 2}+c \\
& \Rightarrow \mathrm{I}=\frac{-2}{\sin \alpha} \cdot \sqrt{\cos \alpha+\cot x \sin \alpha}+c \\
& =\frac{-2}{\sin \alpha} \sqrt{\cos \alpha+\frac{\cos x}{\sin x} \sin \alpha}+c \\
&
\end{aligned}$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\frac{-2}{\sin \alpha} \sqrt{\frac{\sin x \cos \alpha+\cos x \sin \alpha}{\sin x}}+c \\
& =\frac{-2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 19.
$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
Answer.
Let $\mathrm{I}=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$
Putting $\sqrt{x}=t$
$
\begin{aligned}
& \Rightarrow x=t^2 \\
& \Rightarrow d x=2 t d t \\
& \therefore \mathrm{I}=\int \sqrt{\frac{1-t}{1+t}} 2 t d t
\end{aligned}
$
$\begin{aligned}
& =2 \int t \sqrt{\frac{1-t}{1+t}} d t \\
& =2 \int t \sqrt{\frac{1-t}{1+t} \times \frac{1-t}{1-t}} d t \\
& =2 \int \frac{t(1-t)}{\sqrt{1-t^2}} d t \\
& \Rightarrow \mathrm{I}=2 \int \frac{t-t^2}{\sqrt{1-t^2}} d t \ldots \ldots \ldots . .(\mathrm{i}) \\
& \Rightarrow \mathrm{I}=2 \int \frac{\left(1-t^2\right)+t-1}{\sqrt{1-t^2}} d t \\
& =2\left[\int \sqrt{1-t^2} d t+\int \frac{t}{\sqrt{1-t^2}} d t-\int \frac{1}{\sqrt{1-t^2}} d t\right] \\
& \Rightarrow \mathrm{I}=2\left[\frac{t}{2} \sqrt{1-t^2}+\frac{1}{2} \sin ^{-1} t+\int \frac{t}{\sqrt{1-t^2}} d t-\sin ^{-1} t\right]+c \\
& \mathrm{I}=2\left[\frac{1}{2} t \sqrt{1-t^2}-\frac{1}{2} \sin ^{-1} t+\int \frac{t}{\sqrt{1-t^2}} d t\right]+c \ldots \ldots \ldots .(\mathrm{ii})
\end{aligned}$
For evaluating $\int \frac{t}{\sqrt{1-t^2}} d t$, putting $1-t^2=z$
$
\begin{aligned}
& \Rightarrow-2 t d t=d z \\
& \Rightarrow t d t=-\frac{1}{2} d z \\
& \therefore \int \frac{t}{\sqrt{1-t^2}} d t
\end{aligned}
$
$
\begin{aligned}
& =\int \frac{\frac{-1}{2}}{\sqrt{z}} d z \\
& =-\frac{1}{2} \int \frac{1}{\sqrt{z}} d z \\
& =-\frac{1}{2} \int z^{-\frac{1}{2}} d z \\
& =-\frac{1}{2} \cdot \frac{z^{1 / 2}}{1 / 2} \\
& =-\sqrt{1-t^2} \cdots \cdots \cdots
\end{aligned}
$
Putting this value in eq. (ii),
$
\begin{aligned}
& \mathrm{I}=2\left[\frac{1}{2} t \sqrt{1-t^2}-\frac{1}{2} \sin ^{-1} t-\sqrt{1-t^2}\right]+c \\
& =>I=t \sqrt{1-t^2}-\sin ^{-1} t-2 \sqrt{1-t^2}+c \\
& =(t-2) \sqrt{1-t^2}-\sin ^{-1} t+c \\
& \Rightarrow \mathrm{I}=(\sqrt{x}-2) \sqrt{1-x}-\sin ^{-1} \sqrt{x}+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 20.
$\frac{2+\sin 2 x}{1+\cos 2 x} e^x$
$
\begin{aligned}
& \text { Ans. Let } \mathrm{I}=\int \frac{2+\sin 2 x}{1+\cos 2 x} e^x d x \\
& =\int e^x \frac{2+2 \sin x \cos x}{2 \cos ^2 x} d x \\
& =\int e^x\left(\frac{2}{2 \cos ^2 x}+\frac{2 \sin x \cos x}{2 \cos ^2 x}\right) d x
\end{aligned}
$
$\begin{aligned}
& =\int e^x\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos x}\right) d x \\
& =\int e^x\left(\sec ^2 x+\tan x\right) d x \\
& =\int e^x\left(\tan x+\sec ^2 x\right) d x \\
& \left(\int e^x\left(f(x)+f^{\prime}(x)\right)\right) \\
& =e^x \tan x+c
\end{aligned}$
Miscellaneous Exercise Question 21.
$\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$
Answer.
Let $\mathrm{I}=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$
Putting $x=\cos 2 \theta$
$
\begin{aligned}
& \Rightarrow \frac{d x}{d \theta}=-2 \sin 2 \theta \\
& \Rightarrow d x=-2 \sin 2 \theta d \theta \\
& \text { And } \tan ^{-1} \sqrt{\frac{1-x}{1+x}}=\tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}
\end{aligned}
$
And $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}=\tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}$
$
\begin{aligned}
& =\tan ^{-1} \sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}} \\
& =\tan ^{-1} \sqrt{\tan ^2 \theta} \\
& =\tan ^{-1}(\tan \theta)=\theta \\
& \therefore \mathrm{I}=\int \theta(-2 \sin 2 \theta d \theta) \\
& =-2 \int \theta \sin 2 \theta d \theta
\end{aligned}
$
[Applying Product Rule]
$
\begin{aligned}
& =-2\left[\theta\left(\frac{-\cos 2 \theta}{2}\right)-\int 1\left(\frac{-\cos 2 \theta}{2}\right) d \theta\right] \\
& =-2\left[\frac{-1}{2} \theta \cos 2 \theta+\frac{1}{2} \int \cos 2 \theta d \theta\right] \\
& =\theta \cos 2 \theta-\frac{\sin 2 \theta}{2}+c \\
& =\theta \cos 2 \theta-\frac{1}{2} \sqrt{1-\cos ^2 2 \theta}+c \\
& =\theta\left(\cos ^{-1} x\right) x-\frac{1}{2} \sqrt{1-x^2}+c \\
& =\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^2}\right]+c \text { Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 22.
$\text {} \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}$
Answer.
Let $\mathrm{I}=\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$
$
\begin{aligned}
& =\int \frac{\sqrt{x^2+1}}{x^4}\left[\log \left(x^2+1\right)-\log x^2\right] d x \\
& =\int \frac{\sqrt{x^2\left(1+\frac{1}{x^2}\right)}}{x^4} \log \left(\frac{x^2+1}{x^2}\right) d x \\
& =\int \frac{\sqrt{\left(1+\frac{1}{x^2}\right)}}{x^3} \log \left(1+\frac{1}{x^2}\right) d x \\
& =\int \sqrt{\left(1+\frac{1}{x^2}\right)} \log \left(1+\frac{1}{x^2}\right) \frac{d x}{x^3}
\end{aligned}
$
Putting $1+\frac{1}{x^2}=t$
$
\begin{aligned}
& \Rightarrow 1+x^{-2}=t \\
& \Rightarrow \frac{-2}{x^3} d x=d t \\
& \Rightarrow \frac{d x}{x^3}=-\frac{1}{2} d t
\end{aligned}
$
$
\begin{aligned}
& \therefore \mathrm{I}=-\frac{1}{2} \int \sqrt{t} \log t d t \\
& =-\frac{1}{2} \int(\log t) \cdot t^{\frac{1}{2}} d t
\end{aligned}
$
[Applying Product Rule]
$\begin{aligned}
& =-\frac{1}{2}\left[(\log t) \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} d t\right] \\
& \Rightarrow \mathrm{I}=-\frac{1}{3} t^{3 / 2} \log t+\frac{1}{3} \int t^{1 / 2} d t \\
& =-\frac{1}{3} t^{3 / 2} \log t+\frac{1}{3} \cdot \frac{t^{3 / 2}}{3 / 2}+c \\
& \Rightarrow \mathrm{I}=\frac{2}{9} t^{3 / 2}-\frac{1}{3} t^{3 / 2} \log t+c \\
& =\frac{1}{3} t^{3 / 2}\left[\frac{2}{3}-\log t\right]+c \\
& =\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3 / 2}\left[\frac{2}{3}-\log \left(1+\frac{1}{x^2}\right)\right]+c \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 23.
$\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x$
Answer.
Let $\mathrm{I}=\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x$
$
=\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}\right) d x
$
$\begin{aligned}
& =\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1}{2 \sin ^2 \frac{x}{2}}-\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}\right) d x \\
& =\int_{\frac{\pi}{2}}^\pi e^x\left(\frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}-\cot \frac{x}{2}\right) d x \\
& =\int_{\frac{\pi}{2}}^\pi e^x\left(-\cot \frac{x}{2}+\frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}\right) d x \\
& {\left[\because \int^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)\right]} \\
& =\left(-e^x \cot \frac{x}{2}\right)^{\frac{\pi}{2}} \\
& =-e^\pi \cot \frac{\pi}{2}-\left(-e^{\frac{\pi}{2}} \cot \frac{\pi}{4}\right) \\
& =e^{\frac{\pi}{2}} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 24.
$\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$
divide numerator and denominator by $\cos ^4 \mathrm{x}$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{1+\frac{\sin ^4 x}{\cos ^4 x}} d x \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x \\
& =\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{2 \tan x \sec ^2 x}{1+\tan ^4 x} d x \ldots \ldots \ldots . . \text { (i) } \\
& \text { Putting } \tan ^2 x=t \\
& \Rightarrow 2 \tan x \frac{d}{d x}(\tan x)=\frac{d t}{d x} \\
& \Rightarrow 2 \tan x \sec ^2 x d x=d t
\end{aligned}
$
Putting $\tan ^2 x=t$
$
\begin{aligned}
& \Rightarrow 2 \tan x \frac{d}{d x}(\tan x)=\frac{d t}{d x} \\
& \Rightarrow 2 \tan x \sec ^2 x d x=d t
\end{aligned}
$
Limits of integration when $x=0, t=\tan ^2 x=\tan ^2 0^{\circ}=0$ and when
$\begin{aligned}
& x=\frac{\pi}{4}, t=\tan ^2 \frac{\pi}{4}=1 \\
& \therefore \mathrm{I}=\frac{1}{2} \int_0^1 \frac{d t}{1+t^2} \\
& =\frac{1}{2}\left(\tan ^{-1} t\right)_0^1 \\
& =\frac{1}{2}\left(\tan ^{-1} 1-\tan ^{-1} 0\right) \\
& =\frac{1}{2}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{8} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 25.
$\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x d x}{\cos ^2 x+4 \sin ^2 x}$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x$
[Dividing each term by $\cos ^2 x$ ]
$
=\int_0^{\frac{\pi}{2}} \frac{1}{1+4 \tan ^2 x} d x \cdots .(i)
$
Putting $\tan x=t$
$
\begin{aligned}
& \Rightarrow \sec ^2 x=\frac{d t}{d x} \\
& \Rightarrow d x=\frac{d t}{\sec ^2 x}=\frac{d t}{1+\tan ^2 x} \\
& =\frac{d t}{1+t^2}
\end{aligned}
$
Limits of integration when $x=0, t=\tan 0^{\circ}=0$ and when $x=\frac{\pi}{2}, t=\tan \frac{\pi}{2}=\infty$
$\begin{aligned}
&\begin{aligned}
& \therefore \mathrm{I}=\int_0^{\infty} \frac{1}{1+4 t^2} \cdot \frac{d t}{1+t^2} \\
& =\int_0^{\infty} \frac{1}{\left(1+4 t^2\right)\left(1+t^2\right)} d t
\end{aligned}\\
&\Rightarrow \mathrm{I}=\frac{1}{3} \int_0^{\infty} \frac{3}{\left(1+4 t^2\right)\left(1+t^2\right)} d t
\end{aligned}$
$\begin{aligned}
& =\frac{1}{3} \int_0^{\infty} \frac{4\left(t^2+1\right)-\left(4 t^2+1\right)}{\left(1+4 t^2\right)\left(1+t^2\right)} d t \\
& \Rightarrow \mathrm{I}=\frac{1}{3} \int_0^{\infty}\left[\frac{4\left(t^2+1\right)}{\left(1+4 t^2\right)\left(1+t^2\right)}-\frac{\left(4 t^2+1\right)}{\left(1+4 t^2\right)\left(1+t^2\right)}\right] d t \\
& =\frac{1}{3}\left[\int_0^{\infty} 4 \frac{1}{\left(4 t^2+1\right)} d t-\int_0^{\infty} \frac{1}{\left(1+t^2\right)} d t\right] \\
& \Rightarrow \mathrm{I}=\frac{1}{3}\left[\int_0^{\infty} 4 \frac{1}{\left((2 t)^2+1\right)} d t-\tan ^{-1} t\right] \\
& =\frac{1}{3}\left[4 \cdot \frac{\tan ^{-1} 2 t}{2}-\tan ^{-1} t\right]_0^{\infty} \\
& =\frac{1}{3}\left[2\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)-\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)\right] \\
& =\frac{1}{3}\left[2\left(\frac{\pi}{2}-0\right)-\left(\frac{\pi}{2}-0\right)\right] \\
& =\frac{1}{3} \times \frac{\pi}{2} \\
& ==\frac{\pi}{6} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 26.
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
Answer.
Let $\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
Putting $\sin x-\cos x=t$
$\Rightarrow(\cos x+\sin x) d x=d t$
$
\begin{aligned}
& \text { Again }(\sin x-\cos x)^2=t^2 \\
& \Rightarrow \sin ^2 x+\cos ^2 x-2 \sin x \cos x=t^2 \\
& \Rightarrow \sin 2 x=1-t^2
\end{aligned}
$
Limits of integration when $x=\frac{\pi}{6}, t=\sin \frac{\pi}{6}-\cos \frac{\pi}{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{-(\sqrt{3}-1)}{2}=-\alpha$
where $\alpha=\frac{\sqrt{3}-1}{2}$.
$
\begin{aligned}
& \text { when } x=\frac{\pi}{3}, t=\sin \frac{\pi}{3}-\cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{(\sqrt{3}-1)}{2}=\alpha \\
& \therefore \mathrm{I}=\int_{-\alpha}^\alpha \frac{d t}{\sqrt{1-t^2}}=\left[\sin ^{-1} t\right]_{-\alpha}^\alpha \\
& =\sin ^{-1} \alpha+\sin ^{-1} \alpha=2 \sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right) \text { [From eq. (ii) Ans. }
\end{aligned}
$
Miscellaneous Exercise Question 27.
$\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
$
\begin{aligned}
& \text { Ans. Let } \mathrm{I}=\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}} \\
& =\int_0^1 \frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}+\sqrt{x})(\sqrt{1+x}-\sqrt{x})} d x \\
& =\int_0^1 \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^1(\sqrt{1+x}+\sqrt{x}) d x \\
& =\int_0^1(1+x)^{\frac{1}{2}} d x+\int_0^1(x)^{\frac{1}{2}} d x \\
& =\frac{\left[(1+x)^{\frac{3}{2}}\right]_0^1}{\frac{3}{2}(1)}+\frac{\left((x)^{\frac{3}{2}}\right]_0^1}{\frac{3}{2}(1)} \\
& \Rightarrow \mathrm{I}=\frac{2}{3}\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]+\frac{2}{3}\left[(1)^{\frac{3}{2}}-0\right] \\
& =\frac{2}{3}[2 \sqrt{2}-1]+\frac{2}{3}[1-0] \\
& \Rightarrow \mathrm{I}=\frac{4 \sqrt{2}}{3}-\frac{2}{3}+\frac{2}{3}=\frac{4 \sqrt{2}}{3} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 28.
$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$
Answer.
Let $I=\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$
Putting $\sin x-\cos x=t$
$\Rightarrow(\cos x+\sin x) d x=d t$
Again $(\sin x-\cos x)^2=t^2$
$\Rightarrow \sin ^2 x+\cos ^2 x-2 \sin x \cos x=t^2$
$
\Rightarrow \sin 2 x=1-t^2
$
Limits of integration when $x=0, t=0-1=-1$ and when $x=\frac{\pi}{4}, t=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0$
$
\begin{aligned}
& \therefore \mathrm{I}=\int_{-1}^0 \frac{d t}{9+16\left(1-t^2\right)} \\
& =\int_{-1}^0 \frac{d t}{25-16 t^2} \\
& =\int_{-1}^0 \frac{d t}{16\left(\frac{25}{16}-t^2\right)} \\
& =\frac{1}{16} \int_{-1}^0 \frac{d t}{\left(\frac{5}{4}\right)^2-t^2}
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\frac{1}{16} \times\left[\left.\frac{1}{2 \times \frac{5}{4}} \log \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right|_{-1} ^0\right. \\
& =\frac{1}{40}\left[\log 1-\log \frac{1 / 4}{9 / 4}\right]=\frac{1}{40}\left[0-\log \frac{1}{9}\right] \\
& \Rightarrow \mathrm{I}=\frac{1}{40}[-(\log 1-\log 9)]=\frac{1}{40} \log 9 \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 29.
$\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$
$
=\int_0^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x
$
Putting $\sin x=t$
$
\Rightarrow \cos x d x=d t
$
Limits of integration when $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\sin \frac{\pi}{2}=1$
$
\begin{aligned}
& \therefore \mathrm{I}=2 \int_0^1 t \tan ^{-1} t d t \\
& =2 \int_0^1\left(\tan ^{-1} t\right) t d t
\end{aligned}
$
[Applying Product Rule]
$\begin{aligned}
& \Rightarrow \mathrm{I}=2\left[\tan ^{-1} t \cdot \frac{t^2}{2}-\int \frac{1}{1+t^2} \frac{t^2}{2} d t\right] \\
& \Rightarrow \mathrm{I}=2\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{\left(1+t^2\right)-1}{1+t^2} d t\right] \\
& =2\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} \int\left(1-\frac{1}{1+t^2}\right) d t\right]
\end{aligned}$
$\begin{aligned}
& \Rightarrow \mathrm{I}=2\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2}\left(t-\tan ^{-1} t\right)\right] \\
& =2\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t\right]_0^1 \\
& =\left[\left(t^2+1\right) \tan ^{-1} t-t\right]_0^1 \\
& =\left(2 \tan ^{-1} 1-1\right)-(0-0) \\
& =2 \times \frac{\pi}{4}-1 \\
& =\frac{\pi}{2}-1 \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 30.
$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$
$
\begin{aligned}
& \text { Answer. Let } \mathrm{I}=\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x \\
& =\int_0^\pi \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x \\
& =\int_0^\pi \frac{x \sin x}{1+\sin x} d x \ldots \ldots \ldots . \text { (i) } \\
& \Rightarrow \mathrm{I}=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\sin (\pi-x)} d x \\
& =\int_0^\pi \frac{(\pi-x) \sin x}{1+\sin x} d x \ldots \ldots \ldots . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
2 \mathrm{I} & =\int_0^\pi \frac{x \sin x+(\pi-x) \sin x}{1+\sin x} d x \\
& =\int_0^\pi \frac{x \sin x+\pi \sin x-x \sin x}{1+\sin x} d x \\
& =\int_0^\pi \frac{\pi \sin x}{1+\sin x} d x \\
& =\pi \int_0^\pi \frac{\sin x}{1+\sin x} d x \\
& =\pi \int_0^\pi \frac{(1+\sin x)-1}{1+\sin x} d x \\
& =\pi \int_0^\pi\left(1-\frac{1}{1+\sin x}\right) d x \\
& =\pi \int_0^\pi 1 d x-\pi \int_0^\pi \frac{1}{1+\sin x} d x \\
& =\pi(x)_0^\pi-2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\sin x} \\
& =\pi(\pi)-2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\sin \left(\frac{\pi}{2}-x\right)}
\end{aligned}
$
$=\pi^2-2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\cos x}$
$\begin{aligned}
& =\pi^2-2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{2 \cos ^2 \frac{x}{2}} \\
& =\pi^2-\pi \int_0^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\
& =\pi^2-\pi\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]_0^{\frac{\pi}{2}} \\
& \Rightarrow 2 \mathrm{I}=\pi^2-2 \pi(1)=\pi(\pi-2) \\
& \Rightarrow \mathrm{I}=\frac{\pi}{2}(\pi-2) \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 31.
$\int_1^4[|x-1|+|x-2|+|x-3|] d x$
Answer.
Let $\mathrm{I}=\int_1^4(|x-1|+|x-2|+|x-3|) d x$.
If $x-1=0, x-2=0, x-3=0$ we get $x=1, x=2, x=3$
$\Rightarrow x=2,3 \in(1,4)$
$
\begin{aligned}
& \mathrm{I}= \\
& \int_1^2(|x-1|+|x-2|+|x-3|) d x+\int_2^3(|x-1|+|x-2|+|x-3|) d x+\int_3^4(|x-1|+|x-2|+|x-3|) d x \\
& = \\
& \int_1^2\{(x-1)-(x-2)-(x-3)\} d x+\int_2^3\{(x-1)+(x-2)-(x-3)\} d x+\int_3^4\{(x-1)+(x-2)+(x-3)\} d x
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}= \\
& \int_1^2(x-1-x+2-x+3) d x+\int_2^3(x-1+x-2-x+3) d x+\int_3^4(x-1+x-2+x-3) d x \\
& \Rightarrow \mathrm{I}=\int_1^2(4-x) d x+\int_2^3(x) d x+\int_3^4(3 x-6) d x \\
& =\left(4 x-\frac{x^2}{2}\right)_1^2+\left(\frac{x^2}{2}\right)_2^3+\left(\frac{3 x^2}{2}-6 x\right)_3^4 \\
& =(8-2)-\left(4-\frac{1}{2}\right)+\frac{9}{2}-\frac{4}{2}+(24-24)-\left(\frac{27}{2}-18\right) \\
& =6-\frac{7}{2}+\frac{9}{2}-2+0+\frac{9}{2} \\
& =4+\frac{11}{2} \\
& =\frac{19}{2} \text { Ans. }
\end{aligned}$
Miscellaneous Exercise Question 32.
$\int_1^3 \frac{d x}{x^2(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
Answer.
Taking RHS
$
\begin{aligned}
& \text { Let } \mathrm{I}=\int_1^3 \frac{d x}{x^2(x+1)} \\
& =\int_1^3 \frac{1}{x^2(x+1)} d x .
\end{aligned}
$
Let $\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$
$
\begin{aligned}
& \Rightarrow 1=\mathrm{A}(x)(x+1)+\mathrm{B}(x+1)+\mathrm{C}\left(x^2\right) \\
& \Rightarrow 1=\mathrm{A}\left(x^2+x\right)+\mathrm{B} x+\mathrm{B}+\mathrm{C} x^2 \\
& \Rightarrow 1=\mathrm{A} x^2+\mathrm{A} x+\mathrm{B} x+\mathrm{B}+\mathrm{C} x^2
\end{aligned}
$
Comparing coefficients of $x^2 \mathrm{~A}+\mathrm{C}+0$
Comparing coefficients of $x \mathrm{~A}+\mathrm{B}=0$
Comparing constants $\mathrm{B}=1$
On solving eq. (iii), (iv) and (v), we get $A=-1, B=1, C=1$
Putting these values of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in eq. (ii),
$
\begin{aligned}
& \frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{x+1} \\
& \therefore \mathrm{I}=\int_1^3\left(\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\right) d x \\
& =\int_1^3\left(\frac{-1}{x}\right) d x+\int_1^3\left(\frac{1}{x^2}\right) d x+\int_1^3\left(\frac{1}{x+1}\right) d x \\
& =-(\log |x|)_1^3+\int_1^3 x^{-2} d x+(\log |x+1|)_1^3 \\
& =-(\log |3|-\log |1|)+\left(\frac{x^{-1}}{-1}\right)_1^3+(\log |4|-\log |2|) \\
& \Rightarrow \mathrm{I}=-\log 3+0-\left(\frac{1}{x}\right)_1^3+(\log 4-\log 2)
\end{aligned}
$
$=-\log 3-\left(\frac{1}{3}-1\right)+\left(\log 2^2-\log 2\right)$
$\begin{aligned}
& \Rightarrow \mathrm{I}=-\log 3+\frac{2}{3}+2 \log 2-\log 2 \\
& =-\log 3+\frac{2}{3}+\log 2 \\
& =\frac{2}{3}+\log 2-\log 3 \\
& =\frac{2}{3}+\log \frac{2}{3}=\text { LHS Hence Proved }
\end{aligned}$
Miscellaneous Exercise Question 33.
$\int_0^1 x e^x d x=1$
Answer.
Taking RHS
Let $\mathrm{I}=\int_0^1 x e^x d x$
[Applying Product rule]
$
\begin{aligned}
& =\left(x e^x\right)_0^1-\int_0^1 1 \cdot e^x d x \\
& \Rightarrow \mathrm{I}=e-0-\int_0^1 e^x d x \\
& =e-\left(e^x\right)_0^1 \\
& =e-\left(e-e^0\right) \\
& =1 \text { = LHS Hence Proved }
\end{aligned}
$
Miscellaneous Exercise Question 34.
$\int_{-1}^1 x^{17} \cos ^4 x d x=0$
Answer.
Let $\mathrm{I}=\int_{-1}^1 x^{17} \cos ^4 x d x$
Here $f(x)=x^{17} \cos ^4 x$
$
\begin{aligned}
& \Rightarrow f(-x)=(-x)^{17} \cos ^4(-x) \\
& =-x^{17} \cos ^4 x=-f(x)
\end{aligned}
$
$\therefore f(x)$ is an odd function of $x$.
$\left[\because \int_{-a}^a f(x) d x=0\right.$, if $f(x)$ is ann odd funtion of $\left.x\right]$
$=0=$ LHS Hence proved
Miscellaneous Exercise Question 35.
$\int_0^{\frac{\pi}{2}} \sin ^3 x d x=\frac{2}{3}$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \sin ^3 x d x$
$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \frac{1}{4}(3 \sin x-\sin 3 x) d x \\
& =\frac{1}{4}\left[3(-\cos x)-\left(\frac{-\cos 3 x}{3}\right)\right]_0^{-\frac{\pi}{2}}
\end{aligned}
$
$\begin{aligned}
& =\frac{1}{4}\left[-3 \cos x+\frac{1}{3} \cos 3 x\right]_0^{\frac{\pi}{2}} \\
& =\frac{1}{4}\left[\left(-3 \cos \frac{\pi}{2}+\frac{1}{3} \cos \frac{3 \pi}{2}\right)-\left(-3 \cos 0+\frac{1}{3} \cos 0\right)\right] \\
& \Rightarrow \mathrm{I}=\frac{1}{4}\left[-3 \times 0+\frac{1}{3} \times 0+3 \times 1-\frac{1}{3} \times 1\right] \\
& =\frac{1}{4}\left(3-\frac{1}{3}\right) \\
& =\frac{1}{4} \times \frac{8}{3}=\frac{2}{3}=\text { LHS Hence proved }
\end{aligned}$
Miscellaneous Exercise Question 36.
$\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x=1-\log 2$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x$
$
\begin{aligned}
& =2 \int_0^{\frac{\pi}{4}} \tan x \tan ^2 x d x \\
& =2 \int_0^{\frac{\pi}{4}} \tan x\left(\sec ^2 x-1\right) d x \\
& \Rightarrow \mathrm{I}=2 \int_0^{\frac{\pi}{4}}\left(\tan x \sec ^2 x-\tan x\right) d x
\end{aligned}
$
$
=2\left[\int_0^{\frac{\pi}{4}}\left(\tan x \sec ^2 x\right) d x-\int_0^{\frac{\pi}{4}} \tan x d x\right] \ldots \ldots \ldots . \text { (i) }
$$
Let $\mathrm{I}_1=\int_0^{\frac{\pi}{4}}\left(\tan x \cdot \sec ^2 x\right) d x$
Putting $\tan x=t$
$
\Rightarrow \sec ^2 \theta d x=d t
$
Limits of integration when $x=0, t=\tan 0=0$ and when $x=\frac{\pi}{4}, t=\tan \frac{\pi}{4}=1$
$
\begin{aligned}
& \therefore \mathrm{I}_1=\int_0^1 t d t \\
& =\left(\frac{t^2}{2}\right)_0^1 \\
& =\frac{1}{2}-0=\frac{1}{2}
\end{aligned}
$
Putting value of $\mathrm{I}_1$ in eq. (i),
$
\begin{aligned}
& \mathrm{I}=2\left[\frac{1}{2}-\int_0^{\frac{\pi}{4}} \tan x d x\right] \\
& =2\left[\frac{1}{2}-(\log \sec x)_0^{\frac{\pi}{2}}\right] \\
& =1-2\left(\log \sec \frac{\pi}{4}-\log \sec 0\right)
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=1-2(\log \sqrt{2}-\log 1) \\
& =1-2\left(\log 2^{\frac{1}{2}}-0\right) \\
& =1-2\left(\frac{1}{2} \log 2\right) \\
& =1-\log 2 \text { = LHS Hence Proved }
\end{aligned}$
Miscellaneous Exercise Question 37.
$\int_0^1 \sin ^{-1} x d x=\frac{\pi}{2}-1$
Answer.
Let $\mathrm{I}=\int_0^1 \sin ^{-1} x d x$
Putting $x=\sin \theta$
$
\Rightarrow d x=\cos \theta d \theta
$
Limits of integration when $x=0, \theta=0$ and when $x=1, \sin \theta=1$, i.e., $\theta=\frac{\pi}{2}$
$
\begin{aligned}
& \therefore I=\int_0^1 \sin ^{-1} x d x \\
& =\int_0^1 \theta \cos \theta d \theta
\end{aligned}
$
[Integrating by parts]
$
=[\theta \sin \theta]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 1 \cdot \sin \theta d \theta
$
$\begin{aligned}
& \Rightarrow \mathrm{I}=\left(\frac{\pi}{2}-0\right)+[\cos \theta]_0^{\frac{\pi}{2}} \\
& =\frac{\pi}{2}+\left(\cos \frac{\pi}{2}-\cos 0\right) \\
& =\frac{\pi}{2}+(0-1) \\
& =\frac{\pi}{2}-1 \text { = LHS Hence proved }
\end{aligned}$
Miscellaneous Exercise Question 38.
Choose the correct answer:
$\int \frac{d x}{e^x+e^{-x}}$ is equal to:
(A) $\tan ^{-1}\left(e^x\right)+c$
(B) $\tan ^{-1}\left(e^{-x}\right)+c$
(C) $\log \left(e^x-e^{-x}\right)+c$
(D) $\log \left(e^x+e^{-x}\right)+c$
Answer.
Let $\mathrm{I}=\int \frac{d x}{e^x+e^{-x}}$
$=\int \frac{1}{e^x+\frac{1}{e^x}} d x$
$=\int \frac{1}{\left(\frac{e^{2 x}+1}{e^x}\right)} d x$
$=\int \frac{e^x}{e^{2 x}+1} d x$
Putting $e^x=t$
$\Rightarrow e^x d x=d t$
$
\begin{aligned}
& \therefore \text { From eq. (i), I }=\int \frac{d t}{t^2+1} \\
& =\tan ^{-1} t+c \\
& =\tan ^{-1} e^x+c
\end{aligned}
$
Therefore, option (A) is correct.
Miscellaneous Exercise Question 39.
Choose the correct answer:
$\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x$ is equal to:
(A) $\frac{-1}{\sin x+\cos x}+c$
(B) $\log |\sin x+\cos x|+c$
(C) $\log |\sin x-\cos x|+c$
(D) $\frac{1}{(\sin x+\cos x)^2}$
Answer.
Let $\mathrm{I}=\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x$
$
\begin{aligned}
& =\int \frac{\cos ^2 x-\sin ^2 x}{(\sin x+\cos x)^2} d x \\
& =\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\sin x+\cos x)(\sin x+\cos x)} d x \\
& \Rightarrow I=\int \frac{\cos x-\sin x}{\sin x+\cos x} d x
\end{aligned}
$
$
=\log |\sin x+\cos x|+c
$
Therefore, option (B) is correct.
Miscellaneous Exercise Question 40.
Choose the correct answer:
If $f(a+b-x)=f(x)$, then $\int_a^b x f(x) d x$ is equal to:
(A) $\frac{a+b}{2} \int_a^b f(b-x) d x$
(B) $\frac{a+b}{2} \int_a^b f(b+x) d x$
(C) $\frac{b-a}{2} \int_a^b f(x) d x$
(D) $\frac{a+b}{2} \int_a^b f(x) d x$
Answer.
Given: $f(a+b-x)=f(x)$ $\qquad$
Let $\mathrm{I}=\int_a^b x f(x) d y$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_a^b(a+b-x) f(a+b-x) d x \\
& =\int_a^b(a+b-x) f(x) d x \ldots \ldots \ldots . \text { (iii) }
\end{aligned}
$
Adding eq. (ii) and (iii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_a^b(x+a+b-x) f(x) d x \\
&=\int_a^b(a+b) f(x) d x \\
&=(a+b) \int_a^b f(x) d x \\
& \Rightarrow \mathrm{I}=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x
\end{aligned}
$
Therefore, option (D) is correct.
