RD Sharma Class 12 Exercise 19.2 Definite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 19.2 Definite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:22 PM IST

The RD Sharma books have gained popularity and are widely used by class 12 students for their reference purposes. However, chapter 19, the Definite Integrals portion, is always a challenging one for the students when it comes to mathematics. Therefore, to save the students’ time solving and rechecking the sums in the second exercise, Ex 19.2, the RD Sharma Class 12th Exercise 19.2 book will lend a helping hand.

## Definite Integrals Excercise:19.2

Definite integrals exercise 19.2 question 2

$\frac{\log 2}{\log 2e}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$
Solution:$\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$
Put $1+\log x=t$
$\frac{1}{x}dx=dt$
$dx=xdt$
When x=1 then t=1 and when x=2 then t=1+log2
\begin{aligned} &=\int_{1}^{2} \frac{1}{x(1+\log x)^{2}} d x=\int_{1}^{1+\log 2} \frac{1}{x t^{2}} x d t=\int_{1}^{1+\log 2} \frac{1}{t^{2}} d t \\ &=\int_{1}^{1+\log 2} t^{-2} d t=\left[\frac{t^{-2+1}}{-2+1}\right]_{1}^{1+\log 2} \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \end{aligned}
\begin{aligned} &=\left[\frac{t^{-1}}{-1}\right]_{1}^{1+\log 2}=-\left[\frac{1}{t}\right]_{1}^{1+\log 2} \\ &=-\left[\frac{1}{1+\log 2}-1\right]=-\left[\frac{1-1-\log 2}{1+\log 2}\right] \\ &=-\left[\frac{-\log 2}{1+\log 2}\right]=\frac{\log 2}{1+\log 2} \quad[1=\log e] \end{aligned}
\begin{aligned} &=\frac{\log 2}{\log e+\log 2} \\ &=\frac{\log 2}{\log 2 e} \end{aligned}

Definite integrals exercise 19.2 question 3

$\frac{1}{6}\log\left ( \frac{35}{8} \right )$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{1}^{2}\frac{3x}{9x^2-1}\: dx$
Solution: $\int_{1}^{2}\frac{3x}{9x^2-1}\: dx$
Putting $9x^2-1=t$
$\Rightarrow 18x \; dx=dt$
$\Rightarrow dx=\frac{dt}{18x}$
When x=1 then t=9-1=8 and when x=2 then t=36-1=35
\begin{aligned} &=\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x=\int_{8}^{35} \frac{3 x}{t} \frac{d t}{18 x} \\ &=\frac{1}{6} \int_{8}^{35} \frac{1}{t} d t \\ &=\frac{1}{6}[\log |t|]_{8}^{35} \\ &=\frac{1}{6}[\log 35-\log 8] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a-\log b=\log \frac{a}{b}\right] \end{aligned}
$=\frac{1}{6}\log\left ( \frac{35}{8} \right )$

Definite integrals exercise 19.2 question 4

$I =\frac{1}{\sqrt{34}}\left [ \tan^{-1} \frac{5\sqrt{34}}{28}\right ]$
Hint: We use indefinite formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}} \frac{1}{5 \cos x+3 \sin x} d x$
Solution:$\int_{0}^{\frac{\pi}{2}} \frac{1}{5 \cos x+3 \sin x} d x$
Putting $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan 2 \frac{x}{2}}$
$\begin{gathered} \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5\left(1-\tan ^{2} \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}+\frac{3\left(2 \tan \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}} d x \end{gathered}$
$=\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5-5 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right]$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x$
Let$\tan \frac{x}{2}=t$
$\frac{1}{2}\sec^2 \frac{x}{2}dx=dt$
$\sec^2 \frac{x}{2}dx=dt$
When x=0 then $t=\tan0=0$ and when$x=\frac{\pi}{2}$ then $t=\tan \frac{\pi}{4}=1$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+6 \tan \frac{x}{2}-5 \tan \frac{2}{2}} d x$
\begin{aligned} &=\int_{0}^{1} \frac{1}{5+6 t-5 t^{2}} 2 d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{1+\frac{6}{5} t-t^{2}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left(t^{2}-\frac{6}{5} t-1\right)} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[t^{2}-2 t \frac{3}{5}+\left(\frac{3}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}-1\right]} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\frac{9}{25}-1\right]} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\left(\frac{9+25}{25}\right)\right]} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{-\left[\left(t-\frac{3}{5}\right)^{2}-\left(\frac{34}{25}\right)\right]} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(\frac{\sqrt{34}}{5}\right)^{2}+\left(t-\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(\frac{\sqrt{34}}{5}\right)^{2}+\left(t-\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5}\left[\frac{1}{\frac{\sqrt{34}}{5}} \tan ^{-1} \frac{\left(t-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}\right]_{0}^{1} \\ &{\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]} \end{aligned}
\begin{aligned} &=\frac{2}{5} \frac{5}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(1-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}-\tan ^{-1} \frac{\left(0-\frac{3}{5}\right)}{\left(\frac{\sqrt{34}}{5}\right)}\right] \\ &=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{(2)}{(\sqrt{34})}-\tan ^{-1} \frac{(-3)}{(\sqrt{34})}\right] \end{aligned}
$=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(\frac{(2)}{(\sqrt{34})}-\frac{(-3)}{(\sqrt{34})}\right)}{1+\frac{(2)}{(\sqrt{34})} \frac{(-3)}{(\sqrt{34})}}\right]$
\begin{aligned} &=\frac{1}{\sqrt{34}}\left[\tan ^{-1} \frac{\left(\frac{(2+3)}{(\sqrt{34})}\right)}{\frac{34-6}{34}}\right] \\ &=\frac{1}{\sqrt{34}} \mid \tan ^{-1} \frac{\frac{5}{(\sqrt{34})}}{\frac{28}{34}} \end{aligned}
$I =\frac{1}{\sqrt{34}}\left [ \tan^{-1} \frac{5\sqrt{34}}{28}\right ]$

Definite integrals exercise 19.2 question 5

$a(\sqrt{2}-1)$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$
Solution:$\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$
Putting $a^2+x^2=t^2$
$2x dx=2t dt$
$x\; \; dx=t\; \; dt$
When x=0 then t=a and when x=a then t=2a
\begin{aligned} &=\int_{a}^{\sqrt{2} a} \frac{1}{\sqrt{t^{2}}} t d t \\ &=\int_{a}^{\sqrt{2} a} \frac{1}{t} t d t \\ &=\int_{a}^{\sqrt{2} a} 1 d t \\ &=\int_{a}^{\sqrt{2} a} t^{0} d t \\ &=\left[\frac{t^{0+1}}{0+1}\right]_{a}^{\sqrt{2} a} \end{aligned}
$=[t]^{\sqrt{2}a}_{a}$
$=[\sqrt{2}a-a]$
$=a[\sqrt{2}-1]$

Definite integrals exercise 19.2 question 6

$\tan ^{-1}e-\frac{\pi}{4}$
Hint: We use indefinite formula then put limits to solve this integral.
Given: $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$
Solution:$\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$
Put $e^x=t$
$e^x dx=dt$
When x=0 then t=1 and when x=1 then t=e
\begin{aligned} &I=\int_{1}^{e} \frac{1}{1+t^{2}} d t \\ &=\left[\tan ^{-1} t\right]_{1}^{e} \\ &=\tan ^{-1} e-\tan ^{-1} 1\; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}\right] \end{aligned}
$=\tan ^{-1}e-\frac{\pi}{4}$

Definite integrals exercise 19.2 question 7

$\frac{1}{2}(e-1)$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{1} x e^{x^{2}} d x$
Solution:$I=\int_{0}^{1} x e^{x^{2}} d x$
Put $x^2=t$
$2x\; \; dx=dt$
$dx=\frac{dt}{2x}$
When x=0 then t=0 and when x=1 then t=1
\begin{aligned} &I=\int_{0}^{1} x e^{t} \frac{d t}{2 x} \\ &=\frac{1}{2} \int_{0}^{1} e^{t} d t \end{aligned}
$=\frac{1}{2}[e^t]_0^1$ $[\int e^t\; dt =e^t+c]$
$=\frac{1}{2}[e^1-e^0]$
$=\frac{1}{2}(e-1)$

Definite integrals exercise 19.2 question 8

$\sin (\log 3)$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{1}^{3}\frac{\cos (\log x)}{x}dx$
Solution:$\int_{1}^{3}\frac{\cos (\log x)}{x}dx$
Put $\log x=t$
$\frac{1}{x} dx=dt$
$dx=x dt$
When x=1 then t=log1=0 and when x=3 then t=log3
\begin{aligned} &=\int_{0}^{\log 3} \frac{\cos t}{x} x d t \\ &=\int_{1}^{\log 3} \cos t d t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \cos x d x=\sin x+c\right] \end{aligned}
\begin{aligned} &=[\sin t]_{0}^{\log 3} \\ &=[\sin (\log 3)-\sin 0] \\ &=\sin (\log 3) \end{aligned}

Definite Integrals exercise 19.2 question 9.

$\frac{\pi}{4}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{1}\frac{2x}{1+x^4}dx$
Solution:$I=\int_{0}^{1}\frac{2x}{1+x^4}dx$
Put $x^2=t$
$2x\; \; dx=dt$
When x=0 then t=0 and
when x=1 then t=1
$I=\int_{0}^{1}\frac{1}{t^2+1}dt$ $\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} \frac{x}{a}\right]$
\begin{aligned} &=\left[\tan ^{-1}\left(\frac{t}{1}\right)\right]_{0}^{1} \mid \\ &=\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\ &=\frac{\pi}{4}-0=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise 19.2 question 10.

Answer: $\frac{\pi a^2}{4}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{a}\sqrt{a^2-x^2}dx$
Solution: $\int_{0}^{a}\sqrt{a^2-x^2}dx$
Put $x=a\sin \theta$
$dx=a\cos \theta \; d\theta$
When x=0 then $\theta=0$ and
when x=a then $\theta=\frac{\pi}{2}$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}-a^{2} \sin ^{2} \theta} a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}\left(1-\sin ^{2} \theta\right)} a \cos \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\sin ^{2} x=\cos ^{2} x\right] \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} a \sqrt{\cos ^{2} \theta} a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} a \cos \theta a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} a^{2} \cos ^{2} \theta d \theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos ^{2} \theta=\frac{1+\cos 2 \theta}{2}\right] \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} a^{2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \frac{a^{2}}{2}(1+\cos 2 \theta) d \theta \\ &=\frac{a^{2}}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta \end{aligned}
\begin{aligned} &=\frac{a^{2}}{2}\left(\int_{0}^{\frac{\pi}{2}} 1 d \theta+\int_{0}^{\frac{\pi}{2}} \cos 2 \theta d \theta\right) \\ &=\frac{a^{2}}{2}[\theta]_{0}^{\frac{\pi}{2}}+\frac{a^{2}}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{a^{2}}{2}\left[\frac{\pi}{2}-0\right]+\frac{a^{2}}{4}\left[\sin \frac{2 \pi}{2}-\sin 0\right] \end{aligned}
\begin{aligned} &=\frac{a^{2}}{2} \frac{\pi}{2}+\frac{a^{2}}{4}[\sin \pi-\sin 0] \\ &=\frac{a^{2} \pi}{4}+\frac{a^{2}}{4}[0-0] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\sin \pi=\sin 0=0] \end{aligned}
$=\frac{\pi a^2}{4}$

Definite Integrals exercise 19.2 question 11.

Answer: $\frac{64}{231}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi$
Solution:$\int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi$
Put $\sin \phi=t$
$\cos \phi d\phi =dt$
$d\phi =\frac{dt}{cos\phi}$
When $\phi =0$ then $t=0$ and when $\phi =\frac{\pi}{2}$ then $t=1$
\begin{aligned} &=\int_{0}^{1} \sqrt{t} \cos ^{5} \phi \frac{d t}{\cos \phi} \\ &=\int_{0}^{1} \sqrt{t} \cos ^{4} \phi d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(\cos ^{2} \phi\right)^{2} d t \end{aligned}
\begin{aligned} &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-\sin ^{2} \phi\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-t^{2}\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t \end{aligned}
\begin{aligned} &=\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{4+\frac{1}{2}}-2 t^{2+\frac{1}{2}}\right) d t \\ &=\int_{0}^{1} t^{\frac{1}{2}} d t+\int_{0}^{1} t^{\frac{9}{2}} d t-2 \int_{0}^{1} t^{\frac{5}{2}} d t \end{aligned}
\begin{aligned} &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2 t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1} \\ &=\left[\frac{t^{\frac{3}{2}}}{3}\right]_{0}^{1}+\left[\frac{t^{\frac{11}{2}}}{\frac{11}{2}}\right]_{0}^{1}-2\left[\frac{t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{11}\left[t^{\frac{11}{2}}\right]_{0}^{1}-2 \times \frac{2}{7}\left[t^{\frac{7}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]+\frac{2}{11}\left[1^{\frac{11}{2}}-0^{\frac{11}{2}}\right]-\frac{4}{7}\left[1^{\frac{7}{2}}-0^{\frac{7}{2}}\right] \\ &=\frac{2}{3}[1-0]+\frac{2}{11}[1-0]-\frac{4}{7}[1-0] \end{aligned}
$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{154+42-132}{231}$
$=\frac{196-132}{231}$
$=\frac{64}{231}$

Definite Integrals exercise 19.2 question 12.

$\frac{\pi}{4}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin ^2x}dx$
Solution: $\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin ^2x}dx$
Put $\sin x=t$
$\cos x \; \; dx=dt$
$dx=\frac{dt}{\cos x}$
When x=0 then t=0 and when $x=\frac{\pi}{2}$ then t=1
$=\int_{0}^{1} \frac{1}{1+t^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]$
$=[\tan ^{-1}(t)]^{1}_0$
$=[\tan ^{-1}1-\tan ^{-1}0]$
$=\left [\frac{\pi}{4}-0 \right ]$
$= \frac{\pi}{4}$

Definite Integrals exercise 19.2 question 13.

Answer: $2(\sqrt{2}-1)$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta}{\sqrt{1+\cos \theta}}d \theta$
Solution: $I=\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta}{\sqrt{1+\cos \theta}}d \theta$
Put $1+cos\theta =t^2$
$-\sin \theta \; \; d\theta=2t\; dt$
When $\theta=0$ then $t=2$ and
When $\theta=\frac{\pi}{2}$ then $t=1$
\begin{aligned} I &=\int_{\sqrt{2}}^{1} \frac{-2 t d t}{\sqrt{t^{2}}} d t \\ &=-\int_{\sqrt{2}}^{1} \frac{2 t}{t} d t \\ &=-2 \int_{\sqrt{2}}^{1} 1 d t \\ &=-2[t]_{\sqrt{2}}^{1} \\ &=-2[1-\sqrt{2}] \\ &=2(\sqrt{2}-1) \end{aligned}

Definite Integrals exercise 19.2 question 14.

Answer: $\frac{1}{4}\log \left ( \frac{3+2\sqrt{3}}{3} \right )$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\frac{\pi}{3}}\frac{\cos x }{3+4 \sin x}dx$
Solution:$I=\int_{0}^{\frac{\pi}{3}}\frac{\cos x }{3+4 \sin x}dx$
Put $3+4\sin x=t$
$4\cos x \; dx=dt$
$\cos x\; \; dx=\frac{dt}{4}$
When x=0 then t=3 and
when $x=\frac{\pi}{3}$ then $t=3+2\sqrt{3}$
\begin{aligned} &I=\int_{3}^{3+2 \sqrt{3}} \frac{1}{t} \frac{d t}{4} \\ &I=\frac{1}{4} \int_{3}^{3+2 \sqrt{3}} \frac{1}{t} d t \\ &=\frac{1}{4}[\log |t|]_{3}^{3+2 \sqrt{3}} \\ &=\frac{1}{4}[\log (3+2 \sqrt{3})-\log 3] \\ &=\frac{1}{4} \log \left(\frac{3+2 \sqrt{3}}{3}\right) \end{aligned}

Definite Integrals exercise 19.2 question 15.

Answer: $\frac{\pi \sqrt{\pi}}{12}$
Hint: We use indefinite formula then put limits to solve this integral.
Given: $\int_{0}^{1}\frac{\sqrt{\tan ^{-1}x} }{1+x^2}dx$
Solution: $I=\int_{0}^{1}\frac{\sqrt{\tan ^{-1}x} }{1+x^2}dx$
Put $\tan^{-1}x=t$
$\frac{1}{1+x^2}dx=dt$
$dx=(1+x^2)dt$
When x=0 then t=0 and
when x=1 then $t =\frac{\pi}{4}$
\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \frac{\sqrt{t}}{1+x^{2}}\left(1+x^{2}\right) d t \\ &=\int_{0}^{\frac{\pi}{4}} \sqrt{t} d t \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} t^{\frac{1}{2}} d t \\ &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\frac{\pi}{4}} \\ &=\left[\frac{t^{\frac{3}{2}}}{3}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{2}{3}\left[\left(\frac{\pi}{4}\right)^{\frac{3}{2}}-0\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\frac{\pi^{\frac{3}{2}}}{2^{2 \times \frac{3}{2}}}\right] \\ &=\frac{2}{3}\left[\frac{\pi \sqrt{\pi}}{2^{3}}\right] \\ &=\frac{2 \pi \sqrt{\pi}}{3 \times 8} \\ &=\frac{\pi \sqrt{\pi}}{12} \end{aligned}

Definite Integrals exercise 19.2 question 16.

Answer: $\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{2}x\sqrt{x+2}dx$
Solution:$I=\int_{0}^{2}x\sqrt{x+2}dx$
Put $x+2 =t^2$
$dx=2t\; \; dt$
When x=0 then t=2 and
when x=2 then t=2
\begin{aligned} &I=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t \\ &=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t 2 t d t \end{aligned}
\begin{aligned} &=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) 2 t^{2} d t \\ &=2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t \\ &=2 \int_{\sqrt{2}}^{2} t^{4} d t-2 \times 2 \int_{\sqrt{2}}^{2} t^{2} d t \\ &=2\left[\frac{t^{4+1}}{4+1}\right]_{\sqrt{2}}^{2}-4\left[\frac{t^{2+1}}{2+1}\right]_{\sqrt{2}}^{2} \end{aligned}
\begin{aligned} &=\frac{2}{5}\left[t^{5}\right]_{\sqrt{2}}^{2}-\frac{4}{3}\left[t^{3}\right]_{\sqrt{2}}^{2} \\ &=\frac{2}{5}\left[2^{5}-(\sqrt{2})^{5}\right]-\frac{4}{3}\left[2^{3}-(\sqrt{2})^{3}\right] \\ &=\frac{2}{5}[32-2 \times 2 \times \sqrt{2}]-\frac{4}{3}[8-2 \sqrt{2}] \\ &=2\left[\frac{32}{5}-\frac{4 \sqrt{2}}{5}\right]-\frac{4 \times 8}{3}+\frac{8 \sqrt{2}}{3} \end{aligned}
\begin{aligned} &=2\left[\left(\frac{32}{5}-\frac{4 \sqrt{2}}{5}\right)-\frac{16}{3}+\frac{4 \sqrt{2}}{3}\right] \\ &=2\left[\frac{3(32-4 \sqrt{2})-5(16-4 \sqrt{2})}{15}\right] \\ &=2\left[\frac{96-12 \sqrt{2}-80+20 \sqrt{2}}{15}\right] \\ &=2\left[\frac{16+8 \sqrt{2}}{15}\right] \\ &=2 \times \frac{8}{15}[2+\sqrt{2}] \\ &=\frac{16 \sqrt{2}}{15}(\sqrt{2}+1) \end{aligned}

Definite Integrals exercise 19.2 question 17.

Answer:$\frac{\pi}{2}-\log 2$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx$
Solution:$I=\int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx$
Put $x=\tan \theta$
$dx=\sec^2\theta \; d\theta$
When x=0 then $\theta=0$ and
when x=1 then $\theta=\frac{\pi}{4}$
\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}(\tan 2 \theta) \sec ^{2} \theta d \theta \\ &=\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta \end{aligned}
Applying integration by parts method, then
\begin{aligned} &=2\left\{\left[\theta \int \sec ^{2} \theta d \theta\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left[\frac{d(\theta)}{d \theta} \int \sec ^{2} \theta d \theta\right] d \theta\right\} \\ &=2\left\{[\theta \tan \theta]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} 1 \tan \theta d \theta\right\} \quad\left[\int \sec ^{2} x d x=\tan x, \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\ &=2\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-0 \times \tan 0\right]-\int_{0}^{\frac{\pi}{4}} \tan \theta d \theta\right\} \quad\left[\int \tan x d x=-\log |\cos x|, \tan \frac{\pi}{4}=1, \tan 0=0\right] \end{aligned}
\begin{aligned} &=2\left\{\left[\frac{\pi}{4} \cdot 1-0\right]-[-\log |\cos \theta|]_{0}^{\frac{\pi}{4}}\right\} \\ &=\frac{2 \pi}{4}+2\left[\log \frac{1}{\sqrt{2}}-\log 1\right] \; \; \; \; \; \; \; \: \: \: \: \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \\ &=\frac{\pi}{2}+2\left[\log \frac{1}{\sqrt{2}}-0\right] \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log 1=0] \end{aligned}
\begin{aligned} &=\frac{\pi}{2}+2 \log (2)^{-\frac{1}{2}} \\ &=\frac{\pi}{2}+\left(-\frac{1}{2}\right) \times 2 \log 2 \quad\left[\log a^{m}=m \log a\right] \\ &=\frac{\pi}{2}-\log 2 \end{aligned}

Definite Integrals exercise 19.2 question 18.

Answer:$\frac{\pi}{8}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sin^4 x}dx$
Solution:$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sin^4 x}dx$
Put $\sin2x=t$
$2\sin x \cos x=dt$
$\sin x \cos xdx=\frac{dt}{2}$
When x=0 then t=0 and when $x=\frac{\pi}{2}$ then t=1
\begin{aligned} &I=\int_{0}^{1} \frac{1}{1+t^{2}} \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{1+t^{2}} d t \\ &=\frac{1}{2}\left[\tan ^{-1}(t)\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}-0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4} \\ &=\frac{\pi}{8} \end{aligned}

Answer:$\frac{ 2}{3}\tan^{-1}\left (\frac{1}{13} \right )$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
Solution:$I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
Put $\sin x=\frac{2\tan \frac{x}{2} }{1+\tan ^2\frac{x}{2}}$

$I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1}{5+\frac{4\left(2 \tan \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \end{aligned}
$=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1+\tan ^{2} x=\sec ^{2} x\right]$
Put $\tan \frac{x}{2}=t$
$\sec^ 2\frac{x}{2} .\frac{1}{2}dx=dt$
$\sec ^2\frac{x}{2} dx=2t$
When x=0 then t=0 and when $x=\frac{\pi}{2}$ then t=1
\begin{aligned} &l=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \\ &=\int_{0}^{1} \frac{1}{5+5 t^{2}+8 t} 2 d t \\ &=\int_{0}^{1} \frac{1}{5\left(1+t^{2}+\frac{8 t}{5}\right)} 2 d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{1+t^{2}+\frac{8 t}{5}} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{t^{2}+2 t \frac{4}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}-\frac{16}{25}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{25-16}{25}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{9}{25}} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5}\left[\frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &=\frac{2}{5} \times \frac{5}{3}\left[\tan ^{-1}\left(\frac{\frac{5 t+4}{5}}{\frac{3}{5}}\right)\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5 \times 1+4}{3}\right)-\tan ^{-1}\left(\frac{5 \times 0+4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{3-\frac{4}{3}}{1+3 \times \frac{4}{3}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} A-\tan ^{-1} B=\tan ^{-1}\left(\frac{A-B}{1+A B}\right)\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{\frac{9-4}{3}}{1+4}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5}{3 \times 5}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)\right] \end{aligned}

Definite Integrals exercise 19.2 question 21.

Answer:$\frac{\pi}{2}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x +\cos x}dx$
Solution:$\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x +\cos x}dx$
\begin{aligned} &\text { Put } \sin x=k(\sin x+\cos x)+L \frac{d}{d x}(\sin x+\cos x) \; \; \; \; \; \; \; \; \; \quad\left[\frac{d}{d x} \sin x=\cos x, \frac{d}{d x} \cos x=-\sin x\right] \\ &\sin x=k \sin x+k \cos x+L(\cos x-\sin x) \\ &\sin x=k \sin x+k \cos x+L \cos x-L \sin x \end{aligned}
$\sin x(k-L)\sin x+(k+L)\cos x$
Equating coefficient of sinx and cosx respectively, we get
$k-L=1............(i)$
$k+L=1............(ii)$
Solving equation (i) and (ii)
k-L=1
k+L=0
2k=1
k=12
ii=k+L=0
12+L=0
L=-12
\begin{aligned} &\therefore \sin x=\frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x) \\ &I=\int_{0}^{\pi} \frac{\sin x}{\sin x+\cos x} d x \\ &=\int_{0}^{\pi \frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x)}{\sin x+\cos x} d x \\ &=\int_{0}^{\pi \frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x)}{\sin x+\cos x} d x \\ &=\frac{1}{2} \int_{0}^{\pi}\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_{0}^{\pi} 1 d x-\frac{1}{2} \int_{0}^{\pi} \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x \end{aligned}
Put $\sin x+\cos x=t$
\begin{aligned} &(\cos x-\sin x) d x=d t \\ &I=\int_{0}^{\pi} \frac{1}{2} d x-\frac{1}{2} \int_{0}^{\pi} \frac{d t}{t} \\ &=\frac{1}{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\pi}-\frac{1}{2}[\log |t|]_{0}^{\pi} \\ &=\frac{1}{2}[x]_{0}^{\pi}-\frac{1}{2}[\log |\sin x+\cos x|]_{0}^{\pi} \end{aligned}
\begin{aligned} &=\frac{1}{2}[\pi-0]-\frac{1}{2}[\log |\sin \pi+\cos \pi|-\log |\sin 0+\cos 0|] \\ &=\frac{1}{2} \pi-\frac{1}{2}[\log |0+(-1)|-\log |0+1|] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\cos \pi=-1, \cos 0=1, \sin \pi=\sin 0=0] \end{aligned}
\begin{aligned} &=\frac{1}{2} \pi-\frac{1}{2}[\log |-1|-\log |1|] \\ &=\frac{1}{2} \pi-\frac{1}{2}[\log |1|-\log |1|] \quad[\log 1=0] \\ &=\frac{1}{2} \pi-0 \\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise 19.2 question 22.

Answer: $\frac{\pi}{4}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$
Solution:$I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$
Put $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}, \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$
$=\int_{0}^{\pi} \frac{1}{3+2 \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$
$=\int_{0}^{\pi} \frac{1}{\frac{3\left(1+\tan ^{2} \frac{x}{2}\right)+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$
\begin{aligned} &=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{3+3 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \\ &=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{4+2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right] \end{aligned}
Put $\tan \frac{x}{2}=t$
$\sec^2\frac{x}{2}.\frac{1}{2}dx=dt$
$\sec^2\frac{x}{2}dx=2dt$
When x=0 then t=0 and when $x =\pi$ then $t=\infty$
\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{4+2 t^{2}+4 t} 2 d t \\ &=2 \times \frac{1}{2} \int_{0}^{\infty} \frac{1}{2+t^{2}+2 t} d t \\ &=\int_{0}^{\infty} \frac{1}{t^{2}+2 t+1+1} d t \\ &=\int_{0}^{\infty} \frac{1}{\left(t^{2}+2 t+1\right)+1} d t \\ &=\int_{0}^{\infty} \frac{1}{(t+1)^{2}+1^{2}} d t \end{aligned}
\begin{aligned} &=\frac{1}{1}\left[\tan ^{-1}(t+1)\right]_{0}^{\infty} \\ &=\tan ^{-1}(\infty+1)-\tan ^{-1}(0+1) \\ &=\tan ^{-1} \infty-\tan ^{-1} 1 \\ &=\frac{\pi}{2}-\frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 1=\frac{\pi}{4}\right] \\ &=\frac{2 \pi-\pi}{4} \\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise 19.2 question 23.

Answer:$\frac{\pi}{4}-\frac{1}{2}\log 2$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{1}\tan ^{-1}x dx$
Solution:$I=\int_{0}^{1}\tan ^{-1}x dx$
Applying integration by parts method we get
$I=\int_{0}^{1}\tan ^{-1}x dx$
\begin{aligned} &=\left[\tan ^{-1} x \int 1 d x\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d x}\left(\tan ^{-1} x\right) \int 1 d x\right] d x \\ &=\left[\tan ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} x d x \quad\left[\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}\right] \end{aligned}
Put $1+x^2=t$

$2x \; \; dx=dt$
$x \; dx=\frac{dt}{2}$
Then
\begin{aligned} &I=\left[\tan ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{-1} \frac{1 d t}{2} \\ &=\left[1 \cdot \tan ^{-1} 1-0 \cdot \tan ^{-1} 0\right]-\frac{1}{2}[\log |t|]_{0}^{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log |x|\right] \\ &=\left[\frac{\pi}{4}-0\right]-\frac{1}{2}\left[\log \left|1+x^{2}\right|\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\frac{\pi}{4}-\frac{1}{2}\left[\log \left|1+1^{2}\right|-\log \left|1+0^{2}\right|\right] \\ &=\frac{\pi}{4}-\frac{1}{2}[\log (1+1)-\log 1] \\ &=\frac{\pi}{4}-\frac{1}{2}[\log 2-\log 1] \end{aligned}
$=\frac{\pi}{4}-\frac{1}{2}[\log 2-0]$ $[\log 1 =0]$
$=\frac{\pi}{4}-\frac{1}{2}[\log 2]$

Definite Integrals exercise 19.2 question 24.

Answer:$\frac{1}{2}-\frac{\sqrt{3}}{12}\pi$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$I =\int_{0}^{\frac{1}{2}}\frac{x \sin^{-1}x}{\sqrt{1-x^2}}dx$
Solution:$I =\int_{0}^{\frac{1}{2}}\frac{x \sin^{-1}x}{\sqrt{1-x^2}}dx$
Applying integration by parts method we get
$=\left[\sin ^{-1} x \int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^{2}}} d x\right]_{0}^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}}\left(\frac{d}{d x} \sin ^{-1} x \int \frac{x}{\sqrt{1-x^{2}}} d x\right) d x$
In $\int \frac{x}{\sqrt{1-x^2}} dx$ put $1-x^2=t^2$
\begin{aligned} &\rightarrow-2 x d x=2 t d t \\ &\rightarrow x d x=-t d t \\ &\int \frac{x}{\sqrt{1-x^{2}}} d x=-\int \frac{1}{\sqrt{t^{2}}} t d t \\ &=-\int \frac{1}{t} \cdot t d t \\ &=-\int 1 d t \end{aligned}
$=\int t^0 dt$ $\left [ \int x^n dx=\frac{x^n+1}{n+1}+c \right ]$
$=-\left [ \frac{t^0+1}{0+1} \right ]$
$=-t$
\begin{aligned} &\int \frac{x}{\sqrt{1-x^{2}}} d x=-t=-\sqrt{1-x^{2}} \quad \ldots(i) \quad\left[t^{2}=1-x^{2}\right] \\ &\left.I=\left[\sin ^{-1} x \cdot-\sqrt{1-x^{2}}\right]_{0}^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}}\left(\frac{1}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right)\right) d x \quad \text { [using }(i)\right]\left[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right] \end{aligned}
\begin{aligned} &=-\left[\sqrt{1-x^{2}} \cdot \sin ^{-1} x\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}} 1 d x \\ &=-\left[\sqrt{1-\left(\frac{1}{2}\right)^{2}} \cdot \sin ^{-1} \frac{1}{2}-\sqrt{1-0^{2}} \cdot \sin ^{-1} 0\right]+[x]_{0}^{\frac{1}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int 1 d x=x\right] \end{aligned}
\begin{aligned} &=-\left[\sqrt{1-\frac{1}{4}} \cdot \sin ^{-1} \frac{1}{2}-0\right]+\left[\frac{1}{2}-0\right]\\ &\left[\sin ^{-1} 0=0\right]\\ &=-\left[\sqrt{\frac{4-1}{4}} \frac{\pi}{6}\right]+\left[\frac{1}{2}\right] \quad\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right] \end{aligned}
\begin{aligned} &=-\frac{\sqrt{3}}{2} \frac{\pi}{6}+\frac{1}{2} \\ &=\frac{1}{2}-\frac{\sqrt{3} \pi}{12} \end{aligned}

Definite Integrals exercise 19.2 question 25

Answer:$\frac{\pi}{2}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{4}}\left ( \sqrt{\tan x} + \sqrt{\cot x} \right )dx$
Solution:$I=\int_{0}^{\frac{\pi}{4}}\left ( \sqrt{\tan x} + \sqrt{\cot x} \right )dx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{4}}\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\operatorname{cox} x}{\sin x}}\right) d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{\sin x} \sqrt{\cos x}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{2 \sin x \cdot \cos x}} d x \end{aligned}
\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-1+2 \sin x \cdot \cos x}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cdot \cos x}} d x \end{aligned}
\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cdot \cos x\right)}} d x \\ &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^{2}}} d x \end{aligned}

Put $\sin x-\cos x=t$
$\Rightarrow \left [ \cos x -(-\sin x) \right ]dx=dt$
$\Rightarrow \left ( \cos x + \sin x \right )dx=dt$
When x=0 then t=-1 and when $x=\frac{\pi}{4}$ then t=0
\begin{aligned} &I=\sqrt{2} \int_{-1}^{0} \frac{1}{\sqrt{1-t^{2}}} d t \\ &=\sqrt{2}\left[\sin ^{-1} t\right]_{-1}^{0} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\right] \end{aligned}
\begin{aligned} &=\sqrt{2}\left[\sin ^{-1} 0-\sin ^{-1}(-1)\right] \\ &=\sqrt{2}\left[0-\sin ^{-1}(-1)\right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{-1} 0=0\right] \\ &=\sqrt{2}\left[-\left(-\frac{\pi}{2}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{-1}(-1)=-\frac{\pi}{2}\right] \\ &=\sqrt{2}\left[\frac{\pi}{2}\right] \\ &=\frac{\pi}{\sqrt{2}} \end{aligned}

Definite Integrals exercise 19.2 question 26

Answer: $\frac{1}{8}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{1+\cos ^22x}dx$
Solution: $I=\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{1+\cos ^22x}dx$
$=\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{2\cos ^2x}dx$ $\left [ 1+\cos 2\theta =2 \cos^2 \theta \right ]$
\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{\tan ^{3} x}{\cos ^{2} x} d x \\ &I=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \tan ^{3} x \sec ^{2} x d x \end{aligned}
Put $\tan x=t$
$\sec^2 xdx=dt$
When x=0 then t=0 and when $x=\frac{\pi}{4}$ then t=1

\begin{aligned} I &=\frac{1}{2} \int_{0}^{1} t^{3} d t \\ &=\frac{1}{2}\left[\frac{t^{3+1}}{3+1}\right]_{0}^{1} \\ &=\frac{1}{2} \cdot \frac{1}{4}\left[t^{4}\right]_{0}^{1} \\ &=\frac{1}{8}\left[1^{4}-0^{4}\right] \\ &=\frac{1}{8}[1] \\ &=\frac{1}{8} \end{aligned}

Definite Integrals exercise 19.2 question 27

Answer: $\frac{\pi}{4}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$
Solution:$I=\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$
Put $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} d x$
$I=\int_{0}^{\pi}\frac{1}{5+3\cos x}dx$
\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{5+3 \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} d x} \\ &=\int_{0}^{\pi} \frac{1}{\frac{5\left(1+\tan ^{2} \frac{x}{2}\right)+3\left(1-\tan ^{2} \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}} d x \\ &=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+3-3 \tan ^{2} \frac{x}{2}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{8+2 \tan ^{2} \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right] \\ &=\int_{0}^{\pi} \frac{1}{2} \frac{\sec ^{2} \frac{x}{2}}{4+\tan ^{2} \frac{x}{2}} d x \\ &=\frac{1}{2} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{2^{2}+\tan ^{2} \frac{x}{2}} d x \end{aligned}
Put $\tan \frac{x}{2}=t$
$\sec^2 \frac{x}{2}.\frac{1}{2}dx=dt$$\sec^2 \frac{x}{2}dx=2dt$

When x=0 then t=0 and $x=\pi$ then$t=\infty$

\begin{aligned} &I=\frac{1}{2} \int_{0}^{\infty} \frac{1}{2^{2}+t^{2}} 2 d t \\ &=2 \times \frac{1}{2} \int_{0}^{\infty} \frac{1}{2^{2}+t^{2}} d t \\ &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]_{0}^{\infty} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{2}=\infty, \tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 0=0\right] \\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise 19.2 question 28

Answer: $\frac{\pi}{2ab}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$
Solution:
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$
Dividing numerator and denominator by cos2x , we get
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{\cos ^{2} x}}{\frac{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x}{\cos ^{2} x}} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\frac{a^{2} \sin ^{2} x}{\cos ^{2} x}+\frac{b^{2} \cos ^{2} x}{\cos ^{2} x}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{1}{\cos x}=\sec x\right] \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{a^{2} \tan ^{2} x+b^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{\sin x}{\cos x}=\tan x\right] \end{aligned}
Put $\tan x=t$
$\Rightarrow sec^2xdx=dt$
When x=0 then t=0 and when $x=\frac{\pi}{2}$ then $t=\infty$
\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{a^{2} t^{2}+b^{2}} d t \\ &=\frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{t^{2}+\frac{b^{2}}{a^{2}}} d t \\ &=\frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{t^{2}+\left(\frac{b}{a}\right)^{2}} d t \end{aligned}
\begin{aligned} &=\frac{1}{a^{2}}\left[\frac{1}{b} \tan ^{-1}\left(\frac{a t}{b}\right)\right]_{0}^{\infty} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right] \\ &=\frac{1}{a^{2}} \frac{a}{b}\left[\tan ^{-1}\left(\frac{a t}{b}\right)\right]_{0}^{\infty} \\ &=\frac{1}{a b}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &=\frac{1}{a b}\left[\frac{\pi}{2}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 0=0\right] \end{aligned}
$=\frac{1}{ab}\frac{\pi}{2}$
$=\frac{\pi}{2ab}$

Definite Integrals exercise 19.2 question 29

Answer:$\frac{\pi}{2}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}\frac{x+2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos ^2 \frac{x}{2}}dx$
Solution:$I=\int_{0}^{\frac{\pi}{2}}\frac{x+2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos ^2 \frac{x}{2}}dx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2} x+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1+\cos 2 x=2 \cos ^{2} x, \sin 2 x=2 \sin x \cos x\right] \\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{x}{2 \cos ^{2} \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos x=\frac{1}{\sec x}\right] \\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{x \sec ^{2} \frac{x}{2}}{2}+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right) d x \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{x \sec ^{2} \frac{x}{2}}{2}+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \sec ^{2} \frac{x}{2} d x+\int_{0}^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \sec ^{2} \frac{x}{2} d x+\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{\sin x}{\cos x}=\tan x\right] \end{aligned}

Applying integration by parts method to first part of integral then,
\begin{aligned} &=\frac{1}{2}\left\{\left[x \int \sec ^{2} \frac{x}{2} d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{d}{d x}(x) \int \sec ^{2} \frac{x}{2} d x\right) d x\right\}+\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x \\ &=\frac{1}{2}\left\{\left[\frac{x \tan \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 1 \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right\}+\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x \mid \end{aligned}
\begin{aligned} &\left.=\frac{1}{2}\left\{2\left[x \tan \frac{x}{2}\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x\right\}+\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x \quad \int \sec ^{2} \frac{x}{a} d x=\frac{\tan \frac{x}{a}}{\frac{1}{a}}, \frac{d(x)}{d x}=1\right] \\ &=\left[x \tan \frac{x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x+\int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} d x \\ &=\left[\frac{\pi}{2} \tan \frac{\pi}{2 \times 2}-0 \times \tan \frac{0}{2}\right]-0 \end{aligned}
\begin{aligned} &=\left[\frac{\pi}{2} \tan \frac{\pi}{4}-0\right] \\ &=\frac{\pi}{2} \times 1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1, \tan 0=0\right] \\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise 19.2 question 30

Answer:$\frac{\pi ^2}{32}$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given:$\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}dx$
Solution:$I=\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}dx$
Put $\tan^{-1}x=t$
$\Rightarrow \frac{1}{1+x^{2}}dx=dt$
$\Rightarrow dx=\left (1+x \right )^2dt$
When x=0 then t=0 and when x=1 and $t=\frac{\pi}{4}$

\begin{aligned} I &=\int_{0}^{\frac{\pi}{4}} \frac{t}{1+x^{2}}\left(1+x^{2}\right) d t \\ =& \int_{0}^{\frac{\pi}{4}} t d t \\ =&\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\frac{\pi}{4}} \\ =&\left[\frac{t^{2}}{2}\right]_{0}^{\frac{\pi}{4}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[t^{2}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2}-0^{2}\right] \\ &=\frac{1}{2}\left(\frac{\pi^{2}}{16}\right) \\ &=\frac{\pi^{2}}{32} \end{aligned}

Definite Integrals exercise 19.2 question 31

Answer:$\frac{1}{4}\log 3$
Hint: We use indefinite formula then put limits to solve this integral.
Given: $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
Solution:$\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x+1-1} d x$
\begin{aligned} &=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+1-\left(\sin ^{2} x+\cos ^{2} x\right)+\sin 2 x} d x \quad\left[\because 1=\sin ^{2} x+\cos ^{2} x\right] \mid \\ &=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)} d x \quad[\because \sin 2 x=2 \sin x \cdot \cos x] \end{aligned}
\begin{aligned} &=\int_{0}^{\pi / 4}\left(\frac{\sin x+\cos x}{4-(\cos x-\sin x)^{2}}\right) d x \quad\left[\because a^{2}+b^{2}-2 a b=(a+b)^{2}\right] \\ &=\int_{0}^{\pi / 4}\left(\frac{\sin x+\cos x}{2)^{2}-(\cos x-\sin x)^{2}}\right) d x \end{aligned}
Put $\cos x - \sin x = t$
• $\Rightarrow (-\sin x - \cos x)dx = dt$
• $\Rightarrow -(\sin x+\cos x)dx = dt$
• $\Rightarrow (\sin x+\cos x)dx = -dt$
• if x =0 then t = 1 & if$x=\frac{\pi}{4}$ then t = 0
Then,
\begin{aligned} &\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x \\ &=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{(2)^{2}-(\cos x-\sin x)^{2}}\right) d x \\ &=\int_{1}^{0} \frac{1}{(2)^{2}-(t)^{2}}(-d t)=-\int_{1}^{0} \frac{1}{(2)^{2}-(t)^{2}} d t \\ &=\int_{0}^{1} \frac{1}{(2)^{2}-(t)^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right. \end{aligned}
\begin{aligned} &=\frac{1}{2 \times 2}\left[\log \left|\frac{2+t}{2-t}\right|\right]_{0}^{1}\\ &=\frac{1}{4}\left[\log \left|\frac{2+1}{2-1}\right|-\log \left|\frac{2+0}{2-0}\right|\right]\\ &=\frac{1}{4}\left[\log \frac{3}{1}-\log \frac{2}{2}\right]\\ &\left.=\frac{1}{4} \log 3-\log 1\right]\\ &=\frac{1}{4}[\log 3-0]\\ &=\frac{1}{4} \log 3\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log 1=0] \end{aligned}

Definite Integrals exercise 19.2 question 32

Answer:$\frac{\pi}{4}-\frac{1}{2}$
Hint: We use indefinite formula then put limits to solve this integral.
Given: $\int_{0}^1{} x\tan ^{-1}xdx$
Solution: $\int_{0}^1{} x\tan ^{-1}xdx$
Applying integration by parts, then,
$\int_{0}^1{} x\tan ^{-1}xdx$
\begin{aligned} &=\left[\tan ^{-1} x \int x d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right\} d x \\ &=\left[\tan ^{-1} x \cdot \frac{x^{1+1}}{1+1}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} \cdot \frac{x^{2}}{2} \mathrm{dx} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{x^{2}}{1+x^{2}} \mathrm{dx} \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1}\left(\frac{x^{2}+1-1}{1+x^{2}}\right) \mathrm{dx} \end{aligned}
\begin{aligned} &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1}\left(\frac{1+x^{2}}{1+x^{2}}-\frac{1}{1+x^{2}}\right) \mathrm{d} x \\ &=\left[\tan ^{-1} x \cdot \frac{x^{2}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} 1 d x+\frac{1}{2} \int \frac{1}{1+x^{2}} d x \\ &=\left[\tan ^{-1} 1 \cdot \frac{1^{2}}{2}-\frac{0^{2}}{2} \cdot \tan ^{-1} 0\right]-\frac{1}{2}[x]_{0}^{1}+\frac{1}{2}\left[\tan ^{-1} x\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\left[\frac{1}{2} \cdot \tan ^{-1} 1-0\right]-\frac{1}{2}[1-0]+\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2} \cdot 1+\frac{1}{2}\left[\frac{\pi}{4}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2}+\frac{1}{2}, \frac{\pi}{4} \\ &=\frac{\pi}{4}\left(\frac{1}{2}+\frac{1}{2}\right)-\frac{1}{2} \\ &=\frac{\pi}{4}-\frac{1}{2} \end{aligned}

Definite Integrals exercise 19.2 question 33

Answer: $(\log \sqrt{3} )$
Hint: We use indefinite integral formula and the limits to solve this integral.
Given: $\int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x$
Solution: $\int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{x^{2}-1}{x^{4}+x^{2}+1} d x$
Dividing the num. and denom. by x2 than
\begin{aligned} &\int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{x^{2}-1}{x^{2}+x^{2}+1}{x^{2}} d x \mid \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+1+1-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+2-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+2 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}-1} d x \end{aligned}
\begin{aligned} &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{x^{2}+2 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}-1} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-(1)^{2}} d x \\ &=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1^{2}} d x \end{aligned}
put $x+\frac{1}x{}=t$ then, $(1-\frac{1}{x^2}) dx=dt$
\begin{aligned} &\therefore \int_{0}^{1} \frac{1-x^{2}}{x^{4}+x^{2}+1} d x=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1^{2}} d x \\ &=-\int_{0}^{1} \frac{1}{t^{2}-1} d t \\ &=-\frac{1}{2 \times 1}\left[\log \left|\frac{t-1}{t+1}\right|\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int \frac{1}{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2} \log \left|\frac{x-a}{x+a}\right|+c\right] \end{aligned}
\begin{aligned} &=-\frac{1}{2}\left[\log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]_{0}^{1} \quad\left[\because t=x+\frac{1}{x}\right] \\ &=-\frac{1}{2}\left[\log \left|\frac{\frac{x^{2}+1-x}{x}}{\frac{x^{2}+1+x}{x}}\right|\right]_{0}^{1} \\ &=-\frac{1}{2}\left[\log \left|\frac{x^{2}-x+1}{x^{2}+x+1}\right|\right]_{0}^{1} \\ &=-\frac{1}{2}\left[\log \left|\frac{1^{2}-1+1}{1^{2}+1+1}\right|-\log \left|\frac{0^{2}-0+1}{0^{2}+0+1}\right|\right] \\ &=-\frac{1}{2}\left[\log _{3}^{1}-\log \frac{1}{1}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{2}\left[\log _{3}^{1}-\log 1\right] \\ &=-\frac{1}{2}\left[\log _{3}^{1}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log 1=0] \\ &=-\frac{1}{2} \cdot \log _{3}^{\frac{1}{3}} \\ &=-\frac{1}{2} \cdot \log 3^{-1} \\ &=\log 3^{-1\left(\frac{-1}{2}\right)} \end{aligned}
$=\log 3^{\frac{1}{2}}$
$=\log\sqrt{3}$

Definite Integrals exercise 19.2 question 34

Hint: Use indefinite integral formula and the given limits to solve this integral.
Given: $\int_{0}^{1}\frac{24x^2}{(1+x^2)^4}dx$
Solution:$\int_{0}^{1}\frac{24x^2}{(1+x^2)^4}dx$
put $(1+x^2)=t$ than, $2x \; dx = dt$
when x=0 then t=1 & when x=1 then t=2 ,
\begin{aligned} &\int_{0}^{1} \frac{24 x^{3}}{\left(1+x^{2}\right)^{4}} d x=12 \int_{1}^{2} \frac{(t-1)}{(t)^{4}} d t \\ &=12 \int_{1}^{2}\left(\frac{t}{(t)^{4}}-\frac{1}{(t)^{4}}\right) d t \\ &=12 \int_{1}^{2}\left(\frac{1}{(t)^{3}}-\frac{1}{(t)^{4}}\right) d t \\ &=12 \int_{1}^{2}(t)^{-3} d t-12 \int_{1}^{2}(t)^{-4} d t \end{aligned}
\begin{aligned} &=12\left[\frac{t^{-3+1}}{-3+1}\right]_{1}^{2}-12\left[\frac{t^{-4+1}}{-4+1}\right]_{1}^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=12\left[\frac{t^{-2}}{-2}\right]_{1}^{2}-12\left[\frac{t^{-3}}{-3}\right]_{1}^{2} \\ &=\frac{-12}{2}\left[\frac{1}{t^{2}}\right]_{1}^{2}-\frac{12}{(-3)}\left[\frac{1}{t^{3}}\right]_{1}^{2} \end{aligned}
\begin{aligned} &=-6\left[\frac{1}{2^{2}}-1\right]+4\left[\frac{1}{2^{3}}-\frac{1}{1^{3}}\right] \\ &=-6\left[\frac{1}{4}-1\right]+4\left[\frac{1}{8}-1\right] \\ &=-6\left[\frac{1-4}{4}\right]+4\left[\frac{1-8}{8}\right] \\ &=-6\left[\frac{-3}{4}\right]+4\left[\frac{-7}{8}\right] \\ &=\frac{9}{2}-\frac{7}{2} \\ &=\frac{9-7}{2} \\ &=\frac{2}{2}=1 \end{aligned}

Definite integrals exercise 19.2 question 35

Answer: $\frac{720}{7}$
Hint: Use indefinite integral formula and the given limits to solve this integral.
Given: $\int_{2}^{12}x(x-4)^{\frac{1}{4}}dx$
Solution: $\int_{2}^{12}x(x-4)^{\frac{1}{4}}dx$
Put $x - 4 = t^3 \Rightarrow dx = 3t^2dt$
When x = 4 then t =0 & when x =12 then t = 2
\begin{aligned} &\therefore \int_{4}^{12} x(x-4)^{1 / 3} d x \\ &=\int_{0}^{2}\left(t^{3}+4\right)\left(t^{3}\right)^{1 / 3} \cdot 3 t^{2} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because x-4=t^{3} \Rightarrow t^{3}+4=x\right] \\ &=\int_{0}^{2}\left(t^{3}+4\right) t \cdot 3 t^{2} d t \\ &=\int_{0}^{2}\left(t^{3}+4\right) 3 t^{3} d t \end{aligned}
\begin{aligned} &=3 \int_{0}^{2}\left(t^{3+3}+4 t^{3}\right) d t=3 \int_{0}^{2}\left(t^{6}+4 t^{3}\right) d t\\ &=3 \int_{0}^{2} t^{6} d t+12 \int_{0}^{2} t^{3} d t\\ &=3\left[\frac{t^{6+1}}{6+1}\right]_{0}^{2}+12\left[\frac{t^{3+1}}{3+1}\right]_{0}^{2}\ \! \! \! \! \! \! &\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]\\ &=\frac{3}{7}\left[t^{7}\right]_{0}^{2}+\frac{12}{4}\left[t^{4}\right]_{0}^{2} \end{aligned}
\begin{aligned} &=\frac{3}{7}\left[2^{7}-0^{7}\right]+3\left[2^{4}-0^{4}\right] \\ &=\frac{3}{7}[128-0]+3[16-0] \\ &=3\left[\frac{128}{7}+16\right] \\ &=3\left[\frac{128+112}{7}\right] \\ &=3 \times \frac{240}{7} \\ &=\frac{720}{7} \end{aligned}

Definite integrals exercise 19.2 question 36

Answer: $\pi -2$
Hint: use indefinite integral formula and the limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}x^2\sin x dx$
Solution: $\int_{0}^{\frac{\pi}{2}}x^2\sin x dx$
Applying integration by parts method, then,
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} x^{2} \sin x \mathrm{~d} x=\left[x^{2} \int \sin x \mathrm{~d} x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d\left(x^{2}\right)}{d x} \int \sin x d x\right\} d x \\ &=\left[x^{2}(-\cos x)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x(-\cos x) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left(\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \int \sin a x d x=\frac{-\cos a x}{a}\right) \\ &=-\left[x^{2} \cos x\right]_{0}^{\frac{\pi}{2}}+2 \int_{0}^{\frac{\pi}{2}} x \cos x d x \end{aligned}
On applying integration by parts method
\begin{aligned} &=-\left[\left(\frac{\pi}{2}\right)^{2} \cos \frac{\pi}{2}-0^{2} \cos 0\right]+2\left[\left\{x \int \cos x d x\right\}_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \cos x d x\right\} d x\right] \\ &=-\frac{\pi^{2}}{4} \cdot 0-0+2\left\{[x \sin x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 1 \cdot \sin x d x\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ \int \cos x d x=\sin x \end{array}\right] \\ &=2\left[\frac{\pi}{2} \cdot \sin \frac{\pi}{2}-0 \cdot \sin 0\right]-2[-\cos x]_{0}^{\frac{\pi}{2}} \\ &=\pi .1+2\left[\cos \frac{\pi}{2}-\cos 0\right]=\pi+2[0-1] \\ &=\pi-2 \end{aligned} \quad\left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \left.\sin \frac{\pi}{2}=1\right] \\ \end{array}\right.

Definite integrals exercise 19.2 question 37

Answer : $\frac{\pi}{2}-1$
Hint : use indefinite integral formula and the limits to solve this integral
Given :$\int_{0}^1{}\sqrt{\frac{1-x}{1+x}}dx$
Solution : $\int_{0}^1{}\sqrt{\frac{1-x}{1+x}}dx$
put $x=\cos 2\theta \Rightarrow dx= -2\sin 2\theta d\theta$
when $x=0 \; \text{then} \; \; \theta =\frac{\pi}{4} \text{ and when}\; x=1 then \theta=0$
therefore ,
\begin{aligned} &\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x=\int_{\frac{\pi}{4}}^{0} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}(-2 \sin 2 \theta) d \theta \\ &=\int_{0}^{\frac{\pi}{4}} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}} 2 \sin 2 \theta d \theta \\ &=\int_{0}^{\frac{\pi}{4}} \sqrt{\left(\frac{\sin \theta}{\cos \theta}\right)^{2}} 2 \sin 2 \theta d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}} \frac{\sin \theta}{\cos \theta} 2 \sin \theta \cos \theta d \theta \end{aligned}
\begin{aligned} &=4 \int_{0}^{\frac{\pi}{4}} \sin ^{2} \theta d \theta=4 \int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}}(1-\cos 2 \theta) d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}} 1 d \theta-2 \int_{0}^{\frac{\pi}{4}} \cos 2 \theta d \theta \\ &=2\left[\frac{\theta^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{4}}-2\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{4}} \\ &=2\left[\frac{\pi}{4}-0\right]-\left[\sin \frac{2 \pi}{4}-\sin 2 \times 0\right] \\ &=2 \frac{\pi}{4}-\left[\sin \frac{\pi}{2}-\sin 0\right] \\ &=\frac{\pi}{2}-\left[\sin \frac{\pi}{2}-0\right] \\ &=\frac{\pi}{2}-1 \end{aligned}

Definite integrals exercise 19.2 question 38

Answer : $\frac{1}{2}$
Hint : use indefinite integral formula and the limits to solve this integral
Given : $\int_{0}^1{}\frac{1-x^2}{(1+x^2)^2}dx$
Solution :$\int_{0}^1{}\frac{1-x^2}{(1+x^2)^2}dx$
$=-\int_{0}^{1} \frac{x^{2}\left(1-\frac{1}{x^{2}}\right)}{x^{2}\left(\frac{1}{x}+x\right)^{2}} d x=-\int_{0}^{1} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}} d x$
put x+1x=t⇒1-1x2dx=dt
when x=0 then t= ∞ ,when x=1 then t=2

Therefore ,
\begin{aligned} &\int_{0}^{1} \frac{1-x^{2}}{\left(1+x^{2}\right)^{2}} d x \\ &=-\int_{\infty}^{2} \frac{1}{t^{2}} d t=\int_{2}^{\infty} t^{-2} d t \mid \\ &=\left[\frac{t^{-2+1}}{-2+1}\right]_{2}^{\infty}=\left[\frac{t^{-1}}{-1}\right]_{2}^{\infty} \end{aligned}
$=-\left[\frac{1}{t}\right]_{2}^{\infty}=-\left[\frac{1}{\infty}-\frac{1}{2}\right]=-\left[0-\frac{1}{2}\right]=\frac{1}{2}$

Definite integrals exercise 19.2 question 39

Answer : $\frac{4\sqrt{2}}{3}$
Hint :use indefinite integral formula and the limits to solve this integral
Given : $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$
Solution :-$\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$
put $x^5+1=t \Rightarrow 5x^4dx=dt$
when x= -1 then t=0 and when x=1 then t=2
therefore,
\begin{aligned} &\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x \\ &=\int_{0}^{2} \sqrt{t} d t=\int_{0}^{2} t^{\frac{1}{2}} d t \\ &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2} \\ &=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2}=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{2}=\frac{2}{3}\left[2^{\frac{3}{2}}-0\right] \\ &=\frac{2}{3}(\sqrt{2})^{2 \times \frac{3}{2}}=\frac{2}{3}(\sqrt{2})^{3}=\frac{2 \times 2 \sqrt{2}}{3}=\frac{4 \sqrt{2}}{3} \end{aligned}

Definite integrals exercise 19.2 question 40

Answer : $\frac{\pi}{6}$
Hint : use indefinite integral formula and the limits to solve this integral
Given : $\int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}$
Solution : $\int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}$
on multiplying and dividing by sec4x , we get
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos ^{2} x \cdot \sec ^{4} x}{\sec ^{4} x\left(1+3 \sin ^{2} x\right)}\right) d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x \cdot \frac{1}{\cos ^{2} x} \cdot \sec ^{2} x}{\sec ^{2} x \cdot \frac{1}{\cos ^{2} x}\left(1+3 \sin ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\frac{1}{\cos ^{2} x}+3 \frac{\sin ^{2} x}{\cos ^{2} x}\right)} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\sec ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+\tan ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x \end{aligned}
put $\tan x=t \Rightarrow \sec^2x dx=dt$
when x=0 then t=0 , when $x =\frac{\pi}{2}$ then $t=\infty$
therefore ,
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x\\ &=\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \end{aligned}
To solve this integral, first we need to find its partial fraction then integrate it using indefinite integral formula then put the limits to get required answer.
therefore ,
\begin{aligned} &\frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{A t+B}{\left(1+t^{2}\right)}+\frac{c t+D}{1+4 t^{2}} \\ &\Rightarrow 1=\frac{(A t+B)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+t^{2}\right)}+\frac{(C t+D)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+4 t^{2}\right)} \\ &\Rightarrow 1=(A t+B)\left(1+4 t^{2}\right)+(C t+D)\left(1+t^{2}\right) \\ &\Rightarrow 1=A t+4 A t^{3}+B+4 B t^{2}+C t+C t^{3}+D+D t^{2} \\ &\Rightarrow 1=(4 A+C) t^{3}+(4 B+D) t^{2}+(A+C) t+(B+D) \end{aligned}
Equating the coefficient of t3,t2,t and constant term respectively then
$0=4A+C$ … a
$0=4B+D$ … b
$0=A+C$ … c
$1=B+D$ …(d)
-4A=C
-4B=D

Since A= -C=0
Substracting (d) by (b), we get
4B+D=0
B+D =13B= -1
B=-13
1= -13+D ? 1+13=D?3+13=D

Therefore $A=C=0, B=-\frac{1}{3} , D=\frac{4}{3}$
Therefore $\frac{1}{\left.(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{0 . t-\frac{1}{3}}{1+t^{2}}+\frac{0 . t+\frac{4}{3}}{1+4 t^{2}}=\frac{-1}{3\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}$
Now (i)?
\begin{aligned} &\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \\ &=\int_{0}^{\infty}\left\{-\frac{1}{3} \frac{1}{\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}\right\} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+4 t^{2}} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+(2 t)^{2}} d t \end{aligned}
put 2t=u ⇒2dt=du ⇒ dt=du2 in second integral, then
\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+3 \sin ^{2} x} d x=\frac{-1}{3}\left[\tan ^{-1} t\right]_{0}^{\infty}+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+u^{2}} \frac{d u}{2} \\ &=\frac{-1}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]+\frac{2}{3}\left[\tan ^{-1} u\right]_{0}^{\infty} \\ &=\frac{-1}{3}\left[\frac{\pi}{2}-0\right]+\frac{2}{3}\left[\tan ^{-1}(2 t)\right]_{0}^{\infty} \\ &=\frac{-1}{3} \frac{7}{2}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 2 \times 0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \end{aligned}
\begin{aligned} &=\frac{-\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}-0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3} \frac{\pi}{2} \\ &=\frac{-\pi}{6}+\frac{\pi}{3}=\frac{-\pi+2 \Pi}{6} \\ &=\frac{\pi}{6} \end{aligned}

Definite integrals exercise 19.2 question 41

Answer : $\frac{1}{8}$
Hint : use indefinite formula and the limit to solve this integral
Given : $\int_{0}^{\frac{\pi}{2}}\sin ^22t.\cos 2tdt$
Solution : $\int_{0}^{\frac{\pi}{2}}\sin ^22t.\cos 2tdt$
\begin{aligned} &\text { Put } \sin 2 t=u \Rightarrow 2 \cos 2 t d t=d u \Rightarrow \cos 2 t d t=\frac{d u}{2}\\ &\text { When } t=\frac{\pi}{4} \text { then } u=1 \text { when } t=0 \text { then } u=0 \end{aligned}
\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cdot \cos 2 t d t \\ &=\int_{0}^{1} u^{3} \frac{d u}{2}=\frac{1}{2} \int_{0}^{1} u^{3} d u \\ &=\frac{1}{2}\left[\frac{u^{3}+1}{3+1}\right]_{0}^{1}=\frac{1}{2}\left[\frac{u^{4}}{4}\right]_{0}^{1} \\ &=\frac{1}{8}\left[u^{4}\right]_{0}{ }^{1} \\ &=\frac{1}{8}\left[1^{4}-0^{4}\right] \\ &=\frac{1}{8} \end{aligned}

Definite Integrals exercise 19.2 question 42.

Answer : $9 \sqrt{3}-1$
Hint : use indefinite formula and the limit to solve this integral
Given $\int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta$
Solution : $\int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta$
Put $(5-4\cos \theta )=t\Rightarrow -4(-\sin \theta )d\theta =dt$
$\rightarrow \sin \theta d\theta = \frac{dt}{4}$
when $\theta =0$ then t=1 and when $\theta =\pi$ then t=9
Therefore
$\int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta$
$=\int_{1}^{9} 5 t^{\frac{1}{4}} \frac{d t}{4}=\frac{5}{4} \int_{1}^{9} t^{\frac{1}{4}} d t$
\begin{aligned} &=\frac{5}{4}\left[\frac{t^{\frac{1}{4}+1}}{\frac{1}{4}+1}\right]_{1}^{9} \\ &=\frac{5}{4}\left[\frac{t^{\frac{5}{4}}}{\frac{5}{4}}\right]_{1}^{9} \\ &=\frac{5}{4} \times \frac{4}{5}\left[9^{\frac{5}{4}}-1^{\frac{5}{4}}\right] \\ &=3^{2 x_{4}^{5}}-1=3^{\frac{5}{2}}-1 \\ &=9 \sqrt{3}-1 \end{aligned}

Definite Integrals exercise 19.2 question 43.

Answer : $\frac{3}{4}$
Hint : use indefinite formula and the limit to solve this integral
Given : $\int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta$
Solution : $I=\int_{0}^{\frac{\pi}{6}}\cos ^{-3}2 \theta \sin 2 \theta d \theta$
\begin{aligned} &I==\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos ^{3} 2 \theta} \cdot \sin 2 \theta d \theta \\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sin 2 \theta}{\cos 2 \theta} \cdot \frac{1}{\cos ^{2} 2 \theta} d \theta \\ &I=\int_{0}^{\frac{\pi}{6}} \tan 2 \theta \cdot \sec ^{2} 2 \theta d \theta \end{aligned}
put $\tan 2\theta =t \Rightarrow \sec^22\theta .2d\theta = dt \Rightarrow \sec^22\theta d\theta = \frac{dt}2{}$
when $\theta =0$ then $t=0$ and when $\theta = \frac{\pi}{6}$ then $t=3$
Therefore ,
\begin{aligned} &I=\int_{0}^{\sqrt{3}} t \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{\sqrt{3}} t d t \\ &=\frac{1}{2}\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\sqrt{3}} \end{aligned}
\begin{aligned} &=\frac{1}{2} \cdot \frac{1}{2}\left[t^{2}\right]_{0}^{\sqrt{3}} \\ &=\frac{1}{4}\left[\sqrt{3}^{2}-0^{2}\right] \\ &I=\frac{1}{4} \times 3 \\ &I=\frac{3}{4} \end{aligned}

Definite Integrals exercise 19.2 question 44.

Answer :$\frac{\pi}{3}$
Hint : use indefinite formula and the limit to solve this integral
Given : $\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$
Solution :$\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$
put $x^{\frac{3}{2}}=t \Rightarrow \frac{3}{2} x^{\frac{3}{2}-1} d x=d t \Rightarrow \frac{3}{2} x^{\frac{1}{2}} d x=d t \Rightarrow d x=\frac{2}{3 \sqrt{x}} d t$
when x=0 then t=0 when $x=(\Pi)^{\frac{2}{3}}$ then $t=\Pi$
Therefore

$\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x$
\begin{aligned} &=\int_{0}^{\pi} \sqrt{x} \cos ^{2} t \frac{2}{3 \sqrt{x}} d t \\ &=\frac{2}{3} \int_{0}^{\pi} \cos ^{2} t d t \\ &=\frac{2}{3} \int_{0}^{\pi}\left(\frac{1+\cos 2 t}{2}\right) d t \\ &=\frac{1}{3} \int_{0}^{\pi}(1+\cos 2 t) d t \\ &=\frac{1}{3}\left[\int_{0}^{\pi} 1 d t\right]+\frac{1}{3}\left[\int_{0}^{\pi} \cos 2 t d t\right] \\ &=\frac{1}{3}[t]_{0}{ }^{\pi}+\frac{1}{3}\left[\frac{\sin 2 t}{2}\right]_{0}^{\pi} \\ &=\frac{1}{3} \Pi-0+\frac{1}{6}[\sin 2 \pi-\sin 2 \times 0] \end{aligned}
$=\frac{1}{3}\Pi+\frac{1}{6}(0-0)$
$=\frac{\Pi}{3}$

Definite Integrals exercise 19.2 question 45.

Answer : $\frac{\log2}{1+\log 2}$
Hint : use indefinite formula and the limit to solve this integral
Given : $\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$
Solution : $\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$
put $1+\log x =t \Rightarrow \frac{ 1}{x}dx=dt \Rightarrow dx=x \; dt$
when x=1 then t=1 when x=2 then t=1+log2

\begin{aligned} &\int_{1}^{2} \frac{1}{x(1+\log x)^{2}} d x=\int_{1}^{1+\log 2} \frac{1}{x \cdot t^{2}} x d t \\ &=\int_{1}^{1+\log 2} \frac{1}{t^{2}} d t \\ &=\int_{1}^{1+\log 2} t^{-2} d t \end{aligned}
\begin{aligned} &=\left[\frac{t^{-2+1}}{-2+1}\right]_{1}^{1+\log 2} \\ &=\left[\frac{t^{-1}}{-1}\right]_{1}^{1+\log 2} \\ &=-\left[\frac{1}{t}\right]_{1}^{1+\log 2} \\ &=-\left[\frac{1}{1+\log 2}-\frac{1}{1}\right] \\ &=1-\frac{1}{1+\log 2} \\ &=\frac{1+\log 2-1}{1+\log 2} \\ &=\frac{\log 2}{1+\log 2} \end{aligned}

Definite Integrals exercise 19.2 question 46.

Answer :$\frac{8}{15}$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{0}^{\frac{\pi}{2}}\cos ^5x \; dx$
Solution:
\begin{aligned} \int_{0}^{\pi / 2} \cos ^{5} x d x &=\int_{0}^{\pi / 2} \cos ^{4} x \cdot \cos x d x \\ &=\int_{0}^{\pi / 2}\left(\cos ^{2} x\right)^{2} \cos x d x \\ &=\int_{0}^{1 / 2}\left(1-\sin ^{2} x\right)^{2} \cos x d x \quad \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \end{aligned}
Put $\sin x=t\Rightarrow \cos x \; dx=dt$

when x=0 then t=0
when$x=\frac{\pi}{2}$ then t=1
\begin{aligned} &\therefore \int_{0}^{\pi / 2} \cos ^{5} x d x=\int_{0}^{1}\left(1-t^{2}\right)^{2} d t \\ &\qquad \begin{aligned} & \int_{0}^{1}\left(1+t^{4}-2 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ =& \int_{0}^{1} t^{0} d t+\int_{0}^{1} t^{4} d t-2 \int_{0}^{1} t^{2} d t \end{aligned} \end{aligned}
\begin{aligned} &=\left[\frac{t^{1+1}}{0+1}\right]_{0}^{1}+\left[\frac{t^{4+1}}{4+1}\right]_{0}^{1}-2\left[\frac{t^{2+1}}{2+1}\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \because \int x^{n} d x=\frac{x^{n+1}}{x+1}+c \\ &=[t]_{0}^{1}+\frac{1}{5}\left[t^{5}\right]_{0}^{1}-\frac{2}{3}\left[t^{3}\right]_{0}^{1} \\ &=[1-0]+\frac{1}{5}\left[1^{5}-0^{5}\right]-\frac{2}{3}\left[1^{3}-0^{3}\right] \\ &=1+\frac{1}{5}-\frac{2}{3} \\ &=\frac{15+3-10}{15} \end{aligned}

Definite Integrals exercise 19.2 question 47

Answer : $\frac{19}{99}$
Hint: Use in definite formula and the given limits to solve this integral
Given:
$\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx$
Solution:
$\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx$
Put $(30-x^{\frac{3}{2}})=t$
\begin{aligned} &\Rightarrow-\frac{3}{2} x^{\left(\frac{3}{2}-1\right)} d x=d t \\ &\Rightarrow-\frac{3}{2} x^{\frac{1}{2}} d x=d t \\ &\Rightarrow-\frac{2}{3 \sqrt{x}} d t=d x \end{aligned}

when x=4 then t=22
when x=9 then t=3 then
\begin{aligned} &\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{3 / 2}\right)^{2}} d x \\ &=\int_{22}^{3} \frac{\sqrt{x}}{t^{2}} \cdot \frac{-2 d t}{3 \sqrt{x}} \\ &=\frac{-2}{3} \int_{22}^{3} t^{2} d t \\ &=-\frac{2}{3}\left[\frac{t^{-2+1}}{-2+1}\right]_{22}^{3} \end{aligned}
\begin{aligned} &=\frac{-2}{3}\left[\frac{t^{-1}}{-1}\right]_{22}^{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{3}\left[\frac{1}{t}\right]_{22}^{3} \\ &=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right] \\ &=\frac{2}{3}\left[\frac{22-3}{66}\right] \\ &=\frac{2}{3} \cdot \frac{19}{66} \\ &=\frac{19}{99} \end{aligned}

Definite Integrals exercise 19.2 question 48.

Answer : $\frac{8}{3}$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx$
Solution:
$\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx$
put $(\cos x)=t$
$\Rightarrow -\sin x\; dx=dt$
$\Rightarrow dx=dt - \sin x$
When x=0 then t=1
when $x=\pi$ then t=-1
\begin{aligned} &\therefore \int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} d x \\ &=\int_{1}^{-1} \sin ^{3} x(1+2 t)(1+t)^{2} \frac{d t}{-\sin x} \\ &=-\int_{+1}^{-1} \sin ^{2} x(1+2 t)\left(1+t^{2}+2 t\right) d t \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 \alpha b\right] \end{aligned}
\begin{aligned} &=-\int_{+1}^{-1}\left(1-\cos ^{2} x\right)\left(1+t^{2}+2 t+2 t+2 t^{3}+4 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \\ &=-\int_{+1}^{-1}\left(1-t^{2}\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}-1\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}+5 t^{4}+4 t^{3}+2 t^{5}-1-5 t^{2}-4 t-2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(2 t^{5}+2 t^{3}+5 t^{4}-4 t^{2}-4 t-1\right) d t \end{aligned}
\begin{aligned} &=2 \int_{1}^{-1} t^{5} d t+2 \int_{1}^{-1} t^{3} d t+5 \int_{1}^{-1} t^{4} d t-4 \int_{.1}^{-1} t^{2} d t-4 \int_{-1}^{-1} t d t-\int_{1}^{-1} d t \\ &=2\left[\left.\frac{t^{5+1}}{5+1}\right|_{1} ^{-1}+2\left[\frac{t^{3+1}}{3+1}\right]_{1}^{-1}+5\left[\left.\frac{t^{4+1}}{4+1}\right|_{1} ^{-1}-4\left[\left.\frac{t^{2+1}}{2+1}\right|_{1} ^{-1}-4\left[\frac{t^{1+1}}{1+1}\right]_{1}^{-1}-\left[\left.t\right|_{1} ^{-1}\right.\right.\right.\right. \\ &=\frac{2}{6}\left[t^{6}\right]_{1}^{-1}+\frac{2}{4}\left[t^{4}\right]_{1}^{-1}+\frac{5}{5}\left[t^{5}\right]_{1}^{-1}-\frac{4}{3}\left[t^{3}\right]_{1}^{-1}-4\left[\frac{t^{2}}{2}\right]_{1}^{-1}-[t]_{1}^{-1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \end{aligned}
\begin{aligned} &=\frac{2}{6}\left[(-1)^{6}-1^{6}\right]+\frac{2}{4}\left[(-1)^{4}-1^{4}\right]+\left[(-1)^{5}-1^{5}\right]-\frac{4}{3}\left[(-1)^{3}-1\right]-\frac{4}{2}\left[(-1)^{2}-1\right]-[-1-1] \\ &=\frac{1}{3}[1-1]+\frac{1}{2}[1-1]+[-1-1]-\frac{4}{3}[-1-1]-2[1-1]-(-2) \\ &=0+0+(-2)-\frac{4}{3}(-2)-2 \times 0+2 \\ &=\frac{8}{3} \end{aligned}

Definite Integrals exercise 19.2 question 49.

Answer : $\frac{\pi}{2}-1$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{0}^{\frac{\pi}{2}}\sin x.\cos x\tan^{-1}(\sin x)dx$
Solution:
$\int_{0}^{\frac{\pi}{2}}\sin x.\cos x\tan^{-1}(\sin x)dx$
Put $t=\sin x\Rightarrow \cos xdx=dt$
when x=0 then t=0 and
when $x=\frac{\pi}{2}$ , then t=1
$\therefore \int_{0}^{\pi / 2} 2 \sin x \cdot \cos x \tan ^{-1}(\sin x) d x=\int_{0}^{1} 2 t \cdot \tan ^{-1}(t) d t$
Applying integration by parts, method then
\begin{aligned} &=2\left\{\left[\tan ^{-1} t \int t \mathrm{~d} t\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d t}\left(\tan ^{-1} t\right) \int t d t\right] d t\right\} \\ &=2\left\{\left[\tan ^{-1} t \cdot \frac{t^{2}}{2}\right]_{0}^{1}-\int_{0}^{1}\left[\frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right]\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{2}\left[t^{2} \tan ^{-1} t\right]_{0}^{1}-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t \end{aligned}
\begin{aligned} &=\frac{2}{2}\left[t^{2} \tan ^{-1} t\right]_{0}^{1}-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t \\ &=\left[1^{2} \tan ^{-1}(1)-0 \times \tan ^{-1} 0\right]-\int_{0}^{1} \frac{t^{2}+1-1}{1+t^{2}} d t \\ &=\left[1 \cdot \frac{\pi}{4}-0\right]-\int_{0}^{1}\left(\frac{1+t^{2}}{1+t^{2}}-\frac{1}{1+t^{2}}\right) d t \\ &=\frac{\pi}{4}-\int_{0}^{1}\left[1-\frac{1}{1+t^{2}}\right] d t \end{aligned}
\begin{aligned} &=\frac{\pi}{4}-\int_{0}^{1} 1 d t+\int_{0}^{1} \frac{1}{1+t^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \tan ^{-1}=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{\pi}{4}-[t]_{0}^{1}+\left[\tan ^{-1} t\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{0} d x=\frac{x^{0+1}}{0+1}, \int \frac{1}{1+x^{2}}, d x=\tan ^{-1} x\right] \\ &=\frac{\pi}{4}-[1-0]+\left[\tan ^{1} 1-\tan ^{-1} 0\right] \\ &=\frac{\pi}{4}-1+\frac{\pi}{4}-0 \\ &=\frac{2 \pi}{4}-1 \\ &=\frac{\pi}{2}-1 \end{aligned}

Definite Integrals exercise 19.2 question 50.

Answer : $\frac{\pi}{2}-1$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx$
Solution:
$\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx=\int_{0}^{\frac{\pi}{2}}2\sin x .\cos x\tan ^{-1}(\sin x)dx$
put $\sin x= t \Rightarrow \cos xdx=dt$
when x=0 then t=0 &
when $x=\frac{\pi}{2}$ then t=1
$\int_{0}^{\frac{\pi}{2}}\sin 2x \tan ^{-1}(\sin x)dx=\int_{0}^{\frac{\pi}{2}}2\tan ^{-1}(t)dt$
Applying by parts, then
\begin{aligned} &=2\left\{\left[\tan ^{-1} t \int t d t\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d t} \tan ^{-1} t \int t d t\right] d t\right\} \\ &=2\left\{\left[\tan ^{-1} t \cdot \frac{t^{2}}{2}\right]_{0}^{1}-\int_{0}^{1}\left[\frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right]\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{2}\left[t^{2} \tan ^{-1} t\right]_{0}^{1}-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t \\ &=\left[1^{2} \tan ^{-1}(1)-0 \times \tan ^{-1} 0\right]-\int_{0}^{1} \frac{t^{2}+1-1}{1+t^{2}} d t \end{aligned}
\begin{aligned} &=\left[1^{2} \tan ^{-1} 1-0 \times \tan ^{-1} 0\right]-\int_{0}^{1}\left(\frac{1+t^{2}}{1+t^{2}}-\frac{1}{1+t^{2}}\right) d t \\ &=\left[1 \cdot \frac{\pi}{4}-0\right]-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right) d t \\ &=\frac{\pi}{4}-\int_{0}^{1} 1 d t+\int_{0}^{1} \frac{1}{1+t^{2}} d t \end{aligned}
\begin{aligned} &=\frac{\pi}{4}-[t]_{0}^{1}+\left[\tan ^{-1} t\right]_{0}^{1}\ &\left[\because \int 1 d x=x, \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x\right]\\ &=\frac{\pi}{4}-[1-0]+\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\ &\ \left[\therefore \tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right]\\ &=\frac{\pi}{4}-1+\left[\frac{\pi}{4}-0\right]\\ &=\frac{2 \pi}{4}-1\\ &=\frac{\pi}{2}-1 \end{aligned}

### Definite Integrals exercise 19.2 question 51.

Answer : $\pi -2$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{0}^{1}(\cos ^{-1}x)^2dx$
Solution:
$\int_{0}^{1}(\cos ^{-1}x)^2dx$
Applying integration by parts, then
\begin{aligned} &=\left[\left(\cos ^{-1} x\right)^{2} \int 1 d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x}\left(\cos ^{-1} x\right)^{2} \int 1 d x\right\} \mathrm{d} x\\ &=\left[\left(\cos ^{-1} x\right)^{2} \cdot x\right]_{0}^{1}-\int_{0}^{1} 2 \cos ^{-1} x \cdot \frac{-1}{\sqrt{1-x^{2}}} \cdot x d x\; \; \; \; \; \; \; \; \; \; \; \; .........(i).\\ &\therefore \int_{0}^{1} 2 \cos ^{-1} x\left(\frac{-1}{\sqrt{1-x^{2}}} \cdot x\right) d x\\ &\text { put } \cos ^{-1} x=t\\ &\Rightarrow \frac{-1}{\sqrt{1-x^{2}}} d x=d \mathrm{t} \end{aligned}
When x=0 then $t=\frac{\pi}{2}$
and when x=1 then t=0
therefore,
\begin{aligned} &-\int_{0}^{1} 2 \cos ^{-1} x \cdot \frac{x}{\sqrt{1-x^{2}}} d x \\ &=-\int_{\pi / 2}^{0} 2 t \cos t d t \\ &=2 \int_{0}^{\pi / 2} t \cos t d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right] \end{aligned}
\begin{aligned} &=2\left\{\left[t \int \cos t d t\right]_{0}^{\pi / 2}-\int_{0}^{\pi / 2}\left[\frac{d t}{d t} \int \cos t d t\right] d t\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \frac{d(x)}{d x}=1 \\ \int \cos x d x=\sin x \end{array}\right] \\ &=2\left\{\left[\frac{\pi}{2} \cdot \sin \frac{\pi}{2}-0 \cdot \sin 0\right]-[-\cos t]_{0}^{\pi / 2}\right\} \\ &=2\left\{\left[\frac{\pi}{2} \cdot 1-0\right]+\left[\cos \frac{\pi}{2}-\cos 0\right]\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin \frac{\pi}{2}=1, \sin 0=0\right] \\ &=\frac{\pi}{2} \times 2+(0-1) 2 \\ &=\pi-2 \end{aligned}

Definite Integrals exercise 19.2 question 52.

Answer :$a\left [ \frac{\pi}{2} -1\right ]$
Hint: Use indefinite formula and the given limits to solve this integral
Given:
$\int_{a}^{0}\sin^{-1}\sqrt{\frac{x}{a+x}}dx$
Solution:
$\int_{a}^{0}\sin^{-1}\sqrt{\frac{x}{a+x}}dx$
Putting $x=a\tan ^{2}\theta$
$\Rightarrow dx=2a \tan \theta \cdot\sec^ {2}\theta d\theta$
When $x=0\Rightarrow \theta=0$ and
$x=a\Rightarrow \theta =\frac{\pi}{4}$
\begin{aligned} &\therefore \int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}} 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a\left(1+\tan ^{2} \theta\right)}} 2 a \cdot \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{\operatorname{an}^{2} \theta}{\sec ^{2} \theta}} 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right] \end{aligned}
\begin{aligned} &=\int_{0}^{\pi / 4} \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1}\left(\frac{\sin \theta}{\cos \theta \cdot \frac{1}{\cos \theta}}\right) 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \therefore \frac{\sin A}{\cos A}=\tan A \\ \frac{1}{\cos A}=\sec A \end{array}\right] \\ &=\int_{0}^{\pi / 4} \theta \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta\; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \because \sin ^{-1}(\sin A)=A \end{aligned}
Again putting $\tan \theta=u$
$\Rightarrow \sec^2\theta d\theta =du$
When $\theta =0$ then u=0
and $\theta =\frac{\pi}{4}$ then u=1
\begin{aligned} &\therefore \int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=2 a \int_{0}^{1} \tan ^{-1} u \cdot u \cdot d u\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \tan \theta=u \Rightarrow \theta=\tan ^{-1} u\right] \\ &=2 a \int_{0}^{1} u \tan ^{-1} u d u \end{aligned}
Applying integration by parts, method then
\begin{aligned} &=2 a\left\{\left[\tan ^{-1} u \int u d u\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d u}\left(\tan ^{-1} u\right) \int u d u\right] d u\right\} \\ &=2 a\left\{\left[\tan ^{-1} u \frac{u^{2}}{2}\right]_{0}^{1}-\int_{0}^{1}\left[\frac{1}{1+u^{2}} \cdot \frac{u^{2}}{2}\right] d u\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}} \\ \left.\& \int x d x=\frac{x^{2}+1}{x+1}-\frac{x^{2}}{2}\right] \end{array}\right. \end{aligned}
\begin{aligned} &=2 a\left\{\left[\frac{u^{2}}{2} \tan ^{-1} u\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{u^{2}+1-1}{1+u^{2}} d u\right\} \\ &=2 a\left\{\left[\frac{1}{2} \tan ^{-1}(1)-\frac{0}{2} \cdot \tan ^{-1} 0\right]-\frac{1}{2} \int_{0}^{1}\left(\frac{1+u^{2}}{1+u^{2}}-\frac{1}{1+u^{2}}\right) d u\right\} \\ &=2 a\left\{\left[\frac{1}{2} \cdot \frac{\pi}{4}-0\right]-\frac{1}{2} \int_{0}^{1}\left(1-\frac{1}{1+u^{2}}\right) d u\right\} \quad\left[\begin{array}{l} \tan ^{-1} 1=\frac{\pi}{4} \\ \tan ^{-1} 0=0 \end{array}\right] \end{aligned}
\begin{aligned} &=2 a \cdot \frac{1}{2} \cdot \frac{\pi}{4}-a \int_{0}^{1} 1 d u+a \int_{0}^{1} \frac{1}{1+\mathrm{u}^{2}} d u \\ &=\frac{a \pi}{4}-a\left[\frac{u^{0+1}}{0+1}\right]_{0}^{1}+a\left[\tan ^{-1} u\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int 1 d x=x \& \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &=\frac{\pi a}{4}-a[u]_{0}^{1}+a\left[\tan ^{-1}(1)-\tan ^{-1} 0\right] \end{aligned}
\begin{aligned} &=\frac{\pi a}{4}-a[1-0]+a \cdot\left[\frac{\pi}{4}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{n} 1=\frac{\pi}{4}, \tan ^{-1} \theta=0\right] \\ &=\frac{\pi a}{4}-a+\frac{\pi a}{4} \\ &=\frac{2 \pi a}{4}-a \\ &=a\left(\frac{\pi}{2}-1\right) \end{aligned}

Definite Integrals exercise 19.2 question 53.

Hint: Use indefinite integral formula and the limits to solve this integral
Given:$\int_{\frac{x}{3}}^{\frac{\pi}{3}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x$
Solution:
\begin{aligned} &\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2 \cos ^{2} \frac{x}{2}}}{\left(2 \sin ^{2} \frac{x}{2}\right)^{\frac{3}{2}}} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2} \cos \frac{x}{2}}{2^{\frac{3}{2}}\left(\sin ^{3} \frac{x}{2}\right)} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{2} \cos \frac{x}{2}}{2 \sqrt{2}\left(\sin \frac{x}{2}\right)} \cdot \frac{1}{\left(\sin ^{2} \frac{x}{2}\right)} d x \\ &=\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cot \frac{x}{2} \cos e c^{2} \frac{x}{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{\cos \theta}{\sin \theta}=\cot \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] \end{aligned}
\begin{aligned} &\text { Put } \cot \frac{x}{2}=t \Rightarrow-\cos e c^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t\\ &\Rightarrow \operatorname{cosec}^{2} \frac{x}{2} d x=-2 d t\\ &\text { When } x=\frac{\pi}{3} \text { then } t=\sqrt{3} \& \text { when } x=\frac{\pi}{2} \text { then } t=1\\ &\therefore \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x=\frac{1}{2} \int_{\sqrt{3}}^{1} t \cdot(-2) d t \end{aligned}
\begin{aligned} &=-\int_{\sqrt{3}}^{1} t d t=-\left[\frac{t^{1+1}}{1+1}\right]_{\sqrt{3}}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=-\frac{1}{2}\left[t^{2}\right]_{\sqrt{3}}^{1} \\ &=-\frac{1}{2}\left[1^{2}-(\sqrt{3})^{2}\right] \\ &=-\frac{1}{2}[1-3] \\ &=\frac{-1}{2}[-2] \\ &=\frac{2}{2} \\ &=1 \end{aligned}

Definite Integrals exercise 19.2 question 54.

Answer:$a^2\left [ \frac{\pi}{4}-\frac{1}{2} \right ]$
Hint: Use indefinite formula and the limits to solve this integral
Given:$\int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x$
Solution:
$\int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x$
Put $x^2=a^2\cos 2\theta \Rightarrow 2x dx=a^2-\sin2\theta .2d\theta$
$\Rightarrow dx=-a^2\sin 2\theta x\; d\theta$
When x=0 then $\theta =\frac{\pi}{4}$ & when x=a then $\theta =0$
\begin{aligned} &\therefore \int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x=\int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2}(1-\cos 2 \theta)}{a^{2}(1+\cos 2 \theta)}}\left(-a^{2} \sin 2 \theta\right) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2} 2 \sin ^{2} \theta}{a^{2} 2 \cos ^{2} \theta}}(\sin 2 \theta) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) d \theta \mid \end{aligned}\begin{aligned} &=-a^{2} \int_{\frac{\pi}{4}}^{0}\left(2 \sin ^{2} \theta\right) d \theta \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0}(1-\cos 2 \theta) d \theta \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} 1 d \theta+a^{2} \int_{\frac{\pi}{4}}^{0}(\cos 2 \theta) d \theta \\ &=-a^{2}[\theta]_{\frac{\pi}{4}}^{0}+a^{2}\left[\frac{\sin 2 \theta}{2}\right]_{\frac{\pi}{4}}^{0} \\ &=-a^{2}\left[0-\frac{\pi}{4}\right]+\frac{a^{2}}{2}\left[\sin 2(0)-\sin 2 \times \frac{\pi}{4}\right] \\ &=\frac{a^{2} \pi}{4}-\frac{a^{2}}{2} \cdot 1 \\ &=a^{2}\left[\frac{\pi}{4}-\frac{1}{2}\right] \end{aligned}

Definite Integrals exercise 19.2 question 55.

Answer: $\pi a$
Hint: Use indefinite formula and the limits to solve this integral
Given: $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$
Solution:
$\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$
Put $x=a\cos 2\theta \Rightarrow dx=-2a\sin 2\theta d\theta$
When x=-a then $\theta =\frac{\pi}{2}$ & when x=a then $\theta =0$
\begin{aligned} &\therefore \int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} d x=\int_{\frac{\pi}{2}}^{0} \sqrt{\frac{a-a \cos 2 \theta}{a+a \cos 2 \theta}}(-2 a \sin 2 \theta) d \theta \\ &=-2 a \int_{\frac{\pi}{2}}^{0} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}} 2 \sin \theta \cos \theta d x \\ &=-2 a \int_{\frac{\pi}{2}}^{0} \frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) d \theta \\ &=-2 a \int_{\frac{\pi}{2}}^{0}\left(2 \sin ^{2} \theta\right) d \theta \end{aligned}
\begin{aligned} &=-2 a \int_{\frac{\pi}{2}}^{0}(1-\cos 2 \theta) d \theta \\ &=-2 a \int_{\frac{\pi}{2}}^{0} 1 d \theta+2 a \int_{\frac{\pi}{2}}^{0}(\cos 2 \theta) d \theta \\ &=-2 a[\theta]_{\frac{\pi}{2}}^{0}+2 a\left[\frac{\sin 2 \theta}{2}\right]_{\frac{\pi}{2}}^{0} \\ &=-2 a\left[0-\frac{\pi}{2}\right]+a\left[\sin 2 \times(0)-\sin 2 \times \frac{\pi}{2}\right] \\ &=-2 a\left(\frac{-\pi}{2}\right)+a[\sin 0-\sin \pi] \\ &=\pi a+a(0-0) \\ &=\pi a \end{aligned}

Definite Integrals exercise 19.2 question 56

Answer: $\log \frac{9}{8}$
Hint: Use indefinite integral formula and the limits to solve this integral
Given:$\int_{0}^{\frac{\pi}{2}}\frac{\sin x.\cos x}{\cos^2x+3\cos x+2}dx$
Solution:
$\int_{0}^{\frac{\pi}{2}}\frac{\sin x.\cos x}{\cos^2x+3\cos x+2}dx$
\begin{aligned} &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t\\ &\text { Now } x=0 \text { then } \mathrm{t}=1 \& \text { when } x=\frac{\pi}{2} \text { then } \mathrm{t}=0\\ &\therefore \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\cos ^{2} x+3 \cos x+2} d x=\int_{1}^{0} \frac{t}{t^{2}+3 t+2}(-d t)^{\mid}\\ &=-\int_{1}^{0} \frac{t}{t^{2}+3 t+2} d t=-\int_{1}^{0} \frac{t}{(t+1)(t+2)} d t\\ &=\int_{0}^{1} \frac{t}{(t+1)(t+2)} d t \end{aligned}
To solve this integral, first we need to find it’s partial fraction then integrate it by using indefinite formula.
$\frac{t}{(t+1)(t+2)}=\frac{A}{(t+1)}+\frac{B}{(t+2)}$
\begin{aligned} &\Rightarrow t=\frac{A(t+1)(t+2)}{(t+1)}+\frac{B(t+1)(t+2)}{(t+2)} \\ &\Rightarrow t=A(t+2)+B(t+1) \\ &\Rightarrow t=A t+2 A+B t+B=(A+B) t+(2 A+B) \end{aligned}
Equating coefficient of t and constant resp. , then
1=A+B ...a
0=2A+B⇒B=-2A ...b
Put the value of B in a
1=A+-2A=A-2A=-A
A=-1
From a⇒A+B=1⇒-1+B=1⇒B=1+1=2
∴A=-1,B=2
\begin{aligned} &\frac{t}{(t+1)(t+2)}=\frac{-1}{(t+1)}+\frac{2}{(t+2)}\\ &\therefore \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\cos ^{2} x+3 \cos x+2} d x=\int_{0}^{1} \frac{-1}{t+1} d t+\int_{0}^{1} \frac{2}{t+2} d t\; \; \; \; \; \; \; \; \; \; ..........(i) \end{aligned}
\begin{aligned} &=-[\log (t+1)]_{0}^{1}+2[\log (t+2)]_{0}^{1} \\ &=-[\log 2-\log 1]+2[\log 3-\log 2] \\ &=-\log 2+0+2 \log 3-2 \log 2 \\ &=2 \log 3-3 \log 2 \\ &=\log 3^{2}-\log 2^{3} \\ &=\log 9-\log 8 \\ &=\log \frac{9}{8} \end{aligned}

### Definite Integrals exercise 19.2 question 57

Answer :$\left ( \frac{\log m}{m^2-1} \right )$
Hint: Use indefinite integral formula and the limits to solve this integral
Given: $\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{1+m^2\tan^2x}dx$
Solution: $\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{1+m^2\tan^2x}dx$
$=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin x}{\cos x}}{1+m^{2} \frac{\sin ^{2} x}{\cos ^{2} x}} d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin x}{\cos x}}{\frac{\cos ^{2} x+m^{2} \sin ^{2} x}{\cos ^{2} x}} d x \\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{\sin x}{\cos x} \frac{\cos ^{2} x}{\cos ^{2} x+m^{2} \sin ^{2} x}\right) d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos ^{2} x+m^{2} \sin ^{2} x} d x \end{aligned}
put $sin^2x=t \Rightarrow 2\sin x\cos x dx=dt \sin x \cos x=\frac{dt}{2}$
when x=0 then t=0 and $x=\frac{\pi}{2}$ then t=1
Therefore ,
\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\tan x}{1+m^{2} \tan ^{2} x} d x \\ &=\int_{0}^{1} \frac{1}{\left(1-\sin ^{2} x\right)+m^{2} t} \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{(1-t)+m^{2} t} d t \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(m^{2}-1\right) t+1} d t \\ &\text { put } u=\left(m^{2}-1\right) t+1 \Rightarrow d u=\left(m^{2}-1\right) d t \Rightarrow d t=\frac{1}{m^{2}-1} d u \end{aligned}
\begin{aligned} &=\frac{1}{2} \int_{0}^{1} \frac{1}{u} \frac{1}{\left(m^{2}-1\right)} d u=\frac{1}{2\left(m^{2}-1\right)} \int_{0}^{1} \frac{1}{u} d u \\ &=\frac{1}{2\left(m^{2}-1\right)}[\log |u|]_{0}^{1} \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) t+1\right|\right]_{0}^{1} \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) 1+1\right|-\log \left|\left(m^{2}-1\right) 0+1\right|\right] \end{aligned}
\begin{aligned} &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) 1+1\right|-\log \left|\left(m^{2}-1\right) 0+1\right|\right] \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left(m^{2}-1+1\right)-\log 1\right] \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log m^{2}-0\right] \\ &=\frac{1}{2\left(m^{2}-1\right)} \log m^{2} \\ &=\frac{2 \log m}{2\left(m^{2}-1\right)} \\ &=\frac{\log m}{\left(m^{2}-1\right)} \end{aligned}

Definite Integrals exercise 19.2 question 58

Answer : $\frac{1}{\sqrt{2}}\left ( \tan^{-1}\sqrt{\frac{2}{3}} \right )$
Hint: Use indefinite integral formula and the limits to solve this integral
Given:$\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x$
Solution:$\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x$
put $x=\sin u \Rightarrow dx=\cos \: u\; du$
where x=0 then u=0 and when $x=\frac{1}{2}$ then $u=\frac{\pi}{6}$

Therefore,
\begin{aligned} &\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \sqrt{1-\sin ^{2} u}} \cos u d u \\ &=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \sqrt{\cos ^{2} u}} \cos u d u \\ &=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \cos u} \cos u d u \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{6}} \frac{1}{1+\sin ^{2} u} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\frac{1}{\cos ^{2} u}}{\frac{1+\sin ^{2} u}{\cos ^{2} u}} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{\frac{1}{\cos ^{2} u}+\frac{\sin ^{2} u}{\cos ^{2} u}} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{\sec ^{2} u+\tan ^{2} u} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{1+2 \tan ^{2} u} d u \end{aligned}
again , put $\tan u=t \sec^2u\; du=dt$
when u=0 then t=0 and when $u=\frac{\pi}{6}$ then $t=\frac{1}{\sqrt{3}}$
Therefore ,
\begin{aligned} &\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{1+2 \tan ^{2} u} d u \\ &=\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1+2 t^{2}} d t \\ &=\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{2\left(\frac{1}{2}+t^{2}\right)} d t \\ &=\frac{1}{2} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}} d t \end{aligned}
\begin{aligned} &=\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{3}}} \\ &=\frac{1}{2}[\sqrt{2}]\left[\tan ^{-1}(\sqrt{2} t)\right]_{0}^{\frac{1}{\sqrt{3}}} \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{2} \times \frac{1}{\sqrt{3}}-\tan ^{-1} \sqrt{2} \times 0\right] \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{\frac{2}{3}}-\tan ^{-1} 0\right] \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{\frac{2}{3}}-0\right] \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}} \end{aligned}

Definite Integrals exercise 19.2 question 59

Hint: Use indefinite integral formula and the limits to solve this integral
Given: $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$
Solution:
\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x\\ &=\int_{\frac{1}{3}}^{1} \frac{x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x\ \end{aligned}
Put $\frac{1}{x^2}-1=t\Rightarrow \frac{-2}{x^3}dx=dt\Rightarrow dx=\frac{-x^3}{2}dt$
When $x=\frac{1}{3}$ then t=8 and when x=1 then t=0

\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x \\ &=-\int_{8}^{0} \frac{t^{\frac{1}{3}}}{2} d t \\ &=-\frac{1}{2} \int_{8}^{0} t^{\frac{1}{3}} d t=-\frac{1}{2}\left[\frac{t^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{8}^{0}=-\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{8}^{0} \\ &=-\frac{1}{2} \times \frac{3}{4}\left[t^{\frac{4}{3}}\right]_{8}^{0}=\frac{-3}{8}\left[0^{\frac{4}{3}}-8^{\frac{4}{3}}\right] \\ &=\frac{-3}{8}\left[0-2^{4}\right]=\frac{-3}{8}(-16)=2 \times 3=6 \end{aligned}

Definite Integrals exercise 19.2 question 60

Answer : $\frac{1}{6}$
Hint: Use indefinite integral formula and the limits to solve this integral
Given:$\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx$
Solution:$\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x.\cos ^2x}{(\sin ^3x+\cos ^3x)}dx$
Dividing numerator and denominator by $\cos ^6x$

$\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{6} x}}{\frac{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}}{\cos ^{6} x}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin ^{2} x}{\cos ^{2} x} \cdot \frac{\cos ^{2} x}{\cos ^{4} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \frac{1}{\cos ^{2} x}}{\left(\frac{\sin ^{3} x+\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x$
\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\frac{\sin ^{3} x}{\cos ^{3} x}+\frac{\cos ^{3} x}{\cos ^{3} x}\right)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \end{aligned}
Put $\tan x =t \Rightarrow \sec^2x\; dx=dt$
Now x=0 then t=0 & when $x=\frac{\pi}{4}$ then t=1
\begin{aligned} &\therefore \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} x \cdot \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x=\int_{0}^{\frac{\pi}{4}} \frac{\tan ^{2} x \sec ^{2} x}{\tan ^{6} x+1+2 \tan ^{3} x} d x \\ &=\int_{0}^{1} \frac{t^{2}}{1+t^{6}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{1+\left(t^{3}\right)^{2}+2 t^{3}} d t \\ &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t \end{aligned}

Again putting $1+\frac{t}{3}=u\Rightarrow 3t^2dt=du\Rightarrow t^2dt=\frac{du}{3}$
then
\begin{aligned} &=\int_{0}^{1} \frac{t^{2}}{\left(1+t^{3}\right)^{2}} d t=\int_{0}^{1} \frac{1}{u^{2}} \frac{d u}{3}=\frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}} d u \\ &=\frac{1}{3} \int_{0}^{1} u^{-2} d u=\frac{1}{3}\left[\frac{u^{-2+1}}{-2+1}\right]_{0}^{1} \\ &=\frac{1}{3}\left[\frac{u^{-1}}{-1}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{u}\right]_{0}^{1} \\ &=-\frac{1}{3}\left[\frac{1}{1+t^{3}}\right]_{0}^{1}=-\frac{1}{3}\left[\frac{1}{1+1}-\frac{1}{1+0}\right] \\ &=-\frac{1}{3}\left[\frac{1}{2}-1\right]=-\frac{1}{3}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{6} \end{aligned}

Definite Integrals exercise 19.2 question 61

Answer :$\frac{8}{21}$
Hint: Use indefinite integral formula and the limits to solve this integral
Given:$\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^2x}(\sec^2 x-1)\cos^2xdx$
Solution:
$\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^2x}(\sec^2 x-1)\cos^2xdx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x\left(1-\cos ^{2} x\right)}\left(\sec ^{2} x-1\right) \cos ^{2} x d x \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x \sin ^{2} x}\left(\tan ^{2} x\right) \cos ^{2} x d x \mid \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin x\left(\frac{\sin ^{2} x}{\cos ^{2} x}\right) \cos ^{2} x d x \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin ^{3} x d x \end{aligned}

Put $\cos x=t\Rightarrow -\sin xdx=dt\Rightarrow \sin xdx=-dt$
Now x=0 then t=1 & when $x=\frac{\pi}{2}$ then t=0
\begin{aligned} &\therefore \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x-\cos ^{3} x}\left(\sec ^{2} x-1\right) \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin ^{3} x d x \\ &=\int_{1}^{0} \sqrt{t} \sin ^{2} x(-d t) \\ &=-\int_{1}^{0} \sqrt{t}\left(1-\cos ^{2} x\right) d t \end{aligned}
\begin{aligned} &=-\int_{1}^0 \sqrt{t}\left(1-t^{2}\right) d t=-\int_{1}^0\left(\sqrt{t}-t^{2} \sqrt{t}\right) d t \\ &=-\int_{1}^{0}\left(t^{\frac{1}{2}}-t^{2+\frac{1}{2}}\right) d t=-\int_{1}^{0}\left(t^{\frac{1}{2}}-t^{\frac{5}{2}}\right) d t \\ &=-\int_{1}^{0} t^{\frac{1}{2}} d t+\int_{1}^{0} t^{\frac{5}{2}} d t \\ &=-\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{0}+\left[\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{1}^{0} \end{aligned}
\begin{aligned} &=-\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{0}+\left[\frac{t^{\frac{7}{2}}}{7}\right]_{1}^{0} \\ &=\frac{-2}{3}\left[o^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{7}\left[o^{\frac{7}{2}}-1^{\frac{7}{2}}\right] \\ &=\frac{-2}{3}[0-1]+\frac{2}{7}[0-1] \\ &=\frac{2}{3}-\frac{2}{7} \\ &=\frac{14-6}{21}=\frac{8}{21} \end{aligned}

Definite Integrals exercise 19.2 question 62

Answer : $\frac{2}{2-n}\left [ 2^{1-\frac{n}{2}}-1 \right ]$
Hint: Use indefinite formula and the given limits to solve this integral
Given:$\int_{0}^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x$
Solution: $\int_{0}^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x$
\begin{aligned} &=\int_{0}^{\pi / 2} \frac{\cos ^{2} \frac{x}{2}- \sin ^{2} \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right] \\ &=\int_{0}^{\pi / 2} \frac{\left(\cos _{2}^{x}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \quad\left[\left(\boldsymbol{a}^{2}-b^{2}\right)=(a+b)(a-b)\right] \end{aligned}
$=\int_{0}^{\pi / 2} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n-1}} d x$

Put $\cos \frac{x}{2}+\sin \frac{x}{2}=t \Rightarrow\left(-\sin \frac{x}{2} \cdot \frac{1}{2}+\cos \frac{x}{2} \cdot \frac{1}{2}\right) d x=d t$
\begin{aligned} &\Rightarrow \frac{1}{2}\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x=d t \\ &\Rightarrow\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x=2 d t \end{aligned}
when $x=\frac{\pi}{2}\Rightarrow t =\sqrt{2}$
\begin{aligned} &\therefore \int_{0}^{\pi / 2} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n-1}} d x \\ &=\int_{1}^{\sqrt{2}} \frac{1}{t^{n-1}} 2 d t \\ &=2 \int_{1}^{\sqrt{2}} t^{-(n-1)} d t \\ &=2 \int_{1}^{\sqrt{2}} t^{-n+1} d t \end{aligned}
\begin{aligned} &=2\left[\frac{t^{-n+1+1}}{-n+1+1}\right]_{1}^{\sqrt{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=2\left[\frac{t^{-n+2}}{-n+2}\right]_{1}^{\sqrt{2}} \\ &=\frac{2}{(2-n)}\left[\frac{1}{t^{n-2}}\right]_{1}^{\sqrt{2}} \\ &=\frac{2}{(2-n)}\left[\frac{1}{(\sqrt{2})^{n-2}}-\frac{1}{1^{n-2}}\right] \end{aligned}
\begin{aligned} &=\frac{2}{2-n}\left[\frac{1}{2\left(\frac{n-2}{2}\right)}-1\right] \\ &=\frac{2}{2-n}\left[\frac{1}{2\left(\frac{n}{2}-1\right)}-1\right] \\ &=\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right] \end{aligned}

Class 12, mathematics, chapter 19, Definite Integrals, is a challenging portion where the students get frequent confusion and lose marks. Exercise 19.2 consists of the concept of evaluating the Definite Integrals. There are 62 questions to be solved in this exercise. This set of questions is divided into 49 questions in Level 1 and the remaining 13 questions in Level 2. The students can refer to the RD Sharma Class 12 Chapter 19 Exercise 19.2 material to gain ideas and solve the sums without confusion.

Experts from the educational sector and people with in-depth knowledge in the respective domain have contributed to preparing the solutions in RD Sharma books. It follows the NCERT pattern; this is why it is recommended by most of the CBSE board schools to their students. The solutions in the RD Sharma Class 12th Exercise 19.2 consist of problems solve in every possible method. This gives the freedom for the students to select the way that they feel is easy to adapt.

Definite Integrals is not a chapter that can never be solved; all required is proper practice with a good set of reference materials. For example, the Class 12 RD Sharma Chapter 19 Exercise 19.2 Solution book, consists of various practice questions, excluding the solutions given in the textbook. Once a student works on these additional questions, they tend to get familiarised with the concept.

It is a boon that the RD Sharma Class 12 Solutions Definite Integrals Ex 19.2 are available for students for free of cost at the Career 360 website. They need not spend money to purchase other solution books. The RD Sharma books will serve all the needs. Moreover, the RD Sharma Class 12th Exercise 19.2 books can also be downloaded from the same Career 360 website without monetary payment.

The previous batch students have benefitted a lot by using the RD Sharma Solutions Chapter 19 ex 19.2 to prepare this portion for their exams. As there are high chances of sums being asked from the RD Sharma book, it is wise to prepare with these books from the first day of exam preparation.

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1. Where can I find the RD Sharma books for Class 12 Mathematics, chapter 19?

The RD Sharma Class 12th Exercise 19.2 solution is available at the Career 360 website for free of cost. Anyone can access this set of solution books for various classes and subjects.

2. What are the advantages for the students who use the RD Sharma solution books for reference?
• The solutions given in this book are prepared by experts.

• The solutions are given in various methods.

• Numerous practice questions are given.

3. Are the solutions provided in the RD Sharma books verified?

The solutions given in the RD Sharma books are provided by the experts and verified for accuracy. Therefore, the students need not have any hesitation regarding it.

4. Do the RD Sharma books contain solutions for Level and Level 2 questions for chapter 19, mathematics?

The RD Sharma Class 12th Exercise 19.2 solution book consists of answers for the Level 1 and Level 2 questions given in the textbook.

Yes, the option to download the RD Sharma solution books is given on the Career 360 website.

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