#### Need solution for RD Sharma Maths Class 12 Chapter 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}