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