#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 25

Answer:   $\frac{\pi^{2}}{8}$

Hint: You must know the rules of solving definite integral.

Given:

$\int_{\frac{-\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d x$

Solution:

\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x \\ \end{aligned} `

$\sin ^{-1}(\sin x)=\left\{\begin{array}{l} x, \quad \frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \\\\ (\pi-x), \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \end{array}\right\}$

\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x+\int_{\frac{\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d \\ & \end{aligned}

$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x d x+\int_{\frac{\pi}{2}}^{\pi}(\pi-x) d x$                             $\quad\left[\sin ^{-1}(\sin x)=x\right]$

\begin{aligned} &I=\left(\frac{x^{2}}{2}\right)_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\left(\pi x-\frac{x^{2}}{2}\right)_{\frac{\pi}{2}}^{\pi} \\ \end{aligned}

$I=\frac{1}{2}\left(\frac{\pi^{2}}{4}-\frac{\pi^{2}}{4}\right)+\left(\pi \times \pi-\frac{\pi^{2}}{2}-\left(\pi \times \frac{\pi}{2}-\frac{\frac{\pi^{2}}{4}}{2}\right)\right) \\$

$=\pi^{2}-\frac{\pi^{2}}{2}-\frac{\pi^{2}}{2}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{8}$