#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 22

Answer:  $\frac{\pi}{8}+\frac{1}{4}$

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

Given:  $\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x$

Solution:

\begin{aligned} &\mathrm{I}=\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x \\ & \end{aligned}

$I=-\int_{\frac{-\pi}{4}}^{0} \sin ^{2} x d x+\int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$

We will use the formula

So,

\begin{aligned} &x=\frac{1-\cos 2 x}{2} \\ & \end{aligned}

$=-\int_{\frac{-\pi}{4}}^{0} \frac{1-\cos 2 x}{2} d x+\int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x$

\begin{aligned} &=-\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_{\frac{-\pi}{4}}^{0}+\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_{0}^{\frac{\pi}{2}} \\ \end{aligned}

$=-\frac{1}{2}\left(0-\left(-\frac{\pi}{4}+\frac{1}{2}\right)+\frac{1}{2}\left(\frac{\pi}{2}-0\right)\right. \\$

$=-\frac{\pi}{8}+\frac{1}{4}+\frac{\pi}{4} \\$

$=\frac{\pi}{8}+\frac{1}{4}$