#### Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 24

Answer:  $6$

Hint: We will check for the nature of function ( even or odd) then will use the property of definition

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x$

Given:  $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x$

Solution:

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

$\mathrm{f}(\mathrm{x})=2 \sin |\mathrm{x}|+\cos |\mathrm{x}| \\$

\begin{aligned} \mathrm{f}(-\mathrm{x})=2 \sin |-\mathrm{x}|+\cos |-\mathrm{x}| \\ \end{aligned}

$=2 \sin |\mathrm{x}|+\cos |\mathrm{x}|=\mathrm{f}(\mathrm{x})$

This shows that f(x) is an even function.

So we use the property :

\begin{aligned} &\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ & \end{aligned}

$\text { if } f(-x)=f(x)$

\begin{aligned} &I=2 \int_{0}^{\frac{\pi}{2}}(2 \sin |x|+\cos |x|) d x \\ \end{aligned}

$I=2 \int_{0}^{\frac{\pi}{2}}(2 \sin x+\cos x) d x$

\begin{aligned} &=-2(\cos x \times 2)_{0}^{\frac{\pi}{2}}+2(\sin x)_{0}^{\frac{\pi}{2}} \\ & \end{aligned}

$=-4(0-1)+2(1-0)=(4+2)=6$