#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 4 maths

Answer: $\frac{\pi }{2}$

Hint: You must know the integration rules of trigonometric function with its limits

Given: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \; d x$

Solution: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \; d x$

\begin{aligned} &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x \\\\ &=\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 1 . d x+\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos 2 x\; d x \end{aligned}

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

\begin{aligned} &=\frac{1}{2}\left[\frac{2 \pi}{2}\right]+\frac{1}{4}[0+0] \\\\ &=\frac{\pi}{2} \end{aligned}