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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 3

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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}} \sin ^{2} x \; d x

Solution:  \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{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 \cdot d x-\frac{1}{2} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos 2 x\; d x \end{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}}

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


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