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$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x=$
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Let $f(x)=\sin ^{2} x$
$\therefore \mathrm{f}(-\mathrm{x})=[\sin (-\mathrm{x})]^{2}=\sin ^{2} \mathrm{x}$
Thus $\sin ^{2} x$ is an even function.
$\therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$
$=2 \int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1-\cos 2 x) d x=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\left(\frac{\pi}{2}-0\right)-\left(\frac{\sin \pi}{2}-\frac{\sin 0}{2}\right)=\frac{\pi}{2}$
$\therefore \mathrm{f}(-\mathrm{x})=[\sin (-\mathrm{x})]^{2}=\sin ^{2} \mathrm{x}$
Thus $\sin ^{2} x$ is an even function.
$\therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$
$=2 \int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1-\cos 2 x) d x=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\left(\frac{\pi}{2}-0\right)-\left(\frac{\sin \pi}{2}-\frac{\sin 0}{2}\right)=\frac{\pi}{2}$
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