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$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=$
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The correct answer is:
$\frac{\pi}{2}$
$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=2 \int_0^{\pi / 2} \sin ^2 x d x=2 \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2 \Gamma\left(\frac{2+2}{2}\right)}=\frac{\pi}{2}$
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