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प्रश्न
उत्तर
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^2 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 - \cos2x \right)dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 + \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
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