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प्रश्न
उत्तर
\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 - \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{4}\]
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