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प्रश्न
\[\int \sin^2 \frac{x}{2} dx\]
बेरीज
उत्तर
\[\int \sin^2 \frac{x}{2} dx\]
\[ = \int\left( \frac{1 - \cos x}{2} \right)dx \left[ \therefore \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos x \right)dx\]
\[ = \frac{1}{2}\left[ x - \sin x \right] + C\]
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