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Question

\[\int \sin^2 \frac{x}{2} dx\]
Sum
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Solution

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

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Chapter 19: Indefinite Integrals - Exercise 19.06 [Page 36]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.06 | Q 5 | Page 36

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