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Question

\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]
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Solution

\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]

\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]

\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 - \cos2x \right) dx\]

\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]

\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]

\[ = \frac{\pi}{4}\]

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Chapter 19: Definite Integrals - Very Short Answers [Page 115]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Very Short Answers | Q 1 | Page 115

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