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Π ∫ 0 Cos 2 X Log Sin X D X

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Question

\[\int\limits_0^\pi \cos 2x \log \sin x dx\]

Sum
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Solution

\[\int_0^\pi \cos2x \log\sin x d x\]

\[ = \left[ \log\sin x \frac{\sin2x}{2} \right]_0^\pi - \int_0^\pi \frac{\cos x}{\sin x}\frac{\sin2x}{2} dx\]

\[ = \left[ \log\sin x \frac{\sin2x}{2} \right]_0^\pi - \int_0^\pi \cos^2 x dx\]

\[ = \left[ \log\sin x \frac{\sin2x}{2} \right]_0^\pi - \int_0^\pi \frac{1 + \cos2x}{2}dx\]

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

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

\[ = - \frac{\pi}{2}\]

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Chapter 19: Definite Integrals - Revision Exercise [Page 122]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Revision Exercise | Q 49 | Page 122

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