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