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Π ∫ 0 Cos 2 X Log Sin X D X

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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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पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 49 | पृष्ठ १२२

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