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प्रश्न
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उत्तर
\[Let I = \int_0^\frac{\pi}{2} \left( 2 \log \cos x - \log\sin2x \right) d x\]
\[ = \int_0^\frac{\pi}{2} \left[ 2 \log \cos x - \log\left( 2\sin x \cos x \right) \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left( 2\log\cos x - \log2 - \log\sin x - \log\cos x \right)dx\]
\[ = \int_0^\frac{\pi}{2} \left( \log\cos x - \log2 - \log\sin x \right)dx\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\sin x dx\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\sin\left( \frac{\pi}{2} - x \right) dx ..........................\left[\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right]\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\cos x dx\]
\[ = - \log2 \left[ x \right]_0^\frac{\pi}{2} \]
\[ = - \frac{\pi}{2} \log2\]
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