question_answer

Evaluate \[\int_{\pi /4}^{\pi /2}{\cos 2x\log \sin x\,dx.}\]

Answer:

Let \[I=\int_{\pi /4}^{\pi /2}{\underset{II}{\mathop{\cos 2x}}\,\log \underset{I}{\mathop{\sin x}}\,\,dx}\]             \[=\left[ 0-\frac{1}{2}\log \left( \frac{1}{\sqrt{2}} \right) \right]-\int_{\pi /4}^{\pi /2}{{{\cos }^{2}}x\,}dx\] \[\therefore \]      \[I=\frac{1}{4}\log 2-\frac{1}{2}\int_{\pi /4}^{\pi /2}{(1+\cos 2x)\,}dx\]             \[=\frac{1}{4}\log 2-\frac{1}{2}\left[ x+\frac{\sin 2x}{2} \right]_{\pi /4}^{\pi /2}\]             \[=\frac{1}{4}\log 2-\frac{1}{2}\left[ \left( \frac{\pi }{2}+0 \right)-\left( \frac{\pi }{4}+\frac{1}{2} \right) \right]\] \[=\frac{1}{4}\log 2-\frac{\pi }{8}-\frac{1}{4}\]