A) \[-\frac{\pi }{2}\log 2\]
B) \[\frac{\pi }{2}\log 2\]
C) \[\pi \log 2\]
D) \[-\pi \log 2\]
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi /2}{x\cot x\,dx}\] Integrating by parts, we get \[[x(\log \sin x)]_{0}^{\pi /2}-\int_{0}^{\pi /2}{\log \sin x\,dx}\] \[I=-\left( -\frac{\pi }{2}\log 2 \right)=\frac{\pi }{2}\log 2\].
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