A) \[\frac{\pi }{2}\log \frac{1}{2}\]
B) \[\frac{{{\pi }^{2}}}{2}\log \frac{1}{2}\]
C) \[\pi \log \frac{1}{2}\]
D) \[{{\pi }^{2}}\log \frac{1}{2}\]
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi }{x\log \sin x\,dx}\] ?..(i) = \[\int_{0}^{\pi }{(\pi -x)\log \sin (\pi -x)\,dx}\] ?..(ii) By adding (i) and (ii), we get \[2I=\int_{0}^{\pi }{\pi }\log \sin x\,dx\Rightarrow I=\frac{2\pi }{2}\int_{0}^{\pi /2}{\log \sin \,x\,dx}\] \[=\pi \left( \frac{\pi }{2}\log \frac{1}{2} \right)=\frac{{{\pi }^{2}}}{2}\log \frac{1}{2}\].
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