A) \[2\pi {{\log }_{e}}\left( \frac{1}{2} \right)\]
B) \[\pi {{\log }_{e}}2+c\]
C) \[\frac{\pi }{2}{{\log }_{e}}\left( \frac{1}{2} \right)+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{0}^{\pi }{2\log \sin xdx=2\int_{0}^{2\frac{\pi }{2}}{\log \sin xdx=4\int_{0}^{\pi /2}{\log \sin x\,dx}}}\] \[=4\times \left( -\frac{\pi }{2}\log 2 \right)=-2\pi {{\log }_{e}}2=2\pi {{\log }_{e}}\left( \frac{1}{2} \right)\].
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