A) \[-\frac{{{\pi }^{2}}}{2}.log2\]
B) \[\frac{{{\pi }^{2}}}{2}.log2\]
C) \[\pi .log2\]
D) \[{{\pi }^{2}}.log2\]
Correct Answer: A
Solution :
[a] \[\because I=\int\limits_{0}^{\pi }{x.log\left( sinx \right).dx}\] \[=\int\limits_{0}^{\pi }{\left( \pi -x \right)\log \sin \left( \pi -x \right).dx}\] [By property of definite integral] \[=\int\limits_{0}^{\pi }{\pi .\log \sin -.dx}-\int\limits_{0}^{\pi }{x\log x.dx}\,\,\,\,\,\,\,\left[ \therefore \sin (\pi -x)=\sin x \right]\] \[I=\pi \int\limits_{0}^{\pi }{\log \sin x.dx-1}\] \[2I=\int\limits_{0}^{\pi }{\log \sin x.dx}=2\pi \int\limits_{0}^{\frac{\pi }{2}}{\log \sin x}=2\pi (-\frac{\pi }{2}.\log 2)\] \[\therefore I=-\frac{{{\pi }^{2}}}{2}.\log 2\] Hence, option [a] is correct.
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