A) \[-\left( \frac{\pi }{2} \right)\log 2\]
B) \[\pi \log \frac{1}{2}\]
C) \[-\pi \log \frac{1}{2}\]
D) \[\frac{\pi }{2}\log 2\]
Correct Answer: A
Solution :
\[\int_{0}^{\pi /2}{\log \sin x\,dx=\int_{0}^{\pi /2}{\,\,\log \cos x\,dx}}\] Þ \[2I=\int_{0}^{\pi /2}{\log \sin x\cos x\,dx}=\int_{0}^{\pi /2}{\log \sin 2x\,dx}-\int_{0}^{\pi /2}{\,\,\log 2dx}\] \[=\frac{1}{2}\int_{0}^{\pi }{\log \sin tdt-\frac{\pi }{2}\log 2}\], (Putting \[2x=t\]) \[=\frac{1}{2}.2\int_{0}^{\pi /2}{\log \sin t\,dt-\frac{\pi }{2}\log 2}\] \[\Rightarrow 2I=I-\frac{\pi }{2}\log 2\Rightarrow I=\frac{-\pi }{2}\log 2\],\[\left\{ \because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(t)dt}} \right\}\].
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