MCQ
$\int_0^{\pi /2} {} \log \sin x\,dx = $
- ✓$ - \left( {\frac{\pi }{2}} \right)\log 2$
- B$\pi \log \frac{1}{2}$
- C$ - \pi \log \frac{1}{2}$
- D$\frac{\pi }{2}\log 2$
==> $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 $
$\Rightarrow I = \frac{{ - \pi }}{2}\log 2$.
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