MCQ
$\int_0^1 {\log \sin \left( {\frac{\pi }{2}x} \right)} \,dx = $
- ✓$ - \log 2$
- B$\log 2$
- C$\frac{\pi }{2}\log 2$
- D$ - \frac{\pi }{2}\log 2$
As $x = 0$ to $1,$ $\theta = 0$ to $\frac{\pi }{2}$
Then it reduces to
$\frac{2}{\pi }\int_0^{\pi /2} {\,\,\log \sin \theta \,d\theta = \frac{2}{\pi }\left[ { - \frac{\pi }{2}\log 2} \right]} $
$ = - \log 2$.
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