_0^1 ( 2 x ) ,dx = - Vidyadip

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$

✓

Answer

Correct option: A.

$ - \log 2$

a
(a) Put $\frac{\pi }{2}x = \theta \Rightarrow dx = \frac{2}{\pi }d\theta $;

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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