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

MCQ

$\int_0^1 {\frac{{\log x}}{{\sqrt {1 - {x^2}} }}\,dx = } $

A
$\frac{\pi }{2}\log 2$
B
$\pi \log 2$
✓
$ - \frac{\pi }{2}\log 2$
D
$ - \pi \log 2$

✓

Answer

Correct option: C.

$ - \frac{\pi }{2}\log 2$

c
(c) Put $x = \sin \theta ,$ we get

$\int_0^1 {\frac{{\log x}}{{\sqrt {1 - {x^2}} }}dx }$

$={ \int_0^{\pi /2} {\frac{{\log \sin \theta .\cos \theta }}{{\cos \theta }}} } \,d\theta $

$ = \int_0^{\pi /2} {\,\log \sin \theta } \,d\theta = - \frac{\pi }{2}\log 2$.

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