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

MCQ

$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $

✓
$\frac{\pi }{2} - 2\log \sqrt 2 $
B
$\frac{\pi }{2} + 2\log \sqrt 2 $
C
$\frac{\pi }{4} - \log \sqrt 2 $
D
$\frac{\pi }{4} + \log \sqrt 2 $

✓

Answer

Correct option: A.

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

a
(a) Put $x = \tan \theta ,$

$\therefore $ $dx = {\sec ^2}\theta \,d\theta $

As $x = 1 \Rightarrow \theta = \frac{\pi }{4}$ and

$x = 0 \Rightarrow \theta = 0$, then

$I = 2\int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta = 2[\theta \tan \theta ]_0^{\pi /4} - 2\int_0^{\pi /4} {\tan \theta \,d\theta } } $

$= \frac{\pi }{2} + 2\,[\log \cos x]_0^{\pi /4} = \frac{\pi }{2} - 2\log \sqrt 2 $.

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