_0^ 1/ 2 ^ - 1 x (1 - x^2 ) ^ 3/2 dx = - Vidyadip

MCQ

$\int_0^{1/\sqrt 2 } {\frac{{{{\sin }^{ - 1}}x}}{{{{(1 - {x^2})}^{3/2}}}}dx = } $

A
$\frac{\pi }{4} + \frac{1}{2}\log 2$
✓
$\frac{\pi }{4} - \frac{1}{2}\log 2$
C
$\frac{\pi }{2} + \log 2$
D
$\frac{\pi }{2} - \log 2$

✓

Answer

Correct option: B.

$\frac{\pi }{4} - \frac{1}{2}\log 2$

b
(b) $I = \int_0^{1/\sqrt 2 } {\frac{{{{\sin }^{ - 1}}x}}{{{{(1 - {x^2})}^{3/2}}}}} dx$

Put ${\sin ^{ - 1}}x = t$

$\Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}dx = dt$ and $x = \sin t$

Also $t = 0$ to $\frac{\pi }{4}$

as $x = 0$ to $\frac{1}{{\sqrt 2 }}$

$ \Rightarrow I = \int_0^{\pi /4} {t.{{\sec }^2}t\,dt = \frac{\pi }{4} - \frac{1}{2}\log 2} $.

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