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

MCQ

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

✓
$\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$
B
$\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x - \frac{1}{2}\log (1 - {x^2}) + c$
C
$\frac{1}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x - \frac{1}{2}\log (1 - {x^2}) + c$
D
$\frac{1}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$

✓

Answer

Correct option: A.

$\frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c$

(a) Put $t = {\sin ^{ - 1}}x \Rightarrow \sin t = x \Rightarrow \cos t\,dt = dx,$ then
$\int_{}^{} {\frac{{{{\sin }^{ - 1}}x}}{{{{(1 - {x^2})}^{32}}}}\,dx} = \int_{}^{} {t{{\sec }^2}t\,dt = t\tan t + \log \cos t + c} $
$ = {\sin ^{ - 1}}x\tan ({\sin ^{ - 1}}x) + \log \cos ({\sin ^{ - 1}}x) + c$
$ = \frac{x}{{\sqrt {1 - {x^2}} }}{\sin ^{ - 1}}x + \frac{1}{2}\log (1 - {x^2}) + c.$

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