_ ^ x ^ - 1 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {x{{\sin }^{ - 1}}x\;dx} = $

✓
$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$
B
$\left( {\frac{{{x^2}}}{2} + \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$
C
$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x - \frac{x}{4}\sqrt {1 - {x^2}} + c$
D
$\left( {\frac{{{x^2}}}{2} + \frac{1}{4}} \right){\sin ^{ - 1}}x - \frac{x}{4}\sqrt {1 - {x^2}} + c$

✓

Answer

Correct option: A.

$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$

a
(a)$\int_{}^{} {x{{\sin }^{ - 1}}xdx = \frac{{{x^2}}}{2}{{\sin }^{ - 1}}x - \int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}.\frac{{{x^2}}}{2}dx + c} } $
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x - \frac{1}{2}\int_{}^{} { - \frac{{(1 - {x^2}) + 1}}{{\sqrt {1 - {x^2}} }}} dx + c$
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\int_{}^{} {\sqrt {1 - {x^2}} dx - \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}dx + c} } $
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + \frac{1}{4}{\sin ^{ - 1}}x - \frac{1}{2}{\sin ^{ - 1}}x + c$
$ = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} - \frac{1}{4}{\sin ^{ - 1}}x$
$ = \left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}} + c$.

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