_ ^ ^ - 1 x ;dx is equal to - Vidyadip

MCQ

$\int_{}^{} {{{\sin }^{ - 1}}x\;dx} $is equal to

A
$\frac{1}{{\sqrt {1 - {x^2}} }} + c$
B
$x{\sin ^{ - 1}}x - \sqrt {1 - {x^2}} + c$
C
${\cos ^{ - 1}}x + c$
✓
$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + c$

✓

Answer

Correct option: D.

$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + c$

d
(d) $I = \int_{}^{} {{{\sin }^{ - 1}}x} .1\,dx\,dx$
$I = {\sin ^{ - 1}}x.x - \int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}} \,.\,x\,dx$
Put $1 - {x^2} = {t^2} \Rightarrow - 2xdx = 2tdt$ in the second integral and solve it, therefore $I = x{\sin ^{ - 1}}x. + \sqrt {1 - {x^2}} + c$.

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