$\cfrac{{dy}}{{dx}} = {\sin ^{ - 1}}x$

.: We have, $\cfrac{{dy}}{{dx}} = {\sin ^{ - 1}}x \Rightarrow dy = {\sin ^{ - 1}}xdx$

…(1)
Integrating (1) both sides,

we get $\int d y = \int {{{\sin }^{ - 1}}} xdx$

$\Rightarrow y = {\sin ^{ - 1}}x\int {1dx} - \int {\left( {\cfrac{{dy}}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\int {1dx} } \right)dx}$

(using by parts)
$\Rightarrow y = x{\sin ^{ - 1}} - \int {\cfrac{x}{{\sqrt {1 - {x^2}} }}} dx$

$\Rightarrow y = x{\sin ^{ - 1}}x + \cfrac{1}{2}\int {\cfrac{{( - 2x)dx}}{{\sqrt {1 - {x^2}} }}}$

[Let $1 - {x^2} = t \Rightarrow - 2xdx = dt$]

$\Rightarrow y = x{\sin ^{ - 1}}x + \cfrac{1}{2}\int {\cfrac{1}{{\sqrt t }}} dt \Rightarrow y = x{\sin ^{ - 1}}x + \cfrac{1}{2}\int {{t^{ - 1/2}}} dt$

$\Rightarrow y = x{\sin ^{ - 1}}x + \cfrac{1}{2}\cfrac{{{t^{1/2}}}}{{\cfrac{1}{2}}} + C \Rightarrow y = x{\sin ^{ - 1}}x + \sqrt t + C$

$\Rightarrow y = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C$,

which is the required solution.