The general solution of the differential equation dy/dx=sin ^-1 x is :

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

so $dy = sin^{-1} dx$  →  integrating both sides

→  $\int dy = \int 1 × sin^{-1}x dx$

using $\int uv ~dx$

$= v \int u dx - \int v' \int u dx ~dx$

$y = x sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$

as $\frac{d}{dx}(\sqrt{1-x^2})$

$= \frac{1 × (-2x)}{2 \sqrt{1-x^2}} = \frac{-x}{\sqrt{1-x^2}}$

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