Question

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

Answer 1

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

$dy={\sin }^{-1}xdx$

$\int dy=\int {\sin }^{-1}xdx$

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

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

Let $t=1-x^2$

$-2x.dx=dt$

$xdx=\frac{-dt}{2}$

Now

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

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

$y=x{\sin }^{-1}x+\frac{1}{2}\int t^{-1/2}.dt$

$y=x{\sin }^{-1}x+\frac{1}{2}\frac{t^{-1/2+1}}{\frac{-1}{2}+1}+c$

$y=x{\sin }^{-1}x+\frac{1}{2}\frac{t^{1/2}}{1/2}+c$

$y=x{\sin }^{-1}x+\sqrt{t}+c$

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