Question

y^2(1-x^2)^(1/2) dy =arcsin x dx

Answer 1

divide both sides by (1-x^2)^(1/2), considering its very easy to separate variables here.
\[y^{2}dy = \frac{ arcsinx }{ \sqrt{1-x^{2}} }dx\]Now that y and x are separated, we just integrate. The left integration is easy, it becomes y^3/3. As for the right hand side,
let u = arcsinx
\[du = \frac{ 1 }{ \sqrt{1-x^{2}} }dx\]
\[dx= du \sqrt{1-x^{2}}\]\[\frac{ y^{3} }{ 3 }=\int\limits_{}^{}\frac{ u }{ \sqrt{1-x^{2}} }*\sqrt{1-x^{2}}du\] which becomes:

\[\frac{ y^{3} }{ 3 }=\frac{ u^{2} }{ 2 }+C\]
But since u = arcsinx, we can finally get
\[\frac{ y^{3} }{ 3 }=\frac{ (arcsinx)^{2} }{ 2 }+C\]