Question

Evaluate $\int \frac{x{\sin }^{-1}{x}^2}{\sqrt{1-x^4}}dx$

Answer 1

$\int \frac{x{\sin }^{-1}{x}^2}{\sqrt{1-x^4}}dx$

$put{\sin }^{-1}{x}^2=t$

$\frac{1}{\sqrt{1-x^4}}2x{d}^n=dt$

$=\frac{1}{2}\int tdt$

$=\frac{1}{2}\frac{t^2}{2}+c$

$=\frac{1}{4}{\left({\sin }^{-1}{x}^2\right)}^2+c$