Question

Integrate the function $\frac{x^3}{\sqrt{1-x^8}}$

Answer 1

Let $x^4=t\Rightarrow 4x^{3}dx=dt$
$\Rightarrow \int \frac{x^3}{\sqrt{1-x^8}}dx=\frac{1}{4}\int \frac{dt}{\sqrt{1-t^2}}$
$=\frac{1}{4}{\sin }^{-1}t+C$
$=\frac{1}{4}{\sin }^{-1}\left(x^4\right)+C$