Question

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

Answer 1

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

$Let\ x^4=t ⇒ 4x^3 dx=dt$

⇒∫$$\frac {x^3}{\sqrt {1-x^8}}dx = $\frac 14∫\frac {dt}{\sqrt {1-t^2}}$

                          = $\frac 14sin^{-1}t+C$

                          = $\frac 14sin^{-1}(x^4)+C$