Question

Integrate the following functions w.r.t. x.

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

Answer 1

$LetI=\int \frac{x^3}{\sqrt{1-x^8}}dxPut{x}^4=t\Rightarrow 4x^{3}dx=dt\Rightarrow dx=\frac{dt}{4x^3}\therefore I=\int \frac{x^3}{\sqrt{1-t^2}}\frac{dt}{4x^3}=\frac{1}{4}\int \frac{1}{\sqrt{1-t^2}}dt=\frac{1}{4}si{n}^{-1}t+C[\int \frac{dx}{\sqrt{a^2-x^2}}=si{n}^{-1}\left(\frac{x}{a}\right)]=\frac{1}{4}si{n}^{-1}\left(x^4\right)+C$