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

Answer

Let $I=\frac{x^{3}}{\sqrt{1-x^{8}}}$
Now, let $x^4 = t $
$\Rightarrow 4x^3 dx = dt$
And $x^3 dx =\frac{dt}{4}$
$\Rightarrow \int \frac{\mathrm{x}^{3}}{\sqrt{1-\mathrm{x}^{8}}} \mathrm{dx}=\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}}\left(\frac{\mathrm{dt}}{4}\right)$
$=\frac{1}{4} \int \frac{1}{\sqrt{1^{2}-t^{2}}} \cdot d t$
$= \frac{1}{4} \sin ^{-1} t+C$
$\Rightarrow \mathrm{I}=\frac{1}{4} \sin ^{-1}\left(\mathrm{x}^{4}\right)+\mathrm{C}$

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