Question

${\int }_0^1{x}^2\sqrt{\frac{1}{1-x^3}}dx=$

Answer 1

We have,

$I={\int }_0^1{x}^2\sqrt{\frac{1}{1-x^3}}dx$

Let

$t=1-x^3$

$\frac{dt}{dx}=0-3x^2$

$-\frac{dt}{3}=x^{2}dx$

Therefore,

$I=-\frac{1}{3}{\int }_1^0\sqrt{\frac{1}{t}}dt$

$I=-\frac{1}{3}{\left[2\sqrt{t}\right]}_1^0$

$I=-\frac{2}{3}[\sqrt{0}-\sqrt{1}]$

$I=-\frac{2}{3}[-1]$

$I=\frac{2}{3}$

Hence, this is the answer.