<s>[INST] Prove that $\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\pi}{9\sqrt{3}}$ I have to prove that: 

$$\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\,\pi}{9\sqrt{3}}.$$

I tried substituting $x^3=8u$ but I just got stuck.
Any help would be appreciated.
 [/INST] By setting for first $x=2z$ we have:
$$I=\int_{0}^{2}x(8-x^3)^{1/3}\,dx = 8\int_{0}^{1}z(1-z^3)^{1/3}\,dz \tag{1}$$
and by setting $z=t^{1/3}$ we have:
$$I=\frac{8}{3}\int_{0}^{1}t^{-1/3}(1-t)^{1/3}\,dt = \frac{8}{3}\cdot\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma(2)}\tag{2}$$
through Euler's Beta function. Exploiting the reflection formula for the $\Gamma$ function:
$$ I = \frac{8}{9}\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=\frac{8\pi}{9\sin\frac{\pi}{3}}=\color{red}{\frac{16\,\pi}{9\sqrt{3}}}\tag{3}$$ as wanted.
</s>