$$\int^2_0 x(8-x^3)^{1/3}dx$$ I tried to do integration by substitution and ended up with
$$\int^6_4 (u-4)(u^2+48)^{1/2}(12-u)^{1/2}du$$ where $$x+4=u\,, \ dx=du\,.$$
Then I split it into two integrals hoped to see better a form of it
$$\int^6_4 u (u^2+48)^{1/2}(12-u)^{1/2}du + 4\int^6_4 (u^2+48)^{1/2}(12-u)^{1/2}du$$
Couldn't find the next step, considering I'm on the right track by getting to some 2nd roots from a 3rd root.
If I'm wrong from the start, what is the common or "best" approach to integrals with multiple square roots such as this one ?
Set $y=\frac{x^3}{8}$ so $x=2y^{1/3}$ and the integral is $\int_0^1 2y^{1/3}\cdot 2(1-y)^{1/3}\cdot \tfrac{2}{3}y^{-2/3}dy=\tfrac{8}{3}\mathrm{B}(\tfrac{2}{3},\,\tfrac{4}{3})$. This is$$\tfrac{8}{3}\Gamma (\tfrac{2}{3})\Gamma (\tfrac{4}{3})=\tfrac{8}{9}\Gamma (\tfrac{2}{3})\Gamma (\tfrac{1}{3})=\tfrac{8\pi}{9}\csc\tfrac{\pi}{3}=\tfrac{16\pi}{9}.$$