<s>[INST] Integration gamma and beta: $\int_0^4y^3\sqrt{64-y^3}\,\mathrm dy$ How can we evaluate the following integral? $$\int_0^4y^3\sqrt{64-y^3}\,\mathrm dy$$
I can't find anything to substitute because all of the trigonometric identities are in square form... 
 [/INST] $$\begin{align}\int_0^4 dy \, y^3 (64-y^3)^{1/2} &= \underbrace{2048 \int_0^1 dx \, x^3 (1-x^3)^{1/2}}_{y=4 x} \\ &= \frac{2048}{3} \underbrace{\int_0^1 du \, u^{1/3} \, (1-u)^{1/2}}_{x=u^{1/3}}\\ &= \frac{2048}{3} \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{17}{6}\right)} \end{align}$$
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