Beta Gamma functions How to convert this integral into Beta Gamma function?
Any hints will be appreciated.
$$\int_0^{b} x \sqrt{b^3-x^3} dx$$ Thanks
 A: Set $x =bt$ so that, $dx =bdt$
$$I:=\int_0^{b} x \sqrt{b^3-x^3} dx = \int_0^{1} b^2t \sqrt{b^3-b^3t^3} dt = b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt$$
Setting $u=t^3~~$ i.e $~~t= u^{1/3}~~$ yields $dt =\frac{1}{3}u^{-2/3}du~~~$ therefore,
$$I=  b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt =  \frac{1}{3}b^{7/2}\int_0^{1} u^{-1/3} (1-u)^{1/2} du\\=\frac{1}{3}b^{7/2}\int_0^{1} u^{2/3-1} (1-u)^{3/2-1} du\\=\frac{1}{3}b^{7/2} B\left(\frac{2}{3},\frac{3}{2}\right)$$
But 
$$B\left(\frac{2}{3},\frac{3}{2}\right) = \frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{2}{3}+\frac{3}{2}\right)} =\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(1+\frac{1}{2}\right)}{\Gamma\left(2+\frac{1}{6}\right)}=\frac{\frac{1}{2}\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{2}\right)}{(1+\frac{1}{6})\frac{1}{6}\Gamma\left(\frac{1}{6}\right)}\\=
\color{red}{\sqrt{\pi}}\frac{18}{7}\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)}$$
Since$$\Gamma\left(\frac{1}{2}\right) =\sqrt{\pi}$$
Conclusion 
$$\color{blue}{\int_0^{b} x \sqrt{b^3-x^3} dx =\color{red}{\sqrt{\pi}}b^{\frac{7}{2}}\frac{6}{7}\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)} }$$
A: *

*notice that a change of variable $x=by$ will pull $b$ out of the integral (both integrand and bounds).

*then notice that the change of variable $z=y^3$ will let the binomial $1-z$ appear under the radical, while other factor(s) will be powers of $y$, directly leading to a Beta integral.
$$\int _0^b x\sqrt{b^3-x^3}dx=b^{7/2}\int _0^1 y\sqrt{1-y^3} dy=\frac13b^{7/2}\int _0^1 z^{-1/3}\sqrt{1-z}\, dz.$$
