# Evaluate $\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$.

Evaluate following in terms of Gamma function:

$$\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$$

I don't know how to proceed. So, please tell the intuition behind the solution.

Set $$x =a\sin^{\frac{2}{3}}(t) \implies dx = \frac{2a}{3}\frac{\cos t}{\sqrt[3]{\sin t}} dt$$ giving us: $$I = \int_{0}^{a}\sqrt{\frac{x^3}{a^3-x^3}} dx = \frac{2a}{3}\int_{0}^{\frac{\pi}{2}}\sqrt{\frac{a^3\sin^2t}{a^3\cos^2t}}\frac{\cos t}{\sqrt[3]{\sin t}} dt$$ $$=\frac{2a}{3}\int_{0}^{\frac{\pi}{2}}\sin^{\frac{2}{3}}(t) dt$$

Using the result: $$\int_{0}^{\frac{\pi}{2}}\sin^p x \cos^q x dx = \frac{1}{2}B(\frac{1+p}{2}, \frac{1+q}{2})$$ we get, $$I = \frac{2a}{3}\frac{1}{2}B(\frac{5}{6},\frac{1}{2})= \frac{a}{3}B(\frac{1}{2},\frac{5}{6})$$ and since, $$B(l,m) = \frac{\Gamma(l)\Gamma(m)}{\Gamma(l+m)}$$ we have, $$\boxed{I =\frac{a}{3}\frac{\Gamma(\frac{1}{2})\Gamma(\frac{5}{6})}{\Gamma(\frac{4}{3})}}$$

Assuming $a>0$, $$\int_{0}^{a}\sqrt{\frac{x^3}{a^3-x^3}}\,dx \stackrel{x \mapsto az}{=} a\int_{0}^{1}z^{3/2}(1-z^3)^{-1/2}\,dz \stackrel{z \mapsto u^{1/3}}{=}\frac{a}{3}\int_{0}^{1}u^{-1/6}(1-u)^{-1/2}\,du$$ equals $$\tfrac{a}{3}\,B\left(\tfrac{5}{6},\tfrac{1}{2}\right)=a \cdot\frac{\Gamma\left(\tfrac{5}{6}\right)\sqrt{\pi}}{\Gamma\left(\tfrac{1}{3}\right)}=\frac{2^{4/3}\pi a}{\sqrt{3}\,\Gamma\left(\tfrac{1}{3}\right)^3}=\frac{\text{AGM}(2,\sqrt{2+\sqrt{3}})}{2\pi\cdot 3^{1/4}}\cdot a$$ by Legendre duplication formula and the reflection formula. The last equality follows from the relation between $\Gamma\left(\frac{1}{3}\right)$ and an elliptic integral and allows a very efficient numerical evaluation of the LHS through the arithmetic-geometric mean.