How to show that $\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{1/2}\,d\theta\,d\phi=\pi$?

Question

$$\int_0^{\pi/2}\int_0^{\pi/2}\left(\frac{\sin\phi}{\sin\theta}\right)^{1/2}\,d\theta\,d\phi=\pi$$ Indeed, I tried to solve this integral by complexifying (using Euler's formula) the $\sin\theta$ and $\sin\phi$.But it didn't work because I faced the exponent which would make things difficult to tackle such integral.

I would appreciate any suggestions for solving this integral.

Answer 1

One may use a classic integral representation of the Euler beta function $$ \int_0^{\pi/2}\sin^a(x)\:dx=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac12+\frac{a}2\right)}{2\,\Gamma\left(1+\frac{a}2\right)} $$ giving $$ \int_0^{\pi/2}\sqrt{\sin(\phi)}\:d\phi=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac34\right)}{2\,\Gamma\left(\frac54\right)},\quad\int_0^{\pi/2}\frac1{\sqrt{\sin(\theta)}}\:d\theta=\frac{\Gamma\left(\frac12\right) \Gamma\left(\frac14\right)}{2\,\Gamma\left(\frac34\right)} $$ and $$ \int_0^{\pi/2}\sqrt{\frac{\sin(\phi)}{\sin(\theta)}}\:d\phi\:d\theta=\Gamma\left(\frac12\right)\cdot\Gamma\left(\frac12\right)=\pi $$ as announced.