Evaluate this:

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The first integral has a well-known beta function representation, so the second one should too. The beta function itself is defined as

$B(x,y) = \displaystyle \int_0^1 t^{x-1} (1-t)^{y-1} \, dt$

and satisfies the identity

$\displaystyle B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$

Later on, we'll also use the so-called reflection formula for the gamma function; for non-integer z,

$\Gamma(z) \Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}$

as well as the identity

$\dfrac{\Gamma(z+1)}{\Gamma(z)} = z$

Replace $x\to\sin^{-1}(x)$ in both integrals, so that

$\displaystyle \int_0^{\frac\pi2} \sqrt{\sin(x)} \, dx = \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx$

$\displaystyle \int_0^{\frac\pi2} \frac{dx}{\sqrt{\sin(x)}} = \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}}$

Now replace $x\to\sqrt x$ :

$\displaystyle \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx = \frac12 \int_0^1 x^{-\frac14} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac34, \frac12\right)$

$\displaystyle \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}} = \frac12 \int_0^1 x^{-\frac34} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac14, \frac12\right)$

So, the original integral (which I condense here to a double integral) is

$\displaystyle \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sqrt{\frac{\sin(x)}{\sin(y)}} \, dx \, dy = \frac14 B\left(\frac34, \frac12\right) B\left(\frac14, \frac12\right)$

$\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\Gamma\left(\frac54\right) \Gamma\left(\frac34\right)}$

$\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\frac14 \Gamma\left(\frac14\right) \Gamma\left(\frac34\right)}$