The first integral has a well-known beta function representation, so the second one should too. The beta function itself is defined as
[tex]B(x,y) = \displaystyle \int_0^1 t^{x-1} (1-t)^{y-1} \, dt[/tex]
and satisfies the identity
[tex]\displaystyle B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}[/tex]
Later on, we'll also use the so-called reflection formula for the gamma function; for non-integer z,
[tex]\Gamma(z) \Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}[/tex]
as well as the identity
[tex]\dfrac{\Gamma(z+1)}{\Gamma(z)} = z[/tex]
Replace [tex]x\to\sin^{-1}(x)[/tex] in both integrals, so that
[tex]\displaystyle \int_0^{\frac\pi2} \sqrt{\sin(x)} \, dx = \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx[/tex]
[tex]\displaystyle \int_0^{\frac\pi2} \frac{dx}{\sqrt{\sin(x)}} = \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}}[/tex]
Now replace [tex]x\to\sqrt x[/tex] :
[tex]\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) [/tex]
[tex]\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)[/tex]
So, the original integral (which I condense here to a double integral) is
[tex]\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)[/tex]
[tex]\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\Gamma\left(\frac54\right) \Gamma\left(\frac34\right)}[/tex]
[tex]\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)}[/tex]
[tex]\displaystyle = \Gamma\left(\frac12\right)^2 = \frac{\pi}{\sin\left(\frac\pi2\right)} = \boxed{\pi}[/tex]
