A calculus problem by Jose Sacramento

For a function $f$ and real numbers $a < b$ ,

$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.$ Therefore we have:

$I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}\: dx$

$=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{(0+\frac{\pi}{2}-x)}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}\: dx$

$I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}\: dx$

$\therefore 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}\: dx$

$=\int_{0}^{\frac{\pi}{2}}1\: dx$

$=\frac{\pi}{2}.$

$\therefore I=\frac{\pi}{4}=\boxed{0.785}\, ,$

to 3 decimal places.