Stanford Mathematics Tournament.

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {dx}{(\sqrt{\sin x}+\sqrt{\cos x})^4} \\ & = \int_0^\frac \pi 2 \frac {dx}{\cos^2 x(\sqrt{\tan x}+1)^4} & \small \color{#3D99F6} \text{Let }u = \tan x \implies du = \sec^2 x \ dx \\ & = \int_0^\infty \frac {du}{(\sqrt u+1)^4} & \small \color{#3D99F6} \text{Let }u = t^2 \implies du = 2t \ dt \\ & = \int_0^\infty \frac {2t \ dt}{(t+1)^4} \\ & = 2 \int_0^\infty \left(\frac 1{(t+1)^3} - \frac 1{(t+1)^4}\right) dt \\ & = 2 \left[- \frac 1{2(t+1)^2} + \frac 1{3(t+1)^3}\right]_0^\infty \\ & = \frac 13 \approx \boxed{0.333} \end{aligned}$