TITLE: Evaluate: $I = \int^{\pi/2}_0 (\sqrt{\sin x}+\sqrt{\cos x})^{-4}dx$
QUESTION [8 upvotes]: Evaluate : $$I = \int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx$$ 
Attempt :
\begin{align}
I&=\int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx\\
&=\int_{0}^{\Large\frac\pi2} \left( \frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x - \cos x}\right)^{4}\ dx
\end{align}
Is it the right way to proceed it? Please guide me. Thanks.

REPLY [18 votes]: \begin{align}
\int_0^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx&=\int_0^{\Large\frac\pi2}\frac1{(\sqrt{\cos x})^4\ (1+\sqrt{\tan x})^4}\ dx\\
&=\int_0^{\Large\frac\pi2}\frac{\sec^2x}{(1+\sqrt{\tan x})^4}\ dx\\
&\stackrel{\color{red}{[1]}}=\int_0^\infty\frac{du}{(1+\sqrt{u})^4}\\
&\stackrel{\color{red}{[2]}}=2\int_0^\infty\frac{t}{(1+t)^4}\ dt\\
&\stackrel{\color{red}{[3]}}=2\cdot\text{B}(2,2)\\
&=\large\color{blue}{\frac13}.
\end{align}

Notes :
$\color{red}{[1]}\;\;\;u=\tan x$
$\color{red}{[2]}\;\;\;t=\sqrt{u}$
$\color{red}{[3]}\;\;\;$Beta function : $\displaystyle\text{B}(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\ dt$ for $\Re(x)>0$ and $\Re(y)>0$