Integrate $\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $ I found a challenge problem and am confused$$\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $$
$u=\frac{\pi}{2}-x$ is no good and square or 4th power the denominator does not help?  Suggestion?
 A: \begin{align}
\int_0^{\frac{\pi}{2}} \frac{1}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4}{\rm d}x &= \int_0^{\pi/2}\dfrac{\sec^2 x}{(\sqrt{\tan x} + 1)^4}{\rm d}x.
\end{align}
Denote the upper integral by $I$.
Put $u = \tan x$ to get
$$I = \int_0^\infty \dfrac{1}{(1 + \sqrt u)^4}{\rm d}u.$$
Put $u = t^2$ to get
\begin{align}
I &= \int_0^\infty\dfrac{2t}{(1 + t)^4}{\rm d}t.
\end{align}
Puting $v = t+1$ to get
\begin{align}
I &= \int_1^\infty \dfrac{2(v - 1)}{v^4}{\rm d}v\\
&= 2\int_1^\infty \left(\dfrac{1}{v^3} - \dfrac{1}{v^4}\right){\rm d}v\\
&= 2\left(\dfrac{1}{2} - \dfrac{1}{3}\right)\\
&= \boxed{\dfrac{1}{3}}.
\end{align}
A: Define
$$I:=\int_0^{\pi/2} \frac{dx}{\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^4}=\int_0^{\pi/2} \frac{\sec^2 x}{\left(\sqrt{\tan x}+1\right)^4} \:dx$$
Use the change of variable $t=\tan x$ to transform the integral into the following integral:
$$I=\int_0^\infty \frac{dt}{(1+\sqrt{t})^4}$$
Now put $t=y^2$ to obtain
$$I=\int_{0}^\infty \frac{2y}{(1+y)^4}\:dy=2\cdot B(2,2)=\frac{1}{3}$$
where $$B(m,n)=\int_0^\infty \frac{t^{m-1}}{(1+t)^{n-1}} \:dt =\frac{\Gamma(m)\cdot\Gamma(n)}{\Gamma(m+n)}$$
is the Euler integral of the first kind, also known as the Beta function.
