Integrate $\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $

Question

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?

Answer 1

\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}

Answer 2

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.