\begin{align}I&=\int_{0}^{\frac{\pi}{2}}\frac{dx}{\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^2}\\
&=\int_{0}^{\frac{\pi}{4}}\frac{dx}{\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^2}
+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{dx}{\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^2}\\\end{align}
In the second integral perform the change of variable $y=\frac{\pi}{2}-x$,
\begin{align}I&=2\int_{0}^{\frac{\pi}{4}}\frac{dx}{\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^2}\\
\end{align}
Perform the change of variable $y=\tan x$,
\begin{align}
I&=2\int_{0}^{1}\frac{1}{(1+\sqrt{x})^2\sqrt{1+x^2}}\,dx\\\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}I&=\sqrt{2}\int_{0}^{1}\frac{1-\sqrt{1-x^2}}{x^2\sqrt{1+x^2}}\,dx\\
&=-\sqrt{2}\Big[\frac{1-\sqrt{1-x^2}}{x\sqrt{1+x^2}}\Big]_0^1-\sqrt{2}\int_0^1 \frac{\sqrt{1-x^2}-2}{\sqrt{1-x^2}(1+x^2)^{\frac{3}{2}}}\,dx\\
&=-1-\sqrt{2}\int_0^1 \frac{1}{(1+x^2)^{\frac{3}{2}}}\,dx+2\sqrt{2}\int_0^1 \frac{1}{\sqrt{1-x^2}(1+x^2)^{\frac{3}{2}}}\,dx\\
&=-1-\sqrt{2}\left[\frac{x}{\sqrt{1+x^2}}\right]_0^1+2\sqrt{2}\int_0^1 \frac{1}{\sqrt{1-x^2}(1+x^2)^{\frac{3}{2}}}\,dx\\
&=2\sqrt{2}\int_0^1 \frac{1}{\sqrt{1-x^2}(1+x^2)^{\frac{3}{2}}}\,dx-2\\
\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}I&=\int_0^1 \frac{x^2+1+2x}{\sqrt{x}(1+x^2)^{\frac32}}\,dx-2\\
&=\int_0^1 \frac{1}{\sqrt{x}\sqrt{1+x^2}}\,dx+2\int_0^1 \frac{\sqrt{x}}{(1+x^2)^{\frac32}}\,dx-2\\
\end{align}
Perform the change of variable $y=\sqrt{x}$ in both integrals,
\begin{align}I&=2\int_0^1 \frac{1}{\sqrt{1+x^4}}\,dx+4\int_0^1 \frac{x^2}{(1+x^4)^{\frac32}}\,dx-2\end{align}
\begin{align}A&=\int_0^1 \frac{1}{\sqrt{1+x^4}}\,dx\end{align}
Perform the change of variable $y=\frac{1}{x}$,
\begin{align}A&=\int_1^\infty \frac{1}{\sqrt{1+x^4}}\,dx=\int_0^\infty \frac{1}{\sqrt{1+x^4}}\,dx-\int_0^1 \frac{1}{\sqrt{1+x^4}}\,dx\\
&=\int_0^\infty \frac{1}{\sqrt{1+x^4}}\,dx-A
\end{align}
Therefore,
\begin{align}A&=\frac{1}{2}\int_0^\infty \frac{1}{\sqrt{1+x^4}}\,dx\end{align}
In the same manner one obtains,
\begin{align}\int_0^1 \frac{x^2}{(1+x^4)^{\frac32}}\,dx&=\frac{1}{2}\int_0^\infty \frac{x^2}{(1+x^4)^{\frac32}}\,dx\end{align}
Therefore,
\begin{align}I&=\int_0^\infty \frac{1}{\sqrt{1+x^4}}\,dx+2\int_0^\infty \frac{x^2}{(1+x^4)^{\frac32}}\,dx-2\end{align}
Perform the change of variable $y=x^4$,
\begin{align}I&=\frac{1}{4}\int_0^\infty \frac{x^{-\frac34}}{(1+x)^{\frac12}}\,dx+\frac{1}{2}\int_0^\infty \frac{x^{-\frac14}}{(1+x)^{\frac32}}\,dx-2\\
&=\frac{1}{4}\text{B}\left(\frac{1}{4},\frac{1}{4}\right)+\frac{1}{2}\text{B}\left(\frac{3}{4},\frac{3}{4}\right)-2\\
&=\frac{1}{4}\times \frac{\Gamma^2\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{2}\right)}+\frac{1}{2}\times \frac{\Gamma^2\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{2}\right)}-2\\
&=\frac{1}{4}\times \frac{\Gamma^2\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{2}\right)}+\frac{1}{2}\times \frac{\Gamma^2\left(\frac{3}{4}\right)}{\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}-2\\
&=\frac{1}{4}\times \frac{\Gamma^2\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{2}\right)}+\frac{\Gamma^2\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{2}\right)}-2\\
\end{align}
It is well known (Euler's reflection formula) that,
\begin{align}\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\end{align}
Therefore,
\begin{align}\boxed{I=\frac{\Gamma^2\left(\frac{1}{4}\right)}{4\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4}\right)}{\sqrt{\pi}}-2}\end{align}
NB:
$\text{B}$ is the Euler beta function.
Best Answer
$\displaystyle \prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right) =\prod_{n=0}^\infty\left(1-\frac{z}{4n+1}\right)\prod_{n=1}^\infty\left(1+\frac{z}{4n-1}\right)=$
$\displaystyle =(1-z)\prod_{n=1}^\infty\left(\frac{4n(1-\frac{z-1}{4n})}{4n(1+\frac{1}{4n})}\right)\prod_{n=1}^\infty\left(\frac{4n(1+\frac{z-1}{4n})}{4n(1-\frac{1}{4n})}\right)= (1-z)\frac{\prod_{n=1}^\infty\left(1-(\frac{z-1}{4n})^2\right)}{\prod_{n=1}^\infty\left(1-(\frac{1}{4n})^2\right)}$
$\displaystyle =(1-z)\frac{\sin(\frac{z-1}{4}\pi)}{\frac{z-1}{4}\pi}\frac{\frac{1}{4}\pi}{\sin(\frac{1}{4}\pi)}=-\sqrt{2}\sin\left(\frac{\pi z-\pi}{4}\right)= -2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi z-\pi}{4}\right)$
$\displaystyle = \cos\left(\frac{\pi z-\pi}{4}+\frac{\pi}{4}\right) - \cos\left(\frac{\pi z-\pi}{4} -\frac{\pi}{4}\right) = \cos\left(\frac{\pi z}{4}\right) - \sin\left(\frac{\pi z}{4}\right)$
because of $\enspace\displaystyle \sin x\sin y=-\frac{1}{2}\left(\cos(x+y)-\cos(x-y)\right)$
Note:
The decisive step was $\enspace\displaystyle \frac{1}{\sin(\frac{1}{4}\pi)}=2\sin\left(\frac{\pi}{4}\right)\,$ .