Compute $\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$ 
Compute$$\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$$

I think there's no closed form antiderivative of it. I tried WolframAlpha but it didn't help much.
 A: HINT:
Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$$\displaystyle I+I=\int_a^b\{f(x)+f(a+b-x)\}\ dx$$
Now as $\displaystyle\int(\cos x+\sin x)dx=\sin x-\cos x, $ set  $\sin x-\cos x=u\implies1-\sin2x=u^2$ 
A: By setting $x=\frac{t}{2}$, then exploiting symmetry:
$$ I=\int_{0}^{\pi/2}\frac{\cos(x)}{\left(1+\sqrt{\sin(2x)}\right)^2} = \frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1+\cos(x)}+\sqrt{1-\cos(x)}}{\left(1+\sqrt{\sin(x)}\right)^2}\,dx $$
and the last integral can be written in the following form:
$$ I = \frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sin x+2\sqrt{\sin x}}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}}\cdot\frac{dx}{(1+\sqrt{x})^2} $$
or the following one:
$$ I = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-x}(1+\sqrt{x})^2} = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{x}(1+\sqrt{1-x})^2}=\int_{0}^{1}\frac{dx}{\left(1+\sqrt{1-x^2}\right)^2}$$
that also equals
$$ I = \int_{0}^{\pi/2}\frac{\cos t}{(1+\cos t)^2}\,dt = \int_{0}^{2}\frac{\frac{1}{4}-\frac{u^4}{64}}{1+\frac{u^2}{4}}\,du = \frac{1}{16}\int_{0}^{2}\left(4-u^2\right)\,du = \color{red}{\frac{1}{3}}$$
by Weierstrass substitution $t=2\arctan\frac{u}{2}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{}$.

\begin{align}
&\int_{0}^{\pi/2}{\cos\pars{x} \over \bracks{1 + \root{\sin\pars{2x}}}^{2}}
\,\dd x \,\,\,\,\,{\large\substack{x\ \mapsto\ x\ +\ \pi/4 \\[1mm] =}}\,\,\,
\int_{-\pi/4}^{\pi/4}{\cos\pars{x}\cos\pars{\pi/4} - \sin\pars{x}\sin\pars{\pi/4} \over \bracks{1 + \root{\cos\pars{2x}}}^{2}}
\,\,\dd x
\\[5mm] = &\
\root{2}\int_{0}^{\pi/4}{\cos\pars{x}  \over
\bracks{1 + \root{1 -2\sin^{2}\pars{x}}}^{2}}\,\dd x
\,\,\,\stackrel{\sin\pars{x}\ \mapsto\ x}{=}\,\,\,
\root{2}\int_{0}^{\root{2}/2}{\dd x  \over
\bracks{1 + \root{1 - 2x^{2}}}^{2}}
\end{align}

With Euler Substituion
  $\ds{t \equiv  \root{1 - 2x^{2}} - \root{2}x\ic}$ :

\begin{align}
&\int_{0}^{\pi/2}{\cos\pars{x} \over \bracks{1 + \root{\sin\pars{2x}}}^{2}}
\,\dd x =
2\ic\int_{1}^{-\ic}{t^{2} + 1 \over \pars{t + 1}^{4}}\,\dd t =
2\ic\int_{2}^{1 - \ic}
\pars{{1 \over t^{2}} - {2 \over t^{3}} + {2 \over t^{4}}}\,\dd t = \bbx{1 \over 3}
\end{align}
