Can someone help me with this integral? (It involves use of phase shifts) 
Evaluate
  $I=\displaystyle \int_{0}^{\pi/2}{\frac{\sin(x)}{1+\sqrt{\sin(2x)}}}\,dx$.
  $y=\frac{\pi}{2}-x, x=\frac{\pi}{2}-y,dy=-dx$  

$\displaystyle \int_{0}^{\pi/2}{\frac{\sin(x)}{1+\sqrt{\sin(2x)}}}\,dx= \displaystyle \int_{0}^{\pi/2}{\frac{\cos(u)}{1+\sqrt{\sin(2u)}}}\,du$  
$2I=\displaystyle \int_{0}^{\pi/2}{\frac{\cos(u)+\sin(u)}{1+\sqrt{\sin(2u)}}}\,du=\displaystyle \int_{0}^{\pi/2}{\frac{\cos(u)+\sin(u)}{1+\sqrt{\sin(2u)}}}\frac{\cos(u)-\sin(u)}{\cos(u)-\sin(u)}\,du
$  
$=\displaystyle \int_{0}^{\pi/2}{\frac{\cos^2(u)-\sin^2(u)}{(1+\sqrt{\sin(2u)})(\cos(u)-\sin(u))}}\,du\, , v=\cos(u)-\sin(u), v^2=1-\sin(2u)
$  
$2v\,dv=-2\cos(2u)\,du
$  
$
$
 A: Like Integral $\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$
$$I+I=\int_0^{\pi/2}\dfrac{\cos x+\sin x}{1+\sqrt{\sin2x}}dx$$
Set $\sin x-\cos x=u\implies du=?,u^2=?$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{\pi/2}{\sin\pars{x} \over 1 + \root{\sin\pars{2x}}}\,\dd x & =
{\root{2} \over 2}\int_{-\pi/4}^{\pi/4}{\sin\pars{x} + \cos\pars{x} \over 1 + \root{\cos\pars{2x}}}\,\dd x =
\root{2}\int_{0}^{\pi/4}{\cos\pars{x} \over 1 + \root{\cos\pars{2x}}}\,\dd x
\\[5mm] & =
\root{2}\int_{0}^{\pi/4}{\cos\pars{x} \over 1 + \root{1 - 2\sin^{2}\pars{x}}}\,\dd x =
\root{2}\int_{0}^{\root{2}/2}\!\!\!\!\!\!\!\!{\dd x \over 1 + \root{1 - 2x^{2}}}
\end{align}

With
  $\ds{x = {\root{2} \over 4}\,{t^{2} - 1 \over t}\,\ic}$:

\begin{align}
\int_{0}^{\pi/2}{\sin\pars{x} \over 1 + \root{\sin\pars{2x}}}\,\dd x & =
\root{2}\bracks{%
{\root{2} \over 2}\,\ic\int_{1}^{-\ic}{1 + t^{2} \over t\pars{1 + t}^{2}}
\,\dd t} =
\ic\int_{1}^{-\ic}{\dd t \over t} -
2\ic\int_{1}^{-\ic}{\dd t \over \pars{1 + t}^{2}}
\\[5mm] & =
\ic\ln\pars{-\ic} - 2\ic\pars{-\,{1 \over 1 - \ic} + {1 \over 2}} =
\ic\pars{-\,{\pi \over 2}\,\ic} + 2\ic\pars{{1 + \ic \over 2} - {1 \over 2}}
\\[5mm] & =
\bbx{{\pi \over 2} - 1} \approx 0.5708
\end{align}

$\ds{\ln}$ is the $\ds{\log}$-Principal Branch.

