Can someone help me with this integral? (It involves use of phase shifts)

Question

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 $

$ $

Answer 1

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 $\sin x-\cos x=u\implies du=?,u^2=?$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.