Evaluate $\int_0^1 \frac{1}{\sqrt x+\sqrt {(1-x)}}dx$ Evaluate $I=\int_0^1 \frac{1}{\sqrt x+\sqrt {(1-x)}}dx$.
I applied $x=\sin^2\theta$,that makes $I=\int_0^{\pi/2} \frac{\sin2\theta}{\sin\theta+\cos\theta}d\theta$,but the further proceedings makes $I$ quite tedious.
I need to know is there some elegant transformation which can simplify the calculations.
Any suggestions are heartily welcome
 A: Hint:
$$\sin2\theta=(\sin\theta+\cos\theta)^2-1$$
and $\sin\theta+\cos\theta=\sqrt2\sin\left(\dfrac\pi4+\theta\right)$
Use Integral of $\csc(x)$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over
\root{x} + \root{1 - x}}}
\,\,\,\stackrel{x\ =\ \sin^{2}\pars{\theta}}{=}\,\,\,
\int_{0}^{\pi/2}{\sin\pars{2\theta} \over
\sin\pars{\theta} + \cos\pars{\theta}}\,\dd\theta
\\[5mm] = &\
\int_{0}^{\pi/2}{\sin\pars{2\theta} \over
\sin\pars{\theta} + \tan\pars{\pi/4}\cos\pars{\theta}}
\,\dd\theta =
{\root{2} \over 2}\int_{0}^{\pi/2}{\sin\pars{2\theta} \over
\sin\pars{\theta + \pi/4}}\,\dd\theta
\\[5mm] = &\
{\root{2} \over 2}\int_{-\pi/4}^{\pi/4}{\cos\pars{2\theta} \over
\cos\pars{\theta}}\,\dd\theta =
\root{2}\int_{0}^{\pi/4}\bracks{2\cos\pars{\theta} - \sec\pars{\theta}}\,\dd\theta
\\[5mm] = &\
\root{2}\bracks{\vphantom{\Large A}2\sin\pars{\theta} - \ln\pars{\sec\pars{\theta} + \tan\pars{\theta}}}_{\ 0}^{\ \pi/4}
\\[5mm] = &\
\bbx{2 - \root{2}\ln\pars{1 + \root{2}}}
\approx 0.7536
\end{align}
