Evaluate $\int_0^1 \frac{1}{\sqrt x+\sqrt {1-x}}dx$

Question

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

Answer 1

Hint:

$$\sin2\theta=(\sin\theta+\cos\theta)^2-1$$

and $\sin\theta+\cos\theta=\sqrt2\sin\left(\dfrac\pi4+\theta\right)$

Use Integral of $\csc(x)$

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} &\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}

Answer 3

Alternatively, substitute $t=\sqrt x-\sqrt{1-x}$. Then, $dt=\frac{\sqrt x+\sqrt {1-x}}{2 \sqrt {x(1-x)}}$ \begin{align} & \int_0^1 \frac{1}{\sqrt x+\sqrt {1-x}}dx\\ =&\int_0^1 \frac{\sqrt x+\sqrt {1-x}}{2\sqrt {x(1-x)}} - \frac{1}{2 \sqrt {x(1-x)}( \sqrt x+\sqrt {1-x})}\ dx\\ =&\int_{-1}^1 \left(1- \frac1{2-t^2}\right)dt =2 - {\sqrt2}\coth^{-1}\sqrt2 \end{align}