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integrate $$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$

I've started by dividing this into two integrals: $$\int_0^{1/2} \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$ and $$\int_{1/2}^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$

Then I'm trying to find a primitive to $$\int \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$ using substitution. However I don't succeed with this. Using the integral from wolframalpha: https://www.wolframalpha.com/input/?i=integrate+1%2F%28sqrt%28x%281-x%29%29%29

I still fail to find the answer that should be $\pi$.

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    $\begingroup$ Alpha loves to travel through the complex numbers. An unfortunate choice of strategy. $\endgroup$ – André Nicolas May 6 '14 at 7:00
  • $\begingroup$ I find the selection of answers which were downvoted odd. $\endgroup$ – blue May 6 '14 at 17:59
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There is a magic substitution: let $x=\sin^2\theta$. Then $dx=2\sin\theta\cos\theta\,d\theta$, and the bottom is $\sin\theta\cos\theta$. So we want $$\int_{\theta=0}^{\pi/2} 2\,d\theta.$$

Remark: The original integral is actually a convergent improper integral. In the calculation above, we were deliberately sloppy and forgot about that.

If we want to be careful, we will find the limit as $\delta$ and $\epsilon$ approach $0$ from the right of $$\int_\delta^{\pi/2-\epsilon} 2\,d\theta.$$

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  • $\begingroup$ I've seen the substitution $x=sin^2 \theta$ work in many cases where we need to integrate over some $\sqrt{poly(x)}$. $\endgroup$ – WorldSEnder Jul 29 '15 at 16:14
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It's one of the forms of Beta function,

\begin{align} \int_0^1\, x^a\, (1-x)^b\, dx=\mathrm B(a+1,b+1)=\frac{\Gamma{(a+1)}\Gamma{(b+1)}}{\Gamma{(a+b+2)}} \end{align}

For $a=b=-\frac{1}{2}$

$$\mathrm B\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\Gamma{(1/2)}\, \Gamma{(1/2)}}{\Gamma{(1)}}=\pi$$

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$$x(1-x)=\frac14-\left(\frac12-x\right)^2$$

So

$$I=\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}=\int_0^1 \frac{\mathrm{d}x}{\sqrt{\frac14-\left(\frac12-x\right)^2}}$$

Write $\frac{u}2=\frac12-x$, then $\mathrm{d}u=-2\mathrm{d}x$ and

$$I=-\frac12\int_{1}^{-1} \frac{\mathrm{d}u}{\sqrt{\frac14-\frac{u^2}4}}= \int_{-1}^{1} \frac{\mathrm{d}u}{\sqrt{1-u^2}}=[\arcsin u]_{-1}^{1}=\pi$$

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    $\begingroup$ +1 I always like this method for with completion of polynomial. If one does not see the right substitution is kind of helpful. $\endgroup$ – Bman72 May 6 '14 at 14:22
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Integrals of the form $\int \frac{f(x)}{\sqrt{x}} dx$ are often handled with the substitution $u = \sqrt{x}, \frac{dx}{\sqrt{x}}=2\,du$. In this case this yields $\int \frac{2\, du}{\sqrt{1-u^2}} dx$, which you can integrate using $\arcsin$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\dd x \over \root{x\pars{1 - x}}}:\ {\large ?}}$

\begin{align} &\color{#00f}{\large\int_{0}^{1}{\dd x \over \root{x\pars{1 - x}}}} =\overbrace{\int_{0}^{1/2}{\dd x \over \root{x\pars{1 - x}}}} ^{\ds{x = t^{2}}}\ +\ \overbrace{\int_{1/2}^{1}{\dd x \over \root{x\pars{1 - x}}}}^{\ds{x = 1 - t^{2}}} \\[3mm]&=\int_{0}^{\root{2}/2}{2\,\dd t \over \root{1 - t^{2}}} +\int_{\root{2}/2}^{0}{-2\,\dd t \over \root{1 - t^{2}}} =4\ \overbrace{\int_{0}^{\root{2}/2}{\dd t \over \root{1 - t^{2}}}}^{\ds{t = \sin\pars{\theta}}} =4\int_{0}^{\pi/4}\dd\theta =\color{#00f}{\LARGE\pi} \end{align}

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You should be able to integrate $\displaystyle\frac{1}{\sqrt{1-u^2}}$.

Write $x(1-x)$ as $a(x-h)^2+k$ (a standard intermediate algebra task) and then proceed.

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$$x(1-x)=\frac{-(4x^2-4x)}4=\frac{1-(2x-1)^2}4$$

Set $2x-1=\sin\theta$

$$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$

$$=2\int_{-\frac\pi2}^\frac\pi2\frac{\cos\theta}{\cos\theta}\frac{d\theta}2$$

$$=\frac\pi2-\left(-\frac\pi2\right)$$

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    $\begingroup$ @Downvoter, Please pinpoint the mistake $\endgroup$ – lab bhattacharjee May 6 '14 at 16:17

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