Integrate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$ 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 $\pi$.
 A: 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
$$\int_\delta^{\pi/2-\epsilon} 2\,d\theta.$$
A: 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 $a=b=-\frac{1}{2}$
$$\mathrm B\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\Gamma{(1/2)}\, \Gamma{(1/2)}}{\Gamma{(1)}}=\pi$$
A: $$x(1-x)=\frac14-\left(\frac12-x\right)^2$$
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
$$I=-\frac12\int_{1}^{-1} \frac{\mathrm{d}u}{\sqrt{\frac14-\frac{u^2}4}}= \int_{-1}^{1} \frac{\mathrm{d}u}{\sqrt{1-u^2}}=[\arcsin u]_{-1}^{1}=\pi$$
A: $\newcommand{\+}{^{\dagger}}
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$\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}

A: 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 $\arcsin$.
A: $$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}=\int_0^1 \frac{2 d(\sqrt x)}{\sqrt{1-x}}=2\sin^{-1}\sqrt x\ \bigg|_0^1=\pi
$$
A: 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.
A: $$x(1-x)=\frac{-(4x^2-4x)}4=\frac{1-(2x-1)^2}4$$
Set $2x-1=\sin\theta$
$$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x(1-x)}}$$
$$=2\int_{-\frac\pi2}^\frac\pi2\frac{\cos\theta}{\cos\theta}\frac{d\theta}2$$
$$=\frac\pi2-\left(-\frac\pi2\right)$$
