Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$ I came across this as one of the shortcuts in my textbook without any proof.
When  $b\gt a$, 

$$\int\limits_a^b \dfrac{dx}{\sqrt{(x-a)(b-x)}}=\pi$$


My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
 A: It's called an Abel Integral ( at least in my language ). You can write that
$$
\frac{1}{\sqrt{\left(x-a\right)\left(b-x\right)}}=\frac{2}{a-b}\frac{1}{\sqrt{1-\left(\frac{2}{a-b}\left(x-\frac{b+a}{2}\right)\right)^2}}$$
that goes into arcsinus

$$\int_{a}^{b}\frac{\text{d}x}{\sqrt{\left(x-a\right)\left(b-x\right)}}=\text{arcsin}\left(\frac{2}{b-a}\frac{b-a}{2}\right)+\text{arcsin}\left(\frac{2}{a-b}\frac{a-b}{2}\right)=2\text{arcsin}\left(1\right)=\pi$$

A: Let $m = \frac{b+a}{2}$ and $r = \frac{b-a}{2}$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y \geq 0$ is given by $y = \sqrt{r^2 - (x-m)^2} = \sqrt{(x-a)(b-x)}$ for $a \leq x \leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ \frac{dy}{dx} = -\frac{x-m}{y}. $$
So the length of the upper-circular arc is
$$ \pi r = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx = \int_{a}^{b} \sqrt{\frac{(x-m)^2 + y^2}{y^2}} \, dx = \int_{a}^{b} \frac{r}{\sqrt{(x-a)(b-x)}} \, dx. $$
Dividing both sides by $r$ gives the desired answer.
A: \begin{align}
  \tan^2 \theta &= \frac{x-a}{b-x} \\
  2\tan \theta \sec^2 \theta \, d\theta &=
  \frac{b-a}{(b-x)^2} \, dx \\
  2\sqrt{\frac{x-a}{b-x}} \times \frac{(x-a)+(b-x)}{b-x} \, d\theta &=
  \frac{b-a}{(b-x)^2} \, dx \\
  2\, d\theta &= \frac{dx}{\sqrt{(x-a)(b-x)}} \\
  \int \frac{dx}{\sqrt{(x-a)(b-x)}} &=
  2\tan^{-1} \sqrt{\frac{x-a}{b-x}}
\end{align}
The singularity in Wolfram Alpha comes from the upper limit $b$.

Geometrical interpretation
Considering circular arc $(x,y)=(\sqrt{b-u},\sqrt{u-a})$
\begin{align}
  ds &= \frac{\sqrt{b-a} \, du}{2\sqrt{(u-a)(b-u)}} \\
  \tan \theta &= \sqrt{\frac{u-a}{b-u}} \\
  \begin{pmatrix}
    x \\ y
  \end{pmatrix} &=
  \begin{pmatrix}
    \sqrt{b-a} \cos \theta \\
    \sqrt{b-a} \sin \theta
  \end{pmatrix} \\
  ds &= \sqrt{b-a} \, d\theta
\end{align}

See also another integral here.
A: I was taught to use the substitution $x=a \sin^2 \theta+b \cos^2 \theta$
A: Other way is substitution $t=\sin^2\theta$ so
$$\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}=\int\limits_0^\frac{\pi}{2} 2dt=\pi$$
