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 ?

\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.
