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 ?

Other way is substitution $t=\sin^2\theta$ so $$\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}=\int\limits_0^\frac{\pi}{2} 2dt=\pi$$

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

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

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.