Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$

Question

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 ?

It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case. — Angina Seng, Jul 16 '18 at 17:59
Ohk.. mother wolfram is rarely wrong, I'll check.. I'm trying to use @LordSharktheUnknown idea, but I don't see any easy way to change bounds right away. I feel kinda stuck, Lord may I know the substitution ? — AgentS, Jul 16 '18 at 18:08
$\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}=\beta(\dfrac14,\dfrac14)=\pi$ — Nosrati, Jul 16 '18 at 18:16
maybe try the substitution $t=\cos \theta$ or $t=\sin\theta$? I haven't tried it but the denominator would simplify — usr0192, Jul 16 '18 at 18:18
@rsadhvika It was my bad! I missed the square root in the denominator when I entered it into Wolframalpha. — SlipEternal, Jul 16 '18 at 18:58

score 23 · Accepted Answer · answered Jul 16 '18 at 18:19

23

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

answered Jul 16 '18 at 18:19

Nosrati

29,995

score 7 · Answer 2 · answered Jul 17 '18 at 07:01

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.

score 6 · Answer 3 · answered Jul 16 '18 at 18:22

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

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

Ng Chung Tak · Answer 4 · 2018-07-17T03:37:12.377

\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}

Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$

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