11
$\begingroup$

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 ?

$\endgroup$
  • 2
    $\begingroup$ It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case. $\endgroup$ – Angina Seng Jul 16 '18 at 17:59
  • $\begingroup$ @InterstellarProbe here is an example $\endgroup$ – AgentS Jul 16 '18 at 18:04
  • $\begingroup$ @InterstellarProbe So much the worse for Wolfie! $\endgroup$ – Angina Seng Jul 16 '18 at 18:06
  • 1
    $\begingroup$ $\int\limits_0^1 \dfrac{dt}{\sqrt{t(1-t)}}=\beta(\dfrac14,\dfrac14)=\pi$ $\endgroup$ – Nosrati Jul 16 '18 at 18:16
  • 1
    $\begingroup$ maybe try the substitution $t=\cos \theta$ or $t=\sin\theta$? I haven't tried it but the denominator would simplify $\endgroup$ – usr0192 Jul 16 '18 at 18:18
22
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

\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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

I was taught to use the substitution $x=a \sin^2 \theta+b \cos^2 \theta$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.