1

I was looking into rotational motion in physics and encountered this integral where a and b are constants.
$$\int\frac{d\theta}{\sqrt{a-b \sin \theta}}$$ I have found that it looks like an elliptic integral, but I am not familiar with them. How would you evaluate this integral?

Travis Willse
  • 108,056
user1244595
  • 21
  • See here. – J.G. Oct 31 '23 at 23:53
  • 2
    Elliptic Integrals don’t have an elementary closed form so in some sense solving it is just transforming it to a standard form and saying “now we have a standard elliptical integral” and call it a day there. Example instead of trying to find a rational number that squares to 2, you just call it $\sqrt{2}$ and leave it at that. Travis’s answer does exactly this. – Sidharth Ghoshal Nov 01 '23 at 02:06

1 Answers1

3

For general $a, b$, the antiderivative cannot be expressed in closed form using only elementary functions.

Substituting $\phi = \frac{\pi}{4} - \frac\theta2$ transforms the integral to $$-2 \int \frac{d\phi}{\sqrt{(a - b) + 2 b \sin^2 \phi}} = -2 \sqrt{a - b} \int \frac{d\phi}{\sqrt{1 - \frac{2 b}{b - a} \sin^2 \phi}} .$$ It really amounts just to giving a name to the antiderivative, but we can write this expression in terms of the incomplete elliptical integral of the first kind, $$F(x; k) := \int_0^x \frac{dt}{\sqrt{1 - k^2 \sin^2 t}},$$ as $$-2 \sqrt{a - b} F\left(\phi, \sqrt{-\frac{2 b}{a - b}}\right) + C = \boxed{-2 \sqrt{a - b} F\left(\frac{\pi}{4} - \frac{\theta}{2}, \sqrt{-\frac{2 b}{a - b}}\right) + C} .$$

N.b. in the special cases $a = \pm b$ we can write a closed-form antiderivative for the integrand:

$$\sqrt{\frac{2}{a}} \operatorname{artanh} \sqrt{\frac{1 \pm \sin \theta}{2}} + C .$$

Travis Willse
  • 108,056
  • 2
    Depending on the skill level of the OP (I’m assuming they are a beginner), It may be useful to write down the expression $-2\sqrt{a-b}F(\phi,\sqrt{\frac{2b}{b-a}})$ and put the limits of integration to drive the point home of how to use elliptic integral syntax. – Sidharth Ghoshal Nov 01 '23 at 02:09
  • @SidharthGhoshal That's a great suggestion; I'll update my answer in the morning. -T – Travis Willse Nov 01 '23 at 03:56
  • Thanks! How did you find the substitution, is there a good way of thinking about it? – user1244595 Nov 01 '23 at 20:19
  • @user1244595 You're welcome. As for the substitution, the key idea is that you want to replace an expression with $\sin$ or $\cos$ with on with $\sin^2$ or $\cos^2$, and a double angle identity achieves exactly that. – Travis Willse Nov 01 '23 at 22:24