Integrating $\int \frac{d\theta}{\sqrt{a-b \sin \theta}}$

Question

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?

Answer 1

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:

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