Applications of integrals of rational functions of sine and cosine

Question

I earlier asked this question about conformal equivalence of flat tori with embedded tori.

In the ensuing thread the integral $\displaystyle\int\frac{dx}{R+\cos x}$ occurred. If I'm not mistaken, it was for just that integral that Euler first introduced the tangent half-angle substitution sometimes (probably incorrectly?) called the Weierstrass substitution.

So there is an application to geometry, of an integral of a rational function of sine and/or cosine.

$\Large\mathbb Q$uestion: What other applications of such integrals exist?

Answer 1

Problem 1.17 of Fetter & Walecka, Theoretical Mechanics of Particles and Continua states the following interesting problem:

A uniform beam of particles with energy $E$ is scattered by an attractive central potential

$$V(r) = \begin{cases}0 & r \gt a\\ -V_0 & r \lt a \end{cases}$$

Show that the orbit of a particle is identical with that of a light ray refracted by a sphere of radius $a$ and index of refraction $n=[(E+V_0)/E]^{1/2}$. Prove that the differential cross-section for $\cos{\theta/2} \gt 1/n$ is

$$\frac{d\sigma}{d\Omega} = \frac{n^2 a^2}{4 \cos{\frac12 \theta}} \frac{(n\cos{\frac12 \theta}-1)(n-\cos{\frac12 \theta})}{(1+n^2-2 n \cos{\frac12 \theta})^2} $$

What is the total cross-section?

In order to find the total scatter cross-section, one has to integrate over all solid angles; since the expression for the differential cross-section is independent of azimuth angle $\phi$, you end up with total cross-section

$$\sigma = \frac{n^2 a^2}{2} \int_0^{2 \arccos{(1/n)}} d\theta \, \sin{\frac{\theta}{2}} \frac{(n\cos{\frac12 \theta}-1)(n-\cos{\frac12 \theta})}{(1+n^2-2 n \cos{\frac12 \theta})^2}\\ = n^2 a^2 \int_0^{\arccos{(1/n)}} du \, \sin{u} \frac{(n \cos{u}-1)(n-\cos{u})}{(1+n^2-2 n \cos{u})^2}$$

So here is an example in physics of an integral over a rational function of sines and cosines. In this case, however, the integral is pretty easy because one may substitute $v=\cos{u}$ and turn this into a simple integral over a rational function

$$\sigma = n^2 a^2 \int_{1/n}^1 dv \frac{(n v-1)(n-v)}{(1+n^2-2 n v)^2}$$

One may evaluate this integral using the substitution $w=1+n^2-2 n v$; the result is

$$\sigma = \frac12 a^2$$

Answer 2

To get the total cross section, you also have to integrate over the azimuthal angle. Since there is no azimuthal dependence, that will just result in a factor of $$ 2\pi $$ to the total cross section, for a final result of $$ \sigma = \frac{a^2}{2}\cdot2\pi = \pi a^2$$ This is also the cross-sectional area of the scattering potential.