# Prove $\int_0^\pi\frac{\cos(n\theta)}{\cos\theta - \cos\theta_0}d\theta = \frac{\pi\sin(n\theta_0)}{\sin\theta_0}$

I ran across the following formula in a textbook but can't figure out how to prove it. How would I go about solving this?

$$\int_0^\pi\frac{\cos(n\theta)}{\cos\theta - \cos\theta_0}d\theta = \frac{\pi\sin(n\theta_0)}{\sin\theta_0}$$

where $$\theta_0\in[0,\pi]$$.

My initial attempt was to use integration by parts for $$n=1$$ and then try induction, but it didn't work out for me. It's also clear that the integral will be improper at $$\theta=\theta_0$$, so we should handle that somehow.

• That integral is divergent due to the singularity, $\frac1{\cos(\theta)-\cos(\theta_0)}=-\frac1{\sin(\phi)(\theta -\theta_0)}$, where $\phi\in (0,\pi)$. – Mark Viola Mar 6 at 20:06

## 1 Answer

This problem appears in Paul J. Nahin, Inside Interesting Integrals, Springer, 2015, p. 60. I'll write the full details later.

Let $$I_n(\theta_0) = \int_0^\pi \frac{\cos(n\theta)}{\cos \theta - \cos \theta_0} \;\mathrm{d}\theta$$ Then using trigonometric identities one can show that $$I_{n + 1}(\theta_0) - 2\cos\theta_0 \cdot I_n(\theta_0) + I_{n - 1}(\theta_0) = \int_0^\pi \frac{2\cos(n\theta) \cdot (\cos\theta - \cos\theta_0)}{\cos\theta - \cos\theta_0} \;\mathrm{d}\theta = \frac{\sin(n\pi)}{n} = 0$$ since $$n$$ is integer.

The conclusion now easily follows by induction.

• Nahin's is a great book, but it does tend to make proofs unnecessarily long. +1 for condensing it. – J.G. Mar 6 at 19:38