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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.

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

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

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