Show that $\int_0^{2\pi} \frac{1 - \cos nt}{1 - \cos t} \,dt = 2\pi n$

Question

I have derived (by Fourier methods) that, for nonnegative integers $n$, $$\int_0^{2\pi} \frac{1 - \cos nt}{1 - \cos t} \,dt = 2\pi n.$$ I can prove this using complex analysis by the substitution $z = e^{it}$, which gives $$ \oint_{|z|=1} \frac{(1 - z^n)(1 - z^{-n})}{(1-z)(1-z^{-1})} \frac{dz}{z} $$ to which we can apply the residue theorem, but I am interested in other proofs.

Answer 1

It may look similar to the solution using the residue theorem, we use $$ \int_0^{2\pi} \frac{(1-e^{int})(1-e^{-int})}{(1-e^{it})(1-e^{-it})}dt = \int_0^{2\pi} (1+e^{it} + \cdots + e^{i(n-1)t})(1+e^{-it}+\cdots + e^{-i(n-1)t})dt. $$