How do I prove $\int^{\pi}_{0} \frac{\cos nx}{1+\cos\alpha \cos x} \mathrm{d}x = \frac{\pi}{\sin \alpha} (\tan \alpha - \sec \alpha)^n $? The result I wish to show is that for $n \in \mathbb{Z}$, $$\int^{\pi}_{0} \frac{\cos nx}{1+\cos\alpha \cos x} \mathrm{d}x  = \frac{\pi}{\sin \alpha} (\tan \alpha - \sec \alpha)^n $$
I have made a few attempts through the first techniques that came to my mind but I have not made any meaningful progress.

Hint:
Use the following relation
\begin{equation}
\frac{\sin\alpha}{1+\cos\alpha\cos x}=1+2\sum_{k=1}^\infty \left(\frac{\sin\alpha-1}{\cos\alpha}\right)^k\cos(kx)
\end{equation}
It can be obtained by writing cosine in the denominator of the integrand out in exponential form as the following
\begin{equation}
\frac{A^2-B^2}{A^2-2AB\cos x+B^2}=\frac{A}{A-Be^{ix}}+\frac{Be^{-ix}}{A-Be^{-ix}}
\end{equation}
where we use $A^2+B^2=1$, $-2AB=\cos\alpha$, and the geometric series in form of $\frac{1}{1-y}$. Then use the following relations
\begin{equation}
\int_0^\pi\cos nx\cos mx\ dx=\begin{cases} 
   0&, & \text{if}\ \ n\ne m \\[20pt]
   \dfrac{\pi}{2}&,       & \text{if}\ \ n=m
  \end{cases}
\end{equation}
