Integral $\int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx $

I am not entirely sure the binomial theorem can be avoided, but also the Weierstrass substitution ($(t = \tan x/2)$ transforms your integral into an integral of a rational function from $0$ to infinity, at which point the residue theorem is your friend.

EDIT A similar method is the following: first note that your integrand is even, so the integral is half of the integral from $-\pi$ to $\pi.$ Now, make the substitution $z = \exp(i x),$ so that $\cos x = \frac12\left(z + \frac1z\right),$ and so your integral is (half of) the integral over the unit circle:

$$ \begin{multline}-i \int_C ((1+ a (1+z^2)/2z)(1+z^2)/2z)^n \frac{dz}z\\ = \frac{-i}{2^{2n}} \int_C((az^2 + 2 z + a)(1+z^2))^n/ z^{2n+1} dz \\= \frac{-i}{2^{2n}}\int_C (a + 2 z + + 2az^2 + 2 z^3 + az^4)^n/z^{2n+1} d z.\end{multline}$$

So, your goal in life is to find the coefficient of $z^{2 n}$ in $(a + 2 z + + 2az^2 + 2 z^3 + az^4)^n,$ since that will give you the residue at $0.$

This is a proof through small steps without really explaining my thought process, because at this point I spent way too much time on it and my thoughts are lost in the wind. The end result is a series of coefficients for $\alpha$.

$$ I_n(\alpha) = \frac{1}{\pi} \int_0^\pi (1+\alpha \cos(x))^n \cos(x)^n dx$$

$$ I_n(\alpha) = \frac{1}{\pi} \int_0^\pi \sum_{k=0}^n\binom{n}{k}(\alpha\cos(x))^k \cos(x)^n dx$$

$$ I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}\alpha^k\int_0^\pi \cos(x)^{k+n} dx$$

$$ I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}\alpha^k\int_0^\pi 2^{-k-n}(e^{ix}+e^{-ix})^{k+n} dx$$

$$ I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}2^{-k-n}\alpha^k\int_0^\pi (e^{ix}+e^{-ix})^{k+n} dx$$

From this point on $k+n$ must be even, otherwise the integral is zero:

$$ I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n \frac{1+(-1)^{k+n}}{2}\binom{n}{k}2^{-k-n}\alpha^k \pi \binom{k+n}{\frac{k + n}{2}}$$

$$ I_n(\alpha) = \sum_{k=0}^n\frac{1+(-1)^{k+n}}{2}2^{-n-k}\binom{n}{k}\binom{k+n}{\frac{k + n}{2}} \alpha^k $$