For $t \in [-1,1]$ and a nonnegative integer $n$, define
$$
I_n(t) := \int_{-\pi}^\pi (\cos\theta)^n(t\cos\theta + \sqrt{1-t^2}\sin\theta)^n\mathrm{d}\theta = \int_{-\pi}^\pi (\cos\theta\sin(\theta + \alpha))^n \mathrm{d}\theta,
$$
where $\alpha := \arcsin(t)$.
Question. What is an analytic formula for $I_n(t)$, perhaps in terms of special functions ?
I'd also be interested in a recursion formula for $I_n$.
Observations. $I_0=2\pi$, $I_1(t) = \pi t$, $I_2(t) = \dfrac{\pi}{4}(2t^2 + 1)$, $I_3(t) = \dfrac{\pi t}{8}(2t^2 + 3)$.
Update
This question has been fully answered by user Azlif below. Here is a follow up question Simplify via special functions, a certain sum involving binomial coefficients
Best Answer
Using the trigonometric identity for multiplication on your last integral we have $$\cos(\theta)\sin(\theta + \alpha) = \frac{1}{2}(\sin (2\theta + \alpha) + \sin \alpha).$$
Thus, the integral can be written as $$\frac{1}{2^n} \int_{-\pi}^\pi (\sin (\alpha + 2\theta) + t)^nd\theta.$$ We can then expand this integral using binomial expansion as follow \begin{align*} \int_{-\pi}^\pi \sum_{k = 0}^n {n \choose k} \sin^{k}(\alpha + 2\theta) t^{n - k}d\theta &= \sum_{k = 0}^n t^{n -k}{n\choose k}\int_{-\pi}^\pi \sin^k(\alpha + 2\theta) d\theta\\ &= \sum_{k = 0}^n t^{n -k} {n\choose k} \int_{-\pi}^\pi(t \sin 2\theta + \sqrt{1 - t^2} \cos 2\theta)^k d\theta \end{align*}
To continue further, we need the following facts:
$$\int_{-\pi}^\pi \sin^{2m} 2\theta \cos^{2n} 2\theta \, d\theta = \frac{\pi}{2^{2m +2n - 1}}S(m,n) $$
If either $m$ or $n$ is odd, then $$\int_{-\pi}^\pi \sin^m 2\theta \cos^n 2\theta \, d\theta = 0.$$
Where $$S(m,n) = \frac{(2m)!(2n)!}{m!n!(m+n)!}$$
We can evaluate the last integral as follow. If $k$ is odd then the integral equals zero, so we only need to find the value when $k$ is even.
For even $k$, we have
\begin{align*} &\int_{-\pi}^\pi(t \sin 2\theta + \sqrt{1 - t^2} \cos 2\theta)^k d\theta\\ &= \sum_{j = 0}^{k/2} {k\choose 2j}\int_{-\pi}^{\pi} t^{2j} (1 - t^2)^{k/2 - j} \sin^{2j} 2\theta \cos^{k - 2j}2\theta \, d\theta\\ &= \sum_{j = 0}^{k/2} {k\choose 2j} S(j,k/2 - j) t^{2j} (1 - t^2)^{k/2 - j} \frac{\pi}{2^{k - 1}}\\ \end{align*} Then, $$I_n(t)= \pi\sum_{k = 0}^{\lfloor n/2 \rfloor} \sum_{j = 0}^{k } {n\choose 2k}{2k\choose 2j} S(j, k - j) t^{n - 2k}t^{2j}(1 - t^2)^{k - j}. \frac{1}{2^{n + 2k -1}}$$