Analytic formula for integral $\int_{-\pi}^\pi (\cos\theta)^n(t\cos\theta + \sqrt{1-t^2}\sin\theta)^n\mathrm{d}\theta $ in terms of special functions

Question

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. $I_0=2\pi$, $I_1(t) = \pi t$, $I_2(t) = \dfrac{\pi}{4}(2t^2 + 1)$, $I_3(t) = \dfrac{\pi t}{8}(2t^2 + 3)$.

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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

Answer 1

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:

1. $$\int_{-\pi}^\pi \sin^{2m} 2\theta \cos^{2n} 2\theta \, d\theta = \frac{\pi}{2^{2m +2n - 1}}S(m,n) $$

2. 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, $$I_n(t)= \pi\sum_{k = 0}^{\lfloor n/2 \rfloor} \sum_{j = 0}^{k } {n\choose 2k}{2k\choose 2j} S(j, k - j) t^{n - 2k}t^{2j}(1 - t^2)^{k - j}. \frac{1}{2^{n + 2k -1}}$$

Answer 2

Too long for comments.

As usual, there are two patterns depending on the parity of $n$.

For even values of $n$,

$$\color{blue}{I_{2n}=\frac \pi {a_n} P_{2n}(t)}$$ the $a_n$ corresponding to sequence $A224446$ in $OEIS$.

The very first polynomials are

$$\left( \begin{array}{cc} 2n & P_{2n}(t) \\ 0 & 2 \\ 2 & 2 t^2+1 \\ 4 & 8 t^4+24 t^2+3 \\ 6 & 16 t^6+120 t^4+90 t^2+5 \\ 8 & 128 t^8+1792 t^6+3360 t^4+1120 t^2+35 \\ 10 & 256 t^{10}+5760 t^8+20160 t^6+16800 t^4+3150 t^2+63 \\ 12 & 1024 t^{12}+33792 t^{10}+190080 t^8+295680 t^6+138600 t^4+16632 t^2+231 \end{array} \right)$$ For odd values of $n$, $$\color{blue}{I_{2n+1}=\frac \pi {b_n} t\,Q_{2n}(t)}$$ the $b_n$ corresponding to sequence $A061549$ in $OEIS$.

The very first polynomials are

$$\left( \begin{array}{cc} 2n+1 & Q_{2n}(t) \\ 1 & 1 \\ 3 & 2 t^2+3 \\ 5 & 8 t^4+40 t^2+15 \\ 7 & 16 t^6+168 t^4+210 t^2+35 \\ 9 & 128 t^8+2304 t^6+6048 t^4+3360 t^2+315 \\ 11 & 256 t^{10}+7040 t^8+31680 t^6+36960 t^4+11550 t^2+693 \\ 13 & 1024 t^{12}+39936 t^{10}+274560 t^8+549120 t^6+360360 t^4+72072 t^2+3003 \end{array} \right)$$

In both series of polynomials, there are clear patterns which are worth to explore using $OEIS$.

In fact, a CAS produces an ugly antiderivative which involves several hypergeometric functions. I have obtained some expressions but they are so long that they will not fit in the page.

To be contiuned (I hope and wish)