Let two non-negative integers $k,l \in \mathbf{N}$ be given and consider $f(\theta) = \cos^k\theta\sin^l\theta$. The (complex) Fourier series expansion of $f$ is \begin{align*} f(\theta) = \sum_{n\ge0}c_ne^{in\theta} \end{align*} where \begin{align*} c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)e^{-in\theta}\ d\theta. \end{align*} Is there any known elementary formula for $c_n$?
I have tried playing around with various $l$, keeping $k$ fixed. After a long, tedious computation, I started to suspect that this is not trivial. My conjecture so far is that if $n \le k+l$ and that the parity of $n$ and $k+l$ are equal, then \begin{align*} c_n = \frac{i^l}{2^{k+l}}\displaystyle\binom{k+l}{\frac{k+l+n}{2}} 2^{4l} \frac{P_l(n)}{(k+l)(k+l-1)\cdots(k+1)} \end{align*} for some polynomial $P_l(n)$ of degree $l$ whose coefficients depend on $k$. Otherwise, $c_n = 0$.
The conjectured formula above is found in a rather straightforward manner: \begin{align*} c_n &= \frac{1}{2\pi} \int_{-\pi}^\pi \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^k \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^l e^{-in\theta}\ d\theta\\ &= \frac{1}{2^{k+l}i^l}\int_{-\pi}^\pi \sum_{a=0}^k \sum_{b=0}^l (-1)^{l-b}\binom{k}{a}\binom{l}{b} e^{\left(2(a+b)-(k+l+n)\right)i\theta}\ d\theta\\ &= \frac{i^l}{2^{k+l}} \sum_{a=0}^k \sum_{b=0}^l (-1)^{b}\binom{k}{a}\binom{l}{b} \left(\frac{1}{2\pi}\int_{-\pi}^\pi e^{\left(2(a+b)-(k+l+n)\right)i\theta}\ d\theta\right). \end{align*}
We note that since \begin{align*} \frac{1}{2\pi} \int_{-\pi}^\pi e^{im\theta}\ d\theta = \begin{cases}1, &\mbox{if $m = 0$,} \\ 0, &\mbox{otherwise,} \end{cases} \end{align*} we can reduce the double sum above into a single sum but I am having difficulties simplifying the single sum.
Here is the list of $P_l(n)$ for $1 \le l \le 7$: \begin{align*} P_1(n) &= n,\\ P_2(n) &= n^2 - (k+2),\\ P_3(n) &= n^3 - (3k+7)n,\\ P_4(n) &= n^4 - (6k+16)n^2 + 3(k+4)(k+2),\\ P_5(n) &= n^5 - 10(k+3)n^3 + (15k^2 + 100k + 149)n,\\ P_6(n) &= n^6 - (15k+50)n^4 + (45k^2 + 330k + 544)n^2 - 15(k+6)(k+4)(k+2),\\ P_7(n) &= n^7 - (21k+77)n^5 + (105k^2 + 840k + 1519)n^3 - (105k^3 + 1365k^2 + 5439k + 6483)n. \end{align*}
Any idea of whether $P(n)$ is one of the known special polynomials? I tried plugging in the coefficients to OEIS but I didn't get anything. On the other hand, I would be very surprised if people haven't asked this question in the past.
Any help would be greatly appreciated. Thank you!