I wanna to evaluate the Fourier Series of
$$f(t)=\frac{\sin(\alpha\sin(\omega t))}{\alpha\sin(\omega t)}\exp(-j\beta\sin(\omega t))$$
where $\alpha>0$, $\beta>0$ and $\omega>0$ are constants.
It means that the definite integral $$c_n = \frac{\omega}{2\pi}\int_0^{2\pi/\omega}f(t)e^{-jn\omega t}dt$$ need to be evaluated.
From ref we know the definite integral of $\int_0^{2\pi}\frac{\sin(\sin(x))}{\sin(x)}$, which is corresponding the first term of $f(t)$. And it can be easily found that the definite integral of the second term of $f(t)$, i.e., $\exp(-j\beta\sin(\omega t))$, will lead to the Bessel function. However, when the two terms combine, it seems to be a bit difficult...
Denote $f(t) = x(t)+iy(t)$. Then we have that
$$x(t)\pm y(t) = \frac{\sin(\alpha\sin(\omega t))\cos(\beta\sin(\omega t)) \mp \cos(\alpha\sin(\omega t))\sin(\beta\sin(\omega t))}{\alpha \sin(\omega t)} = \frac{\sin((\alpha\mp \beta)\sin(\omega t))}{\alpha \sin(\omega t)}$$
$$= \frac{1}{\alpha}\sum_{n=0}^\infty\frac{(-1)^n(\alpha\mp\beta)^{2n+1}}{(2n+1)!}\left(\frac{e^{i\omega t}-e^{-i\omega t}}{2i}\right)^{2n} = \frac{\alpha\mp\beta}{\alpha}\sum_{n=0}^\infty\sum_{k=0}^{2n}\frac{\left(\frac{\alpha\mp\beta}{2}\right)^{2n}(-1)^k}{(2n+1)!}{2n \choose k}e^{i2(n-k)\omega t}$$
Immediately the last representation implies that $\hat{f}_l = 0$ for $l$ odd. So now consider $l=2m$. The only term to survive the Fourier series integral is the term $k=n-m$
$$\hat{x}_{2m}\pm\hat{y}_{2m} = \frac{\alpha\mp\beta}{\alpha}\sum_{n=m}^\infty\frac{(-1)^{n-m}}{(2n+1)!}{2n \choose n-m}\left(\frac{\alpha\mp\beta}{2}\right)^{2n}$$
Adding and subtracting these series gives the coefficients for $f$ by linearity.