Define $f:\mathbb{R}^2 \to \mathbb{R}$ by:
$$f(x,y) = \int_{0}^{2 \pi} \frac{\exp(\cos(x \theta+y))-1}{\cos(x \theta+y)\exp(\cos(x \theta+y))}d \theta $$
I was very much hoping to be able to write $f(x,y)$ without the integral there, any chance this is able to be simplified somehow? Wolfram has trouble doing definite integrals with symbolic manipulations.
I don't even need a completely closed form expression, I would be happy to know $f$ is a Bessel function or some such thing.
(Too long for a comment)
A slightly more general observation is as follows: If we define
$$A(x,z):=\int_{0}^{x}\frac{e^{z\cos\theta}-1}{z\cos\theta}\,d\theta,$$
then $A(x,z)=\sum_{a,b=0}^{\infty}\frac{(z/2)^{a+b}}{a!b!(a+b+1)}x\operatorname{sinc}((a-b)x)$. From this, we can verify that $A$
$$A(x+\pi,z)=A(\pi,z)+A(x,-z).$$
In particular, using $A(\pi,-z)=\pi\sum_{n=0}^{\infty}\frac{(z/2)^{2n}}{n!^2(2n+1)}=A(\pi,z)$, we obtain $A(n\pi,z)=nA(\pi,z)$ for any integer $n$. This can be used to derive a special value of
$$f(x,y)=\frac{A(2\pi x+y,-1)-A(y,-1)}{x}, $$
such as $f(\frac{n}{2}, 0) = 2A(\pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $y\in\mathbb{R}$.