I was trying to solve this integral
$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt \, .$$
I checked on WolframAlpha and Gradshteyn and Ryzhik but didn't find anything useful.
To get a first impression I plotted the integrand for $v=1$. So, for $f_{\theta}(t) = e^{\cos \theta \cos t} \cosh(\sin \theta \sin t)$ we get the following plots:
So far, nothing too weird about the plot. I proceeded by trying to numerically evaluate the integral itself.
Turns out, the integral is invariant under $\theta$! Even better, since the integral is independent of $\theta$ we can just set $\theta=0$ and obtain
$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt = \int_0^\pi e^{v \cos t} dt = \pi I_0(v)\quad \forall \theta \in [-\pi,\pi]\; ,$$
where the second equality is a known identity of the modified Bessel function of the first kind. Since I only stumbled upon this identity numerically, I was wondering if someone has some analytical insights into this. Put into a question:
Does someone know, why this identity holds?
Bonus
I now face the same integral but with an additional linear term, that is
$$\int_0^{\pi} tf_{\theta}(t)dt \; ,$$ with $f_{\theta}$ as defined above. I am hoping that the techniques that illuminate the identity above will also shed some light at this new integral, which by the way is not independent in $\theta$ anymore.
For the main problem, a bit of algebra: \begin{align*}e^{v\cos\theta\cos t}\cosh(v\sin\theta\sin t) &= e^{v\cos\theta\cos t}\left(e^{v\sin\theta\sin t}+e^{-v\sin\theta\sin t}\right)\\ &= \frac12\left(e^{v\cos\theta\cos t+v\sin\theta\sin t}+e^{v\cos\theta\cos t-v\sin\theta\sin t}\right)\\ &= \frac12\left(e^{v\cos(\theta-t)}+e^{v\cos(\theta+t)}\right)\end{align*} Now we integrate that: \begin{align*}\int_0^{\pi}e^{v\cos\theta\cos t}\cosh(v\sin\theta\sin t)\,dt &= \frac12\int_0^{\pi}e^{v\cos(\theta-t)}+e^{v\cos(\theta+t)}\,dt\\ &=\frac12\left(\int_0^{\pi}e^{v\cos(\theta+t)}\,dt+\int_{-\pi}^{0}e^{v\cos(\theta+s)}\,ds\right)\\ &=\frac12\int_{-\pi}^{\pi}e^{v\cos(\theta+t)}\,dt = \frac12\int_{-\pi-\theta}^{\pi-\theta}e^{v\cos s}\,ds\end{align*} Flipping $\theta-t$ to $\theta+s$ gives us an integral over the other half of the period - and it's the same function, so we just write it as one integral. Then, in that final integral of $e^{v\cos s}$ over one full period, it doesn't matter where that period is; from $-\pi$ to $\pi$ is the same as from $-\pi-\theta$ to $\pi-\theta$.
That leaves us with the Bessel function identity, that the average value of $e^{v\cos t}$ over a full period is $I_0(\cos v)$. For this, since Bessel functions are defined by a differential equation, we differentiate (under the integral sign): \begin{align*}I(v) &= \frac1{2\pi}\int_0^{2\pi}e^{v\cos\theta}\,d\theta\\ I'(v) &= \frac1{2\pi}\int_0^{2\pi}\cos\theta\cdot e^{v\cos\theta}\,d\theta\\ I''(v) &= \frac1{2\pi}\int_0^{2\pi}\cos^2\theta\cdot e^{v\cos\theta}\,d\theta\\ I'(v) &= \frac1{2\pi}\left[\sin\theta\cdot e^{v\cos\theta}\right]_{\theta=0}^{\theta=2\pi} +\frac1{2\pi}\int_0^{2\pi}\sin\theta\cdot v\sin\theta\cdot e^{v\cos\theta}\,d\theta\\ I'(v) &= \frac{v}{2\pi}\int_0^{2\pi}\sin^2\theta\cdot e^{v\cos\theta}\,d\theta\end{align*} The first three lines are $I$ and its derivatives, calculated the obvious way. Then, in the next two, we apply integration by parts to transform the $I'$ integral into a form that works better with the others. Then, from $\cos^2+\sin^2=1$, we get $vI''(v)+I'(v)-vI(v)=0$, the modified Bessel equation of order zero. Together with the initial condition $I(0)=1$ (since the average value of $1$ is $1$) and $I'(0)=0$, this gives that $I(v)=I_0(v)$. Done.
A brief note on the bonus question: we can apply the same identities, but we run into trouble when we try to fold over and transform the $e^{v\cos(\theta-t)}$ term into an integral over $[-\pi,0]$. The way the $t$ factor transforms, we end up with $$\frac12\int_{-\pi}^{\pi}|t|\cdot e^{v\cos(\theta+t)}\,dt$$ Multiplying by a triangle wave isn't going to come out cleanly. I might look at Fourier series next, but not in this answer.