A peculiar integral identity Here I was, innocently trying to solve this daunting-looking integral
$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt $$
when the inner beauty behind this beast slowly started to disclose itself.

Of course, the first thing I did, was checking if WolframAlpha can help, futilely.
Next on my lookup table was Gradshteyn and Ryzhik. Yet again, the god of integrals had no mercy upon me and I was left to my own device.

To get a first impression I plotted the function, that needs to be integrated, that is for $f_{\theta}(t) = e^{\cos \theta \cos t} \cosh(\sin \theta \sin t)$, we get the following plots:

Okay, nothing too special about that. So I proceeded by trying to numerically evaluate the integral itself. And then, something strange happened...

Turns out the integral is invariant under $\theta$!
Even better, we have

$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt = \int_0^\pi e^{v \cos t} dt = \pi I_0(v)\quad \forall \theta \in [-\pi,\pi]\; ,$$

where for the first equality I just set $\theta = 0$ and the second equality is a known identity of the modified Bessel function of the first kind. Now, I only stumbled upon this identity numerically, and I was wondering if someone can share some analytical wisdom regarding 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 constant in $\theta$ anymore.
 A: 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. 
A: Note that
$$\begin{align*}
I_\theta &= \frac14\int_0^{2\pi}e^{v\cos\theta\cos t}\left(e^{v\sin\theta\sin t}+e^{-v\sin\theta\sin t}\right) \mathrm dt\\&=\frac14\int_0^{2\pi}e^{v\cos(t-\theta)} \mathrm dt+\frac14\int_0^{2\pi}e^{v\cos(t+\theta)} \mathrm dt.
\end{align*}$$ Because $t\mapsto  e^{v\cos t}$ is $2\pi$-periodic, we have
$$
\int_0^{2\pi}e^{v\cos(t-\theta)} \mathrm dt=\int_0^{2\pi}e^{v\cos(t+\theta)} \mathrm dt=\int_0^{2\pi}e^{v\cos t} \mathrm dt
$$ so it follows that
$$
I_\theta = \frac 12\int_0^{2\pi}e^{v\cos t} \mathrm dt =\int_0^\pi e^{v\cos t} \mathrm dt.
$$
