Is there a (general) solution to the following integral?
$$\int_{-\frac\pi2}^{\frac\pi2} e^{a\sin x} \cos(2^n x)I_0(b\cos(x))dx$$
$I_0$ is a modified Bessel function of the first kind. I started out by finding the sequence for $\cos(2^n x)$ such that I could perform a change of variables. (Error here)
I've been looking in Gradshteyn&Ryzhik, and I found a solution to $n=0$: 6.616.5.
$$\int_{-1}^1e^{-ax}I_0(b\sqrt{1-x^2})dx = 2(a^2+b^2)^{-1/2}\sinh(\sqrt{a^2+b^2})$$
I haven't been able to find a general solution. The Marcum Q function and Laguerre polynomials only address infinite integrals. Section 6.59 seems useful. I imagine I need to perform a change of variables and find a better definition of the cosine double sequence before I can find a solution, as the section on trig functions together with Bessel functions doesn't contain the solution. Since the expansion of $C_n$ is a polynomial in $\cos(x)$, I believe that any formula that solved the following equation would solve this problem.
$$\int_{-1}^{-1}e^{-ax}x^bI_0(c\sqrt{1-x^2})dx$$
This result seems like it would be adjacent to 6.616.5, but I have no idea how that result was obtained.
The problem is related to Evaluating a double integral involving exponential of trigonometric functions. However, the addition of the coefficient in the cosine makes it much more difficult.
Any help would be greatly appreciated. Thanks!
Edit: An update:
I have managed to find the solution to
$$\int_{-1}^{1}e^{-ax}x^bI_0(c\sqrt{1-x^2})dx$$
using integration by parts for positive integers. However, I need to solve it for -1. I'm stuck here for the foreseeable future.
$$\int_{-1}^{1}e^{-ax}x^{-1}I_0(c\sqrt{1-x^2})dx$$
or, equivalently,
$$\int_{-\pi/2}^{\pi/2}e^{-a\cos(\theta)}I_0(c\sin(\theta))d\theta$$
Edit 2: Found an error in my formula for doubling cosine. Was very tired apparently. Here is the correct one: $C_0 = \cos(\theta), C_n = 2C_{n-1}^2-1$. It doesn't change tooo much about how to solve the problem. I can still decompose the integral using this polynomial.