$\int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}dx$ for an integer n I'm wondering how to integrate expressions of the following type:
$$\int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}dx$$
For an integer $n\neq 1$. I've tried looking at it every way I know how, I was wondering whether anyone has any clever insight? If there was a way to express this in Bessel functions that would also be great.
(I include the Bessel function tag because if A = 0 this integral can be related to a Bessel function)
 A: What I present is not a closed expression but might be interesting nonetheless: we first of all can notice that it's possible to rewrite the expression $\sin^2(x) + A\sin^2(nx)$ as $$\frac{1+A}{2}-\frac{\cos(2x)+A\cos(2nx)}{2}.$$ Thus we have for the integral in question:
\begin{align*}
    \int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}\,d{x}&=\int_{-\pi}^\pi\exp\left(-\frac{1+A}{2}+\frac{\cos(2x)+A\cos(2nx)}{2}\right)\,d{x}\\&=\exp\left(-\frac{1+A}{2}\right)\int_{-\pi}^\pi\exp\left(\frac{\cos(2x)+A\cos(2nx)}{2}\right)\,d{x}.
\end{align*}
After a change of variable $2x\to\theta$ and use of symmetry, this becomes:
\begin{align*}
    &\frac{1}{2}\exp\left(-\frac{1+A}{2}\right)\int_{-2\pi}^{2\pi}\exp\left(\frac{\cos(\theta)+A\cos(n\theta)}{2}\right)\,d{\theta}\\&=\exp\left(-\frac{1+A}{2}\right)\int_{0}^{2\pi}\exp\left(\frac{A}{2}\cos(n\theta)+\frac{1}{2}\cos(\theta)\right)\,d{\theta}.
\end{align*}
There exists a special relation for integrals like this using modified Bessel functions, namely
$$
    \frac{1}{2\pi }\int_0^{2\pi} \exp\bigl(z \cos (n\theta) + y \cos(\theta)\big) \,d{\theta} = I_0(z)I_0(y) + 2\sum_{m=1}^\infty I_m(z)I_{mn}(y),
$$
which after comparing the coefficients yields as final answer:
$$2\pi\exp\left(-\frac{1+A}{2}\right)\left[ I_0\left(\frac{A}{2}\right)I_0\left(\frac{1}{2}\right) + 2\sum_{m=1}^\infty I_m\left(\frac{A}{2}\right)I_{mn}\left(\frac{1}{2}\right)\right].$$
As the modified Bessel $I_n(x)$ function goes quickly to zero for fixed $x$ and $n\to\infty$, your integral is quite well approximated with
$$\int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}\,d{x}\approx2\pi\exp\left(-\frac{1+A}{2}\right)I_0\left(\frac{A}{2}\right)I_0\left(\frac{1}{2}\right).$$
