$\int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}dx$ for an integer n

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 $$\int_{-\pi}^\pi e^{-\sin^2(x) - A\sin^2(nx)}\,d{x}\approx2\pi\exp\left(-\frac{1+A}{2}\right)I_0\left(\frac{A}{2}\right)I_0\left(\frac{1}{2}\right).$$