How do I solve this integral $\displaystyle\int_{0}^{2\pi}e^{-\sin^{2}(x)}\cos\left(6x-\frac{\sin(2x)}{2}\right)\,dx$? [duplicate]

I have the integral with $\sin()$ sum expression in $\cos()$ argument: $$\int_0^{2\pi}e^{-\sin^2x}\cos\Bigl(6x-\frac{\sin(2x)}{2}\Bigr)dx.$$ Can anyone please explain an algorithm for solving it?

I've tried so far: Weierstrass substitution (if I can use it with $\int \cos(f(x))dx$), Euler formula, integration by parts, formula of $\cos(x-y)$ and $\sin(2x)$ formula and everything not seems to work or I make actions not in the right order. Need some fresh view on the problem.

Thanks!

$$\begin{split} \int_0^{2\pi}e^{-\sin^2x}\cos\Bigl(6x-\frac{\sin(2x)}{2}\Bigr)dx. &= \Re \int_0^{2\pi}e^{-\sin^2x}e^{6ix}e^{-i\frac{\sin(2x)}2}dx \\ &= \Re \int_0^{2\pi}e^{-\frac 1 2}e^{\frac{\cos (2x)}2}e^{6ix}e^{-i\frac{\sin(2x)}2}dx \\ &= e^{-\frac 1 2}\Re \int_0^{2\pi}e^{6ix}e^{\frac{e^{-2ix}}2}dx \\ &= e^{-\frac 1 2}\Re \int_0^{2\pi}e^{6ix}\left (\sum_{n\geq 0} \frac 1 {n!}\left ( \frac{e^{-2ix}} 2\right)^n \right)dx \\ &=e^{-\frac 1 2}\Re \sum_{n\geq 0} \frac 1 {n!}\frac {1}{2^n} \int_0^{2\pi}e^{6ix}e^{-2inx}dx\\ &=e^{-\frac 1 2} \left ( \frac 1 {3!} \frac {1}{2^3} 2\pi\right )\\ &=\frac{e^{-\frac 1 2}\pi}{24} \end{split}$$ (In case you're wondering how the sum disappeared, notice that all terms in the sum are equal to 0 except for when $n=3$).