The $n$ in the integral is an integer, ranging from $0$ to $\infty$
$$ \int_0^{2\pi}{\exp(A \cos(x) + B \sin(x))(\cos(n(y-x))dx}$$
I am wondering if there is a closed form solution for this type of integral?
I am interested to know if there for some n's for which this integral goes to zero.
This integral can be written in terms of special functions. Consider the following extension of the integral above
$$R_n(z,z^*)=e^{iny}\int_{0}^{2\pi}dx ~\exp(z^*e^{ix}+ze^{-ix}-inx)$$
Then it is clear that the integral we want is given by $$Q=\text{Re}\left(R_n\left(\frac{A+iB}{2}, \frac{A-iB}{2}\right)\right)$$
which is manifestly real. Now notice that
$$R_n(z,z^*)=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(ze^{-ix}+z^*e^{-ix})dx=\\=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(A\cos x+B\sin x)dx\\=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(\sqrt{A^2+B^2}\cos(x-x_0))dx\\=2\pi e^{iny}\frac{\partial^n}{\partial z^n}I_0(4zz^*)\\=2\pi~ 2^n ~e^{in(y-\arg z)}(A^2+B^2)^{n/2}I_0^{(n)}(\sqrt{A^2+B^2})$$
and thus
$$Q=2\pi ~2^n(A^2+B^2)^{n/2} ~\cos(y-\arg(A+iB))I_0^{(n)}(\sqrt{A^2+B^2})$$
where $I_0^{(n)}(x)$ stands for the $n$-th derivative of the modified Bessel function $I_0(x)$. This integral can be identically zero only if
$$y-\arg(A+iB)=\frac{p\pi}{2}~,~ p\in \mathbb{Z}$$
One may be interested in a simpler formula of high derivatives of $I_0$. Indeed, one can show that it can be written in terms of higher-order modified Bessel functions as follows:
$$I_0^{(n)}(x)=\begin{Bmatrix}\frac{1}{2^{n-1}}\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose k}I_{n-2k}(x)~~~ &\text{n odd}\\ -\frac{1}{2^{n}}{n\choose n/2}I_0(x)+\frac{1}{2^{n-1}}\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose k}I_{n-2k}(x)~~~ &\text{n even}\end{Bmatrix}$$