two dimensional integration of a trigonometric function
Since, the Taylor series expansion for $e^z$ about $z=0$ is
$$e^z = \sum_{n=0}^{\infty}\dfrac{z^n}{n!}$$
then, ignoring the question of convergence, you can say
$$\begin{align*} \int_a^b \int_c^d e^{A\sin(x)\cos(y-B)}dx dy &amp;= \int_a^b \int_c^d  \sum_{n=0}^{\infty}\dfrac{(A\cos(y-B))^n}{n!}\sin^n x \; dx \, dy\\
\\
&amp;= \int_a^b   \sum_{n=0}^{\infty}\dfrac{(A\cos(y-B))^n}{n!}\int_c^d\sin^n x \; dx \, dy\\
\\
&amp;=   \sum_{n=0}^{\infty}\dfrac{A^n}{n!}\int_a^b \cos^n(y-B)\int_c^d\sin^n x \; dx \, dy\\
\\
&amp;=   \sum_{n=0}^{\infty}\dfrac{A^n}{n!}\left(\int_{a-B}^{b-B} \cos^n u\;du\right)\left(\int_c^d\sin^n x \; dx \right)\\
\\
&amp;=  \left(u\big|_{a-B}^{b-B}\right)\left(x\big|_c^d\right) + A\left(\sin u\big|_{a-B}^{b-B}\right)\left(-\cos x\big|_c^d\right) \\
&amp;\quad + \sum_{n=2}^{\infty}\dfrac{A^n}{n!}\left(\int_{a-B}^{b-B} \cos^n u\;du\right)\left(\int_c^d\sin^n x \; dx \right)\\
\end{align*}$$
Using integration by parts, the reduction formulas for the integrals of $\sin^nx$ and $\cos^nu$ work out to be
$$\int \sin^nx \; dx = -\dfrac{1}{n}\sin^{n-1} x \cos x +\dfrac{n-1}{n}\int \sin^{n-2} x \;dx$$
and
$$\int \cos^nu \;du = \dfrac{1}{n}\cos^{n-1} u \sin u +\dfrac{n-1}{n}\int \cos^{n-2} u\; du$$
For $|A|\gg 0$, you could use the Taylor series expansion of $e^z$ about $z=A$ for a series that converges more rapidly for that value of $A$.