$$\ \int\_0^{\pi} \bigl(\sin(x)\bigr)^{2n-2k+1} e^{a\cos(x)} dx , \qquad a,n,k\in\mathbb Z.$$
I tried to solve this integral by parts, but I didn't get any result. I look forward to your experience.



Integration of the product of sin and exponential with power
$$\int\_0^{\pi} (\sin x)^{2n-2k+1} e^{a\cos x} dx =\int\_{-1}^1 (1-\xi^2)^{n-k}e^{a\xi}\,d\xi$$
$$\qquad\qquad=\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I\_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0,$$
with $I\_\alpha(x)$ the modified Bessel function of the first kind.


