$$\int_{0}^{\pi}(\sin x)^{2n-2k+1}e^{a\cos x}dx$$
To avoid writing $n-k$ all the times let $v=n-k$.
$$u=\cos x \implies du=-\sin x\,dx$$
$$\int_{-1}^{+1}(1-u^2)^{v} e^{au}du$$
Split the integral at 0 and we have,
$$2\int_{0}^{1}(1-u^2)^{v}\cosh(au)du$$
(Because $\cosh x=\frac{e^a+e^{-a}}{2}$)
$t=u^{2}\implies dt=2u\,du$
$$\int_{0}^{1}(1-t)^{v}\cosh(a\sqrt[2]{t})t^{\frac{-1}{2}}dt$$
Taylor series expansion of hyperbolic cosine gives us,
$$\cosh(x)=\sum_{z\ge 0}\frac{x^{2z}}{(2z)!}$$
Placing this in the integrand and bringing the summation outside the integrand,
$$\sum_{z\ge 0}\left(\frac{a^{2z}}{(2z)!}\int_{0}^{1}(1-t)^{(v+1)-1}t^{(z+1)-1}dt\right)$$
$$\sum_{z≥0}\frac{a^{2z}(B(z+1,v+1))}{(2z)!}$$
Taking certain values of v,a we can calculate the value of integrand by using properties of beta function.
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Having the values of $v,a\in\mathbb{Z}$ we can use the above the relation and simplify the summation.