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$$I=\int_0^{\pi} (\sin x)^{2n-2k+1} e^{a\cos x} dx =\int_{-1}^1 (1-\xi^2)^{n-k}e^{a\xi}\,d\xi$$ $$\qquad\qquad=\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0.$$

$$\int_0^{\pi} (\sin x)^{2n-2k+1} e^{a\cos x} dx =\int_{-1}^1 (1-\xi^2)^{n-k}e^{a\xi}\,d\xi$$ $$\qquad\qquad=\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0,$$ with $I_\alpha(x)$ the modified Bessel function of the first kind.

$$\ \int_0^{\pi} \sin^{2n-2k+1} e^{a\cos x} dx =\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0.$$

$$I=\int_0^{\pi} (\sin x)^{2n-2k+1} e^{a\cos x} dx =\int_{-1}^1 (1-\xi^2)^{n-k}e^{a\xi}\,d\xi$$ $$\qquad\qquad=\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0.$$

$$\ \int_0^{\pi} \sin^{2n-2k+1} e^{a\cos x} dx =\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a).$$

$$\ \int_0^{\pi} \sin^{2n-2k+1} e^{a\cos x} dx =\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0.$$