# Trigonometric and exponential integral $\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx$

How can we prove this Integral relation? $$\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{\pi }{e}\cdot \frac{e^{\sqrt{1-a^2}}}{\sqrt{1-a^2}}$$ where $$\text{ }a\in(-1,1)$$.

$$I(a)=\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{i \left(a\sin x\right)}}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}}{1+a\cos x}dx$$ $$=\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}e^{ix}}{ae^{2ix}+2e^{ix}+a}dx=\Re\,(-i)\oint_{|z|=1}\frac{e^{az}}{az^2+2z+a}dz$$ The zeros of the denominator are $$z_{1,2}=-\frac{1}{a}\pm\sqrt{\frac{1}{a^2}-1}$$; only one pole ($$\,z_1=-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\,\,$$) lies inside the closed contour $$|z|=1$$.
Therefore, $$I(a)=\Re \operatorname {Rez}_{z=z_1}\,2\pi i\,(-i)\frac{e^{az}}{az^2+2z+a}=\pi\frac{e^{a\big(-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\big)}}{a\sqrt{\frac{1}{a^2}-1}}=\frac{\pi}{e}\frac{e^\sqrt{1-a^2}}{\sqrt{1-a^2}}$$
Per the generalized result $$\begin{equation*} \int_0^\pi\frac{f(e^{ix})+f(e^{-ix})}{1+2p\cos x+p^2}\mathrm dx=\frac{2\pi}{1-p^2}f(-p) , \end{equation*}$$ set $$f(t)=\frac12e^{at}$$ and $$p = \frac{1-\sqrt{1-a^2}}a$$ to obtain $$\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{\pi e^{\sqrt{1-a^2}-1}}{\sqrt{1-a^2}}$$