How can I obtain $I_n(x)= \frac{1}{2\pi}\int_{0}^{2\pi}e^{in\theta}e^{x \cos\theta}\,d\theta ,$ from the integral $J_n (x) =\frac{1}{2\pi} \int_{0}^{2\pi}e^{-in\theta}e^{ix \sin\theta}\,d\theta .$
I started with $I_n(x) = i^{-n} J_n(ix)$, but when I change the variable accordingly in the integral of $J_n$ above, I cannot get something that resembles the final result. I have tried manipulating the trigonometric identities in several ways, but couldn't get it, either.