Asymptotic Expansion of Bessel $\frac{1}{\pi}\int_0^{\pi}e^{x\cos t}dt$

Question

My question is how to find the asmyptotic expansion of $I(x)=\frac{1}{\pi}\int_0^{\pi}e^{x\cos t}dt$ as $x\rightarrow\infty$.

I already got the expansion of $\int_0^{\pi/2}e^{-x\sin^2t} dt$ by using the substitution $\sin^2 t=u$ but how are they connected?

Answer 1

$$ \int_0^{\pi/2}e^{-x\sin^2t} dt = \int_0^{\pi/2}e^{-x\cdot\frac{1-\cos 2t}{2}} dt= e^{-\frac{x}{2}} \int_0^{\pi/2}e^{\frac{x}{2}\cos 2t}dt = \frac{e^{-\frac{x}{2}}}{2} \int_0^{\pi}e^{\frac{x}{2}\cos u}du = \frac{\pi e^{-\frac{x}{2}}}{2} I(\tfrac{x}{2}) $$ I think...