Let $I(x)$ be
$I(x) = \int _{-\pi} ^\pi e^{x \cos \phi}d\phi$.
When $x \gg 1$, how to get major terms of (how to approximate) $I(x)$?
There must be the solution but I don't have any idea. In the textbook, it is said that "You can use
$\int _{-\infty} ^\infty e^{-\alpha \psi ^2}d\psi = \sqrt{\frac{\pi}{\alpha}}$
for $\alpha > 0$ thought I don't know this is really needed. (The result is used to solve a bachelor-level physics problem. So the process may not be a quite difficult one.)
Note that $$ I(x) = \int_{-\pi}^{\pi} e^{x \cos \phi}\, d\phi = e^x \cdot \int_{-\pi}^{\pi} e^{-2x \sin^2 \left(\frac{\phi}{2}\right)}\, d\phi.$$ As a result, we have the estimation $$ e^x \cdot \int_{-\pi}^{\pi} e^{-2x \sin^2 \left(\frac{\phi}{2}\right)}\cdot \cos\left(\frac{\phi}{2}\right)\, d\phi \leq I(x) \leq e^x \cdot \int_{-\pi}^{\pi} e^{-2x \left(\frac{\phi}{\pi}\right)^2 }\, d\phi, $$ which can be reduced to $$ \frac{2e^x}{\sqrt{x}}\cdot \int_{-\pi\sqrt{x}}^{\pi\sqrt{x}}e^{-2t^2}\,dt \leq I(x) \leq \frac{\pi e^x}{\sqrt{x}}\cdot \int_{-\sqrt{x}}^{\sqrt{x}}e^{-2t^2}\,dt .$$ Thus we have $$ \sqrt{2\pi}\left(1 - 2e^{-2\pi^2 x}\right) \cdot \frac{e^x}{\sqrt{x}} \leq I(x) \leq \sqrt{\frac{\pi^3}{2}} \frac{e^x}{\sqrt{x}},$$ where the first inequality follows from the Chernoff bound.
When $x \gg 1$, we have $\displaystyle I(x) = \Theta\left(\frac{e^x}{\sqrt{x}}\right)$.
Remark: Laplace's method will give $$\lim_{x\to +\infty} \frac{I(x)}{\sqrt{2\pi} e^x\cdot x^{-1/2}} = 1.$$
