large x and small x expansion for gamma-like function Find two approximations for the integral ($x&gt;0$)
  \begin{equation}
    I(x) = \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta
  \end{equation}
  one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).



For the small x expansion, I tried using a Maclaurin series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer.  I know this is likely related to the Gamma/Gaussian function, but am fairly novice at problems of this form.  Any help deeply appreciated!
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.

Let's transform the integral first:

$$ I(x) = \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta =\frac{1}{\pi}\int_0^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta=\frac{e^{x}}{\pi} \int_0^{\frac{\pi}{2}}
      e^{-x \sin^2(\theta)}d\theta= \\ =\frac{e^{x}}{\pi} \int_0^1
      \frac{e^{-x s^2}}{\sqrt{1-s^2}} d s=\frac{e^{x}}{2\pi} \int_0^1
      \frac{e^{-x t}}{\sqrt{t}\sqrt{1-t}} d t$$

Now for the latter integral, the main contribution for $x \to \infty$ will be given by $t \to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).

Using the notation from the article, we have:

$$\phi(t)=\frac{1}{\sqrt{t}\sqrt{1-t}}=t^{-1/2} g(t)$$

Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $\int_0^1 |\phi(t)| dt=\pi&lt;\infty$. So the lemma conditions are satisfied.

Then we can represent:

$$\int_0^1
      \frac{e^{-x t}}{\sqrt{t}\sqrt{1-t}} d t \asymp \sum_{n=0}^\infty \frac{\Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
How about this:
$$I(x)=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}e^{x\cos^2\theta}d\theta=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}e^{x(1-\sin^2\theta)}d\theta$$
$$\theta=\pi/2-\phi$$
$$d\theta=-d\phi$$
$$I(x)=\frac{1}{2\pi}\int_0^\pi e^{x\sin^2\phi}d\phi$$

EDIT:

taking the other approach, I have:
$$I'(x)=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\cos^2\theta e^{x\cos^2\theta}d\theta$$
then using $u=cos\theta$ you can obtain:
$$I'(x)=\frac1\pi\int_0^1\frac{u^2}{\sqrt{1-u^2}}e^{xu^2}du$$
now using $v=\sqrt{1-u^2}$ we can get:
$$I'(x)=\frac1\pi\int_0^1\sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=\frac1\pi\int_0^1\sqrt{1-v^2}e^{-(\sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=\frac{1}{2\pi x^2}\int_0^x\beta e^\beta d\beta$$
so:
$$I(x)=\frac1{2\pi}\iint\frac{(x-1)e^x+1}{x^2}dxdx$$