Asymptotic expansion of integral Find the first two terms in asymptotic expansion (as $x \rightarrow \infty$) of the following integral
$$\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt.$$
 A: $\sin^2t$ is an even function, so consider the integral:
$$\int ^{\pi/2}_{-\pi/2}e^{-x\sin ^2 t}dt$$
For very large values of x, non zero values of $\sin t$ will make the integrand very small. So we are justified in making this approximation:
$$\int ^{\pi/2}_{-\pi/2}e^{-x\, t^2}dt=\frac{\sqrt{\pi } \text{erf}\left(\frac{\pi  \sqrt{x}}{2}\right)}{\sqrt{x}}$$
Since the error function is asymptotic to 1, we can further simplify it as
$$\sqrt{\frac{\pi }{x}}$$
Or we could notice that only very small values of $ t$ makes any contribution so we can use the gaussian integral formula (for $(\infty,-\infty)$) to get the same answer. You want half of this result.
Edit: As M.Wing has suggested:

I like your answer. May I suggest to write the exponential term as $\exp(−xt^2)$ times $\exp(−x(\sin^2(t)−t^2))$. The first term is then used as a Gaussian (as in your calculation); the second term can be expanded in Taylor series of t around zero.

Doing this gives the following integral
$$x\int_{-\infty }^{\infty } e^{-xt^2} \left(\sin ^2t-t^2\right) \, dt=x\frac{\sqrt{\pi } e^{-1/x} \left(e^{1/x} (x-1)-x\right)}{2 x^{3/2}}$$
(Remember the integrand is practically zero beyond $\pi/2$)
The first term is a very good approximation, it is within %5 for $x>5$ but the second term makes it even better. The ratios to the actual result:

A: As Lucian pointed out in a comment $$f(x)=\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt=\frac{ \pi}{2}  e^{-x/2}\, I_0\left(\frac{x}{2}\right)$$ At this point, one could use the asymptotic expansion given by Hankel for the modified Bessel function of the first kind $$I_{\alpha}(z)=\frac{e^z}{\sqrt{2\pi z}}\Big(1-\frac {4\alpha^2-1}{8z}+\frac {(4\alpha^2-1)(4\alpha^2-9)}{2! (8z)^2}-\frac {(4\alpha^2-1)(4\alpha^2-9)(4\alpha^2-25)}{3! (8z)^3}+\cdots\Big)$$ Applied to $\alpha=0$ and $z=\frac x2$, this gives $$f(x)\approx\frac{\sqrt{\pi } }{2
   \sqrt{x}}\left(1+\frac{1}{4 x}+\frac{9}{32 x^2}+\frac{75}{128 x^3}+\cdots\right)$$
