Consider the integral
$$I(x)= \int^{\frac{\pi}{2}}_0 (1+t^2)e^{-x\cosh t}dt$$
Use Laplace's method to show that
$$I(x) = \sqrt{\frac{\pi}{2}}e^{-x}+c\frac{e^{-x}}{x}, as \space x \to \infty$$
where $c$ is a constant you must determine.
Where I have got to so far:
let $\phi(t)=-x\cosh t$
$\phi'(t)=x\sinh t$
$\phi''(t)=x\cosh t$
$$I(x) \sim \int_0^{\epsilon}(1+t^2)e^{-x\cosh t}dt \space + \space \int_{\epsilon}^{\frac{\pi}{2}}(1+t^2)e^{-x\cosh t}dt$$
A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-x\cosh(t) + \log(1+t^2)$ about its maximum at $t=0$ I get $$ I(x) \sim \int_0^\epsilon {\rm e}^{-x-\frac{x-2}{2}\, t^2} \, {\rm d}t = \sqrt{\frac{\pi/2}{x-2}} \, {\rm e}^{-x} \, {\rm erf}\left( \sqrt{\frac{x-2}{2}} \, \epsilon \right) \sim \sqrt{\frac{\pi/2}{x-2}} \, {\rm e}^{-x} \, . $$
If I'm not mistaken I think the next term should come from $$ \int_0^{\frac{\pi}{2}} \left\{ {\rm e}^{-x-\frac{x-2}{2}\,t^2} - (1+t^2) \, {\rm e}^{-x\cosh(t)} \right\} {\rm d}t \sim \frac{\sqrt{2\pi}}{16 \, x^{3/2}}\, {\rm e}^{-x} \, , $$ so we would have $$ \int_0^{\frac{\pi}{2}} (1+t^2) \, {\rm e}^{-x \cosh(t)} \, {\rm d}t \sim \sqrt{\frac{\pi}{2x}} \, {\rm e}^{-x} + \frac{7\sqrt{2\pi}}{16 \, x^{3/2}} \, {\rm e}^{-x} \, . $$