= Solution
For the <Laplace method>, under the usual regularity ($p$ continuous at zero and $q$ sufficiently smooth near its isolated minimum), write $q(t)=q(0)+q''(0)t^2/2+o(t^2)$. On any interval bounded away from zero, <continuity> and uniqueness of the minimum give a positive gap in $q$, making that contribution exponentially smaller. Set $t=s/\sqrt{x}$ near zero. The leading integral is
$$
I(x)=e^{-xq(0)}x^{-1/2}\left[p(0)\int_0^\infty e^{-q''(0)s^2/2}ds+o(1)\right].
$$
A local quadratic lower bound on $q-q(0)$ justifies dominated convergence after this rescaling. Thus, when $p(0)\ne0$,
$$
\boxed{I(x)\sim p(0)e^{-xq(0)}\sqrt{\frac{\pi}{2xq''(0)}}.}
$$
If $p(0)=0$, the displayed coefficient vanishes and further local terms of $p$ determine the first nonzero asymptotic term; one must not interpret a zero coefficient as an <asymptotic equivalence>.
For the first Bessel integral, take $q(\theta)=-\cos\theta$, $p(\theta)=\cos(\nu\theta)$. The endpoint minimum is $q(0)=-1$, $q''(0)=1$, $p(0)=1$, so after dividing by $\pi$ its leading term is $e^x/\sqrt{2\pi x}$.
For the second integral, $\cosh t\geq1+t^2/2$ gives, for fixed real $\nu$,
$$
\left|\int_0^\infty e^{-x\cosh t-\nu t}dt\right|\leq e^{-x}\int_0^\infty e^{-xt^2/2+|\nu|t}dt=O(e^{-x}x^{-1/2}).
$$
Completing the square proves the last bound for large $x$. It is exponentially smaller than the first integral; the fixed prefactor $\sin(\nu\pi)$ does not change this. Therefore
$$
\boxed{I_\nu(x)\sim\frac{e^x}{\sqrt{2\pi x}}\quad(x\to+\infty,\ \nu\text{ fixed}).}
$$
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