Solution
= Solution
Write $\sinh(x\Phi)=(e^{x\Phi}-e^{-x\Phi})/2$. The global maximum of $\Phi$ is the nonstationary endpoint $t=1$, where
$$
\Phi(1)=1,\qquad \Phi'(1)=6.
$$
Endpoint <Laplace's method> therefore gives
$$
\frac12\int_0^1e^{x\Phi(t)}dt\sim\frac{e^x}{12x}.
$$
The second exponential has its maximum at $t=1/2$ and is only of order $x^{-1/2}e^{x/8}$, exponentially smaller. Hence
$$
\boxed{J(x)\sim\frac{e^x}{12x}}.
$$