= Solution
{c}
A fixed small-angle approximation alone is insufficient because its error is multiplied by $x$. Introduce the shrinking cutoff $\eta=x^{-1/3}$, which satisfies $\eta\ll\epsilon$ but $\sqrt x\,\eta\to\infty$.
On $[0,\eta]$, the <Taylor expansion> $\sin^2t=t^2+O(t^4)$ gives $x(\sin^2t-t^2)=O(x\eta^4)=O(x^{-1/3})$ uniformly. Thus
$$
e^{-x\sin^2t}=e^{-xt^2}[1+O(x^{-1/3})]
$$
uniformly there, and the two central integrals are asymptotically equal. On $[\eta,\epsilon]$, choose a fixed $c>0$ with $\sin t\geq ct$ for $0\leq t\leq\epsilon$. The <sine>-integral tail is at most $\epsilon e^{-c^2x\eta^2}=\epsilon e^{-c^2x^{1/3}}$; the Gaussian tail has the same exponential bound with $c=1$. Both are $o(x^{-1/2})$, while the central <Gaussian integral> is asymptotic to $\sqrt\pi/(2\sqrt x)$. Hence
$$
\boxed{\int_0^\epsilon e^{-x\sin^2t}dt\sim\int_0^\epsilon e^{-xt^2}dt.}
$$
This justifies the localized <Laplace method> rather than interchanging a nonuniform approximation with integration.
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