 Solution

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 $\eta \ll ϵ$ but $\sqrt{x}\eta \to \infty$.

On $[0,\eta ]$, the Taylor expansion ${\sin }^{2}t=t^2+O(t^4)$ gives $x({\sin }^{2}t-t^2)=O(x{\eta }^4)=O(x^{-1/3})$ uniformly. Thusuniformly there, and the two central integrals are asymptotically equal. On $[\eta ,ϵ]$, choose a fixed $c>0$ with $\sin t\ge ct$ for $0\le t\le ϵ$. The sine-integral tail is at most $ϵe^{-c^{2}x{\eta }^2}=ϵe^{-c^2{x}^{1/3}}$; 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 $\sqrt{\pi }/(2\sqrt{x})$. HenceThis justifies the localized Laplace method rather than interchanging a nonuniform approximation with integration.