= Solution
{c}
Fix $0<\epsilon<\pi/2$. On $[\epsilon,\pi/2]$, $\sin^2t\geq\sin^2\epsilon>0$, so that part of the integral is bounded by $(\pi/2)e^{-x\sin^2\epsilon}$, exponentially smaller than $x^{-1/2}$.
By the assumed local equivalence, the remaining part is asymptotic to $\int_0^\epsilon e^{-xt^2}dt$. Substitute $s=\sqrt x\,t$:
$$
\int_0^\epsilon e^{-xt^2}dt=x^{-1/2}\int_0^{\epsilon\sqrt x}e^{-s^2}ds
\sim\frac{\sqrt\pi}{2\sqrt x}.
$$
The last step uses the <Gaussian integral> and monotone convergence of the upper limit to infinity. Hence
$$
\boxed{\int_0^{\pi/2}e^{-x\sin^2t}dt\sim\tfrac12\sqrt{\pi/x}.}
$$
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