I wanted to know if I did this calculation correctly. I used the Laplace Method and considered where in the range of integration we have the most contribution to the integral from the integrand.
Here the integral is in the form: $\int^1_0e^{xg(t)}f(t)dt$
The maximum of $g(t)=-t^2$ is at $t=0$, thus we can write the integral as: $\int^{\epsilon}_0 e^{-xt^2} \sin (t)dt$ where $\epsilon$ is a small parameter. Now we need to Taylor expand $f(t) = \sin(t) = t + O(t^3)$.
$$\int^{\epsilon}_0e^{-xt^2}(t+O(t^3))dt \sim \int^{\epsilon}_0 te^{-xt^2}dt $$
We make the change of variables: $s^2=xt^2$ and the integral becomes:
$$x^{-1}\int^{\epsilon \sqrt{x}}_0 se^{-s^2}ds = x^{-1}\int^{\infty}_0 se^{-s^2}ds \sim \frac{1}{2x}$$
Where we extended the limits of integration to infinity and the remaining integral was a known form.
Any ideas on if this is correct?
The calculations are correct. For your interest, a complete asymptotic expansion can also be found.
Substitute $u=t^2$, $$\int_0^1\sin(t)e^{-xt^2}\ dt=\frac{1}{2}\int_0^1\frac{\sin(\sqrt u)}{\sqrt u}e^{-xu}\ du$$ where there is the expansion, $$\frac{\sin(\sqrt u)}{\sqrt u}=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(2n+2)}u^n$$ then by Watson’s Lemma we obtain the complete asymptotic expansion $${\int_0^1\sin(t)e^{-xt^2}\ dt\sim\frac{1}{2x}\sum_{n=0}^\infty(-1)^n\frac{\Gamma(n+1)}{\Gamma(2n+2)}\frac{1}{x^{n}}}.$$ as $x\to+\infty$. Taking the first term from the series verifies $I\sim 1/2x$.