I'm done (a) and (b)(i) but I'm stuck on (b)(ii). I thought that since the dominant contribution is coming in at t=0 that I could approximate the sin in the logarithm and use a Taylor expansion, but then my resulting f(t) is 0 at t=0 so I don't get any contribution at all. Really stuck on how to proceed here.
On
It's true that you can't immediately apply the formula from 1(a), but you can prove a similar result in the same way:
If $\phi(t)$ attains its maximum value on $[a,b]$ at $t=a$, $\varphi'(a) = 0$, $\varphi''(a) \neq 0$, $f(a) = 0$, and $f'(a) \neq 0$, then
$$ \int_a^b e^{x \phi(t)} f(t)\,dt \sim -\frac{f'(a)}{\phi''(a)x} e^{x \phi(a)} $$
as $x \to \infty$.
It's straightforward to generalize this further.
Indeed the contribution comes from $t=0$. We have $\log{(1+\sin{t})} \sim \sin{t} \sim t $ as $t \to 0$, so by Laplace's Method, $$ \begin{align} I(x) &\sim \int_0^{\varepsilon} e^x e^{-xt^2/2} (1+xt^3)t \, dt \\ &\sim \frac{e^x}{x} \int_0^{x^{1/2}\varepsilon} u e^{-u^2/2} \, du \\ &\sim \frac{e^x}{x} \int_0^{\infty} u e^{-u^2/2} \, du = \frac{e^x}{x}. \end{align} $$ using $t=x^{-1/2}u$, so $dt = x^{-1/2} \, du$.