$$ f(t)= \begin{cases} \sin(\pi t) \,\,; 0 \leq t \leq 4 \\ \\ 0 \,\,; t> 4 \end{cases} $$
I am unsure of how to go about doing this. Seems to me the answer should be $$((1-e^{42})\pi)/s^2+\pi^2$$
I keep seeing answers online that tell me I am wrong but I do not know how to even go about solving this.
Well, in general we are looking at:
$$\text{Y}\left(\text{s}\right)=\mathscr{L}_t\left[\text{y}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{y}\left(t\right)\cdot e^{-\text{s}t}\space\text{d}t\tag1$$
In your case we get:
$$\text{F}\left(\text{s}\right)=\mathscr{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(t\right)\cdot e^{-\text{s}t}\space\text{d}t=\int_0^4\sin\left(\pi\cdot t\right)\cdot e^{-\text{s}t}\space\text{d}t+\int_4^\infty0\cdot e^{-\text{s}t}\space\text{d}t$$
$$=\int_0^4\sin\left(\pi\cdot t\right)\cdot e^{-\text{s}t}\space\text{d}t=\frac{\pi-\pi\cdot e^{-4\text{s}}}{\text{s}^2+\pi^2}\tag2$$