Firstly I am posting this from my phone so apologies for the layout.
Just wondering what the significance of multiplying by the step function $u(t)$ on the RHS is. When I carry out the transformation of
$$u(t) + \sin(t)$$
I end up at the same result as the answers that didn't omit the step function on the sin function. It would seem to me that if I were to calculate the transform on
$$u (t) + \sin(t) u (t)$$
I would expect an extra $1/s$ term to be floating around.
Does its meaning lie in the fact that we are taking the transform of $\sin(t)$ for all values $t \ge 0$?
Thank you for any help.
Commonly, the (one-sided) Laplace transform is defined by
$$X(s)=\mathcal{L}(x(t))=\int_0^\infty x(t)\exp(-st)dt$$
i.e. the integration starts at $t=0$. So, it does not change, if you calculate the function multiplied by the unit step or not. In fact, when looking at tables for the Laplace transform, it is implicitely defined, that all functions are multiplied by a unit step (see, e.g. here).