# Am I using the Laplace Transform pair correctly?

I am given a question of Laplace Transform which is as follows:

$$e^{t}sin2t$$ for $$t\leq0$$

Now I know that by using the transform pair we get:

$$[e^{-at}sin\omega_0 t]u(t) \rightarrow \frac{\omega_0}{(s+\alpha)^2 +(\omega_0)^2}$$

Using this pair we:

$$[e^{t}sin2t]u(t) \rightarrow \frac{2}{(s+1)^2 +(2)^2}$$

Now for $$u(-t)$$ what I think should come is: $$-\frac{2}{(s-1)^2 +(2)^2}$$

Is it correct?

Start first at $$f(t)=sin(2t)u(t)$$, which has a Laplace transform of $$\frac{2}{s^2+2^2}$$ then apply time-scaling property $$f(\alpha t)=\frac{1}{|\alpha|}F(\frac{s}{\alpha})$$ with $$a=-1$$. Now, $$f(t)$$ becomes $$sin(-2t)u(-t)$$, but this is just equal to $$-sin(2t)u(-t)$$ because $$sin(x)$$ is odd function.
Lastly, apply $$e^t$$ to $$f(t)=-sin(2t)u(-t)$$, and then remove the negative.