Compute a laplace transform 
Compute $\mathcal{L} [e^{-2x} \sin x]$

So I want to do $\int_0^\infty e^{-2x} \sin x e^{-px} \, dx=\int_0^\infty e^{-(p+2)x}\sin x \, dx$
But due to the way $e$ and $\sin x$ integrates I dont know where to go from here?

Try integration by parts, twice.  When you have done that you find you have gone in a circle...you should have something along the lines of:
$$f(p) - g(p)\int_0^\infty e^{-(p+2)x} \sin x \,dx$$
And you have the same integral that you already know that you didn't know how to solve... but you are not dead yet.
$$\int_0^\infty e^{-(p+2)x} \sin x \, dx = f(p) - g(p)\int_0^\infty e^{-(p+2)x} \sin x \, dx\\
(1+g(p)) \int_0^\infty e^{-(p+2)x} \sin x \, dx = f(p)\\
\int_0^\infty e^{-2x} \sin x e^{-px} \, dx = \frac {f(p)}{1+g(p)}$$
Alternatively say: $\sin x = \frac 1{2i} (e^{ix} - e^{-ix})$
$$\frac 12 \int_0^\infty e^{-(p+2-i)x} - e^{-(p+2+i)x} \, dx$$
