You're going to have to use integration by parts twice. I also like to set my integral to a variable, say a = ∫e^-3t sin wt dt
let
\(\displaystyle u=e^{-3t}\)
\(\displaystyle du=-3e^{-3t}\)
\(\displaystyle dv=\sin{t}\)
\(\displaystyle v=-\cos{t}\)
I am guessing the "w" in your sin wt is a typo? If not, easy enough, just treat as a constant -
\(\displaystyle dv=\sin{wt}\)
\(\displaystyle v=-\frac{\cos{wt}}{w}\)
So now use integration by parts (a = uv - ∫vdu) , and then use it again on the integral you get from your ∫vdu, and you should get your original integral back (∫e^-3t sin wt dt) which of course is your variable "a". So now solve for a and that's what the integral equals. Hope this helps. Im sure one of the hot shots on the forum will correct any errors I made, but this should get you started.