The term you can handle by using line 107 here. The sin function you can handle by line 305 on the same page. It's not pretty, but you can get a closed-form expression for your function, despite not being square-integrable.
Hmm. Check your dv integration. You have to do by-parts twice on functions like that in order to get back to your original integral. You then pull it over to the other side and solve for it. You follow?
[EDIT]: Also, check whether you're computing definite or indefinite integrals, and where you're doing that. Don't mix them up!
Reply to CB at Post # 11:
Well, I was hoping that through the computations being done above, we'd be able to get to the point of recognizing the integrals that are left as the second derivatives of the Dirac Delta function. There is an inverse FT for that function: