∫ e^at. sinωt dt This is the second part of an electrical circuit DE problem from our notes (first part not required to solve the above integral) however in-between this integral and the answer our professor only told us that we would get to the answer by using integration by parts twice. I am curious to understand this missed step and have tried integration by parts twice but can't seem to get the same answer as our professor gave us which should work out as: e^at (asinωt - ωcosωt) / (ω^2 + a^2) I would really appreciate it if someone could help me here.. I basically tried integrating by parts twice and taking the original negative integral from the second integration by parts to the LHS and then using that to half the RHS and ended up with e^at(sintωt - cosωt)/2, which is different to our lecturers result. Thanks to anyone for their time!