How do you integrate ##[(e^(2x))sinx]dx##?
February 20th, 2023
Integrate by parts twice using ##u = e^(2x)## both times.
After the second , you’ll have
##int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx – 4 int e^(2x)sinx dx##
Note that the last integral is the same as the one we want. Call it ##I## for now.
##I = -e^(2x)cosx + 2e^(2x)sinx – 4 I##
So ##I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C##
You may rewrite / simplify / factor as you see fit.
