Question

Im trying to calculate \[\int\limits e ^{ax+b} \sin x dx \] by using partial integration 2 times. I don't wanna use the formula.

I tried it my self but i ended up in an endless loop :S

Answer 1

Ok, you get an answer which contains an integral equal to the original problem but negative in value, now move it to the original problem side and you have "2" times the original problem (you are adding 2 of the same kind, even as they are 2 whole integral problems). Then solve for the "original" integral problem.
\[\int\limits_{a}^{b} e ^{ax+b}\sin x dx = (e ^{ax+b}\sin x)/a - (1/a)\int\limits_{a}^{b}e ^{ax+b}\cos xdx = \] \[e ^{ax+b}(\sin x-\cos x)/a -(1/a)\int\limits_{a}^{b}e ^{ax+b} \sin xdx=\] \[e ^{ax+b}(\sin x-\cos x)/a = \int\limits_{a}^{b}e ^{ax+b} \sin xdx +(1/a)\int\limits_{a}^{b}e ^{ax+b}\sin xdx=\] \[e ^{ax+b}(\sin x+\cos x)/a=(a+1)/a \int\limits_{a}^{b}e ^{ax+b}\sin xdx=\] \[\int\limits_{a}^{b}e ^{ax+b}\sin xdx=(1/a+1)e ^{ax+b}(\sin x-\cos x)+c\] In your case you don't have limits of integration.

Answer 2

thanks! the only thing left is the be able to come up with such smart answers on the exam :P