Starting with
\(\displaystyle \int {e^x } \sin 2x\;dx\)
\(\displaystyle \left[ \matrix{
u = \sin 2x \hfill \cr
dv = e^x dx \hfill \cr
du = 2\cos 2x\;dx \hfill \cr
v = e^x \hfill \cr} \right] \to e^x \sin 2x - \int {2e^x \cos 2x\;dx \to \left[ \matrix{
u = \cos 2x \hfill \cr
du = - 2\sin 2x\;dx \hfill \cr
dv = 2e^x \;dx \hfill \cr
v = 2e^x \hfill \cr} \right]e^x \sin 2x - (2e^x \cos 2x - \int { - 4e^x \sin 2x\;dx} }\)
So where do I go from here? It looks uglier than I started out with if you ask me, so I don't know if I am doing it right or what I need to do next. Any help would be appreciated.
\(\displaystyle \int {e^x } \sin 2x\;dx\)
\(\displaystyle \left[ \matrix{
u = \sin 2x \hfill \cr
dv = e^x dx \hfill \cr
du = 2\cos 2x\;dx \hfill \cr
v = e^x \hfill \cr} \right] \to e^x \sin 2x - \int {2e^x \cos 2x\;dx \to \left[ \matrix{
u = \cos 2x \hfill \cr
du = - 2\sin 2x\;dx \hfill \cr
dv = 2e^x \;dx \hfill \cr
v = 2e^x \hfill \cr} \right]e^x \sin 2x - (2e^x \cos 2x - \int { - 4e^x \sin 2x\;dx} }\)
So where do I go from here? It looks uglier than I started out with if you ask me, so I don't know if I am doing it right or what I need to do next. Any help would be appreciated.