Integration by Parts twice....

jaredld

New member
Joined
Jul 3, 2005
Messages
20
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.
 
Try letting \(\displaystyle \L\\u=e^{x}dx;

dv=sin(2x)dx;

du=e^{x}dx;

v=\frac{-cos(2x)}{2}\)
 
I appreciate the link. That is quite a neat thing. Actually I have a TI-89 so I know what the answer is. I am just having difficulty getting there. Thanks though. That is a good second resource.

Galactus- thanks, I will try that.
 
jaredld said:
It looks uglier than I started out with if you ask me
More like it looks exactly like it did when you started. Stop integration and use algebra.
 
I figured out what happened. I was thinking that when you do integration by parts you get it to a certain point like I showed in the first post and then find a formula to take the derivitive of the final integral.

But you don't. You have to set the whole thing equal to the original and solve for the original integral. Thanks TKbunny. I didn't quite understand what you were telling me with "use algebra" but I know now. DUH! Thanks all....here is the last couple steps for those that are interested.

Set the ORIGINAL= to the "Parts" that you worked so hard to come up with.
\(\displaystyle \[
\int {e^x \sin 2x\;dx = \;} e^x \sin 2x - 2e^x \cos 2x + 4*\int {e^x \sin 2x\;dx \to } 5*\int {e^x \sin 2x\;dx = e^x \sin 2x - 2e^x \cos 2x}
\]\)Add over "Similar Integrals". Then divide to obtain the integral. In this case we divide over the coeficient of 5 to obtain our original integral on the left side equal to the mess on the right.
\(\displaystyle \[
\int {e^x \sin 2x\;dx = \frac{{e^x \sin 2x - 2e^x \cos 2x}}{5}}
\]\)
 
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