# [SOLVED] Integration by parts...I think...

• Sep 7th 2008, 07:50 PM
redman223
[SOLVED] Integration by parts...I think...
$
\int e^{3x} sin(5x)~dx
$

I tried using both sin(5x) and $e^{3x}$ as u and all I ended up doing was coming up with the same problem again, except with a constant pulled out in front. There must be some trick I need to use on this problem that I am not seeing.
• Sep 7th 2008, 07:58 PM
o_O
Integration by parts twice should do the trick. To start you off: $\begin{array}{ll} u = e^{3x} & dv = \sin 5x \ dx\\ du = 3e^{3x} dx & v = -\frac{1}{5}\cos 5x \end{array}$

$\int e^{3x} \sin 5x \ dx = -\frac{1}{5}e^{3x}\cos 5x + \frac{3}{5} {\color{blue}\int e^{3x} \cos 5x \ dx}$

Apply integration by parts again on the blue. Then you should recognize the new integral ;)
• Sep 7th 2008, 08:04 PM
redman223
What should I use for u and dv the next time around though? I keep coming up with the same problem I started with.
• Sep 7th 2008, 08:07 PM
o_O
YES! You end up with the same integral as you started right? Why don't you move it to the other side as if it were an algebra problem and solve for it? ;)
• Sep 7th 2008, 08:22 PM
redman223
Thanks, I never would have thought of that, I just figured I was doing something wrong. But what about the constants out front? In this case I came up with:

$
-\frac {3}{75}\int e^{3x}sin(5x)~dx
$

So what I add it to the other side, I should get:
$
\frac {78}{75}\int e^{3x}sin(5x)~dx
$

Would I then multiply everything on the other side by
$
\frac {75}{78}
$

• Sep 7th 2008, 08:33 PM
redman223
Well I tried that and it didn't work...

I ended up with:
$
\frac {-75cos(5x)e^{3x}}{390} + \frac {225sin(5x)e^{3x}}{1950}
$
• Sep 8th 2008, 12:48 AM
11rdc11
$
\int e^{3x} \sin 5x \ dx = -\frac{1}{5}e^{3x}\cos 5x + \frac{3}{5} \int e^{3x} \cos 5x \ dx$

= $-\frac{1}{5}e^{3x}\cos 5x + \frac{3e^{3x}sin(5x)}{25} - \frac{9}{25}\int e^{3x}sin(5x)$

$\int e^{3x} \sin 5x \ dx = -\frac{1}{5}e^{3x}\cos 5x + \frac{3e^{3x}sin(5x)}{25} - \frac{9}{25}\int e^{3x}sin(5x)$

$\frac{34}{25}\int e^{3x} \sin 5x \ dx = -\frac{1}{5}e^{3x}\cos 5x + \frac{3e^{3x}sin(5x)}{25}$

$\int e^{3x} \sin 5x \ dx = -\bigg(\frac{5e^{3x}cos(5x)}{34}\bigg) + \bigg(\frac{3e^{3x}sin(5x)}{34}\bigg) + C$

= $\frac{e^{3x}\bigg(3sin(5x) - 5cos(5x)\bigg)}{34} + C$