Question

Integration by parts problem,

\[\int\limits_{}^{}e^{2x}\sin(3x)dx\]

Trying to relearn integration, very rusty on it, haven't taken a calc course in almost 2 years.

Totally stuck with this problem, any hints or help would be appreciated

Answer 1

You remember how integration by parts works? \[\int u' v dx = uv - \int u v' dx\]

Here you probably will have to apply the rules twice so that you'll have something like \[ I = uv - (rs + I) \]

then you solve for \(I\) to get the solution.

Answer 2

I remember the rule just fine, but aren't these just functions that repeat endlessly?

\[\int\limits_{}^{} \sin(3x)e^{2x}dx = \frac{\sin(3x)e^{2x}}{2} + \int\limits \frac{e^{2x}\cos(3x)}{6}\]

Answer 3

No, because you'll get an integral of the form
\[- k \int e^{2 x} \sin(3 x} dx \]
if you apply integration by parts again. It's the same integral with which you started so that you can solve for that integral as I tried to describe in my first answer.

Answer 4

Woops, you differentiated the sin(3x) so you get a 3 on top, not on bottom:\[\int\limits\limits_{}^{} \sin(3x)e^{2x}dx = \frac{\sin(3x)e^{2x}}{2} + \int\limits\limits \frac{\color{red}{3}e^{2x}\cos(3x)}{2}\]Integrate one more time.
Make sure you assign your exponential to be your dv again, otherwise you'll end up undoing your work from before.

Answer 5

When memorising the formula for UV-substitution (Integration by Parts), use the mneumonic

"UltraViolet-VooDu"
UV-∫vdu

Answer 6

lol clever :3

Answer 7

I already have it memorized but that is pretty clever

Answer 8

I noticed my mistake Im just tired

Answer 9

As for settng exponential as du do you mean setting u = e^2x

that will just give me du = 2e^2xdx

Answer 10

in the first step you set \(e^{2x}\) as your \(v'\). you need to make sure to do that again

Answer 11

no no. :O
The first time you integrated it looks like you assigned:
u=sin(3x), dv=e^(2x)

So it's important that you do likewise again:
u=cos(3x), dv=e^(2x)

Answer 12

\[\frac{\sin(3x)e^{2x}}{2} + \frac{e^{2x}3\cos(3x)}{4} + \int\limits\limits 9\sin(3x)\frac{e^{2x}}{4}dx\]

Answer 13

oh I think I get what you are saying,

set u = cos(3x)
du = -3sin(3x)dx

then I just go

arccos(u)/3 = x

then I just sub that into e^2x

Answer 14

wait no that will make things even more complicated blah

Answer 15

\[\int\limits\limits\limits_{}^{} \sin(3x)e^{2x}dx = \frac{\sin(3x)e^{2x}}{2} \color{red}{-} \int\limits\limits\limits \frac{\color{red}{3}e^{2x}\cos(3x)}{2}\]Oh I just noticed, shouldn't this first sign have been minus?
You differentiated sin(3x) to positive 3cos(3x), yes?

Answer 16

yeah you are right I made a mistake there

Answer 17

\[\large\rm \color{orangered}{\int\limits\sin(3x)e^{2x}dx}=\frac{1}{2}\sin(3x)e^{2x}-\frac{3}{4}\cos(3x)e^{2x}+\frac{9}{4}\color{orangered}{\int\limits\sin(3x)e^{2x}dx}\]Ok so by integrating twice you've gotten to this point.

Answer 18

If you call it something else, it might be easier to work with.
\[\large\rm \color{orangered}{\mathcal I}=\frac{1}{2}\sin(3x)e^{2x}-\frac{3}{4}\cos(3x)e^{2x}+\frac{9}{4}\color{orangered}{\mathcal I}\]
It's just algebra from this point.
Solve for I.

Answer 19

I- 9/4I = 5/4I

so (4/5)*( 1/sin(3x)e^(2x) - 3/4cos(3x)e^2x )

Answer 20

Yes, you've got the right idea.
Sec, I wanna check your signs a minute...

Answer 21

man Im so tired my signs are going to always be messed up lol, I totally dont remember this trick at all.

Answer 22

pretty neat though

Answer 23

Oh oh, remember we fixed that minus sign a little later on?
So we need to distribute that negative to each term in the `by parts` application.
So the 9/4I term should be negative.

Answer 24

in the second `by parts` application*

Answer 25

\[\large\rm \color{orangered}{\mathcal I}=\frac{1}{2}\sin(3x)e^{2x}-\frac{3}{4}\cos(3x)e^{2x}-\frac{9}{4}\color{orangered}{\mathcal I}\]So easy to lose track of these things :( lol

Answer 26

oh oh damns well its the same principle I'm way too tired at the moment to finish this thanks for showing me this trick though, I will just do it again tomorrow and check it on wolfram thanks for the help

Answer 27

heh :3