Question

Indefinite integral:

Answer 1

\[\LARGE \int~e^{2\theta}~sin(3\theta)~d\theta\]

Answer 2

Oooo super fun problem! :)

Answer 3

So this is one of those nasty integrals where you have to do Parts twice, then do a little magic to fix everything up.

Answer 4

Are you at all familiar with the complex exponential?\[\Large\rm e^{i \theta}=\cos \theta+\mathcal i \sin \theta\]It makes this problem a little easier to work with.
If not, that's fine. We can approach it the normal way.

Answer 5

Uh, nope I don't it >.<

Answer 6

*know it

Answer 7

@zepdrix you can teach us the new way, friend, Please

Answer 8

Ok let's start by calling this integral something,\[\Large\rm \mathcal I= \int\limits~e^{2\theta}~\sin(3\theta)~d\theta\]This will be useful later on.

With a problem like this,
it doesn't matter which part you choose for u and dv.
What WILL matter, is what parts you choose for your second `by parts` application.

Answer 9

Since we have sine (which differentiates nicely, without giving us a negative), let's call that our u, ok?

Answer 10

\[\Large\rm u=\sin(3\theta),\qquad\qquad~~\qquad dv=e^{2\theta}d \theta\]\[\Large\rm du=3\cos(3\theta)d \theta,\qquad\qquad v=\frac{1}{2}e^{2\theta}\]

Answer 11

Mmmm ok that look about right so far? :o

Answer 12

yes, Sir :)

Answer 13

Yup, I got that too~

Answer 14

\[\Large\rm \mathcal I=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\int\limits e^{2\theta}\cos(3\theta)d \theta\]Mmm ok so this is what we get after our first application of IBP.

Answer 15

We need to apply IBP to that integral.
So the important thing here is,
Since we chose the trig function for our U, we have to do that again!
If we did it the other way around, we would simply undo the work we did the first time by parts.

Answer 16

\[\Large\rm u=\cos(3\theta),\qquad\qquad\quad~\qquad dv=e^{2\theta}d \theta\]\[\Large\rm du=-3\sin(3\theta)d \theta,\qquad\qquad v=\frac{1}{2}e^{2\theta}\]Mmmmm ya?

Answer 17

Yes~

Answer 18

So after that we end up with a similar integral as the first, right?

Answer 19

\[\rm \mathcal I=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\color{royalblue}{\int\limits\limits e^{2\theta}\cos(3\theta)d \theta}\]
\[\rm \mathcal I=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\color{royalblue}{\left[\frac{1}{2}e^{2\theta}\cos(3\theta)+\frac{3}{2}\int\limits e^{2\theta}\sin(3\theta)d \theta\right]}\]

Answer 20

Yes very good :)
We end up with the integral that we started with!\[\rm \mathcal I=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\color{royalblue}{\left[\frac{1}{2}e^{2\theta}\cos(3\theta)+\frac{3}{2}\mathcal I\right]}\]

Answer 21

Then from there, just solve for \(\Large\rm \mathcal I\).

Answer 22

So much integration in just 3 days.. it hurts >.<

But I see what you mean zepdrix, thanks again :D

Answer 23

yay team \c:/

Answer 24

You gotta learn the complex variable stuff!! >.<
So much more fun!

Answer 25

So.. Much.. Math..