Question

∫e3tcos(−4t)dt

Answer 1

This is what we have to solve?\[\Large \int\limits e^{3t}\cos(-4t)\;dt\]

Answer 2

yes

Answer 3

Before we get to work on that, are you familiar with how to solve `this` one?
\[\Large \int\limits e^{t}\cos t\;dt\]
If not, our problem is going to be quite a bit tougher to get through.

Answer 4

no

Answer 5

Hmmm.. well ok.. here's the thing.

Answer 6

How comfortable are you with `Integration by Parts` ?
We'll need to apply integration by parts several times to solve this.

It's very easy to get lost in all of the coefficients that are going to pop out due to our 3t and -4t.

Want me to burn through the e^t cos t really quick so you can see the path we'll be taking?
Or is that not necessary?

Answer 7

actually is not necessary

Answer 8

\[\Large \int\limits\limits e^{3t}\cos(-4t)\;dt\]So which one do you want to be your `u`, and `dv` ? :)
It doesn't matter which is which in this case :D so you choose.

Answer 9

u=cos(-4t), dv=e^3t

Answer 10

able to find your `du` and `v` ok? :)

Answer 11

du=-sin(-4t)-4, v=3 e^3t

Answer 12

I'm assuming the -4 is multiplication, not subtraction right?

And we need to fix that v up a second also.
Integrating shouldn't give us a factor of 3.
We should `lose` a factor of 3 when we integrate.

Answer 13

Everything else you did looks good though.
So this will be our substitution block.
\[\Large u=\cos(-4t) \;\qquad\to\qquad du=4\sin(-4t)\;dt\]\[\Large dv=e^{3t}dt \qquad\qquad\to\qquad v=\frac{1}{3}e^{3t}\]
Understand why our v is producing a 1/3?

Answer 14

yes

Answer 15

continue

Answer 16

Try to follow along, I don't wanna do all the work for you hehe.
Important that you actually learn it! XD

These coefficients will get a little tricky, let's be careful.
Applying integration by parts gives us,\[\Large \frac{1}{3}e^{3t}\cos(-4t) \quad-\quad\frac{4}{3}\int\limits e^{3t}\sin(-4t)\;dt\]

Answer 17

With the exponential and the trig function as our parts, it feels like we're going to just keep going in circles doesn't it?

Well it'll work out in a second. We'll be able to do some cool mathemagic.

Answer 18

We'll apply integration by parts again.
The important thing here is:
`We MUST match up our u's and dv's with the same type of function as we did before!`

So in our first IBP, we let the trig function be `u`, so we'll do that again.

Answer 19

\[\Large u=\sin(-4t) \qquad \to\qquad du=?\]\[\Large dv=e^{3t}dt \qquad\to\qquad v=?\]

Answer 20

du= cos(-4t)4, v=1/3e^3t dt

Answer 21

mmm I think we get a -4 on our du right? not positive 4

Answer 22

\[\Large u=\sin(-4t) \qquad \to\qquad du=-4\cos(-4t)\;dt\]\[\Large dv=e^{3t}dt \quad\;\qquad\to\qquad v=\frac{1}{3}e^{3t}\;dt\]

Answer 23

So there is our substitution stuff.

Answer 24

\[\Large \frac{1}{3}e^{3t}\cos(-4t) \quad\quad\color{royalblue}{-\frac{4}{3}\int\limits\limits e^{3t}\sin(-4t)\;dt}\]
So our IBP will give us,\[\large \frac{1}{3}e^{3t}\cos(-4t)\color{royalblue}{-\frac{4}{3}\left[\frac{1}{3}e^{3t}\sin(-4t)+\frac{4}{3}\int\limits e^{3t}\cos(-4t)\;dt\right]}\]

Answer 25

It's starting to get a little messy :)
Following along ok?
Take a minute to make sure that makes sense.

Answer 26

okay

Answer 27

Distributing the -4/3 to each term in the square brackets gives us,\[\large \frac{1}{3}e^{3t}\cos(-4t)-\frac{4}{9}e^{3t}\sin(-4t)-\frac{16}{9}\int\limits\limits e^{3t}\cos(-4t)\;dt\]

Answer 28

Here's where the magic happens.
Refer back to our integral that we started with. Give it a variable name like `I` so we can work with it easier.\[\Large \int\limits e^{3t}\cos(-4t)\;dt \quad=\quad I\]

Answer 29

See how this integral is showing up in our solution work??

Answer 30

Here's what we've determined so far:\[\color{royalblue}{\int\limits e^{3t}\cos(-4t)dt}=\frac{1}{3}e^{3t}\cos(-4t)-\frac{4}{9}e^{3t}\sin(-4t)-\frac{16}{9}\color{royalblue}{\int\limits\limits e^{3t}\cos(-4t)\;dt}\]

Answer 31

With our variable substitution, we can write it like this,\[\Large \color{royalblue}{I}=\frac{1}{3}e^{3t}\cos(-4t)-\frac{4}{9}e^{3t}\sin(-4t)-\frac{16}{9}\color{royalblue}{I}\]