# Homework Help: Integral Help

1. Mar 8, 2008

### ns5032

The integral I need help with is:

integral of:

[3e^(t/3)]*cos(3t)

I know that I can pull out the 3 in front of the integral, but after that what??

2. Mar 8, 2008

### arunbg

Do you know how to http://en.wikipedia.org/wiki/Integration_by_parts" [Broken]

Last edited by a moderator: May 3, 2017
3. Mar 8, 2008

### rocomath

$$3\int e^{\frac t 3}\cos{3t}dt$$

You need to show more effort than that if you want to receive a lot of help.

4. Mar 8, 2008

### EngageEngage

If I'm thinking of the problem correctly, you can solve it by doing integration by parts twice -- the second time you do it you should receive your initial integral again and some other terms (if you made the right substitutions for your u and v terms). Then you just solve for your integral and you're done. I didn't work it out at all, so that could be wrong. However, I think thats usually the case when you have an exponential and a single trig term.

5. Mar 8, 2008

### ns5032

Well if I do integration by parts, the cos switches back and forth between sin and cos, and the e^(t/3) stays in there every time... I don't know what I'm doing wrong...

u = e^(t/3)
du = (1/3)e^(t/3)

v = (1/3)sin(3t)
dv = cos(3t)

so then it's:
uv- integral of v*du
[e^(t/3)*(1/3)sin(3t)]-integral of [(1/3)sin(3t)*(1/3)e^(t/3)]
and by simplifying:
[(1/3)e^(t/3)*sin(3t)]-(1/9)*integral of [sin(3t)*e^(t/3)]

It just seems like I'll keep making circles...?

6. Mar 8, 2008

### dynamicsolo

Have faith -- it's supposed to do that! Integrals of the form exponential times sine or cosine have to go through two levels of integration by parts. What you'll see, after the second time, is that you get, on the right-hand side of the equation, your original integral back multiplied by a negative constant. You now have an exercise in algebra! Move that multiplied integral to the other side and you'll have an expression which reads (constant) times original integral = remaining right-hand side terms. If you divide through by the constant, you'll have the result for your original integral (+C).

7. Mar 8, 2008

### EngageEngage

you will need to make one circle to solve the integral. like i said this will give you your original integral and some other terms. it will look like this:

OrigIntegral = -OrigIntegral + Terms
2OrigIntegral = Terms
OrigIntegral = Terms/2, the final answer.

You need to make sure that the 'Terms' part doesn't get thrown out in your second integration by parts attempt. This will happen if you chose the wrong u and v the second time around, in which case you will be left with: