# Some integrals I just don't know how to do

1. Sep 13, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

Knowing what I do (U-Substitution, beginning Integration by Parts) what would you do for these?

(ln t)^2
(sin t)^2

2. Relevant equations

3. The attempt at a solution

All I have been able to do is change these to (ln t)(ln t) and then try by parts, but I just end up with something more complicated.

2. Sep 13, 2011

### Staff: Mentor

For the second one, sin2(t) = (1 - cos(2t))/2.
For the first, I think you are on the right track with integration by parts. Can you show us what you've tried?

3. Sep 13, 2011

### KingBigness

You are correct with using integration by parts on (lnt)^2

Instead of splitting it into (lnt)(lnt) set your v=log^2(t) and du=dt and find dv and u then sub that into your integration by parts formula.

See how that goes. Keep in mind you will do integration by parts twice in this problem.

4. Sep 13, 2011

### 1MileCrash

Sorry, I'm not sure I understand. Why would I set v to be log^2(t)? Did you mean u instead? Or dv? My understanding is that u and dv are the two parts that compose the actual function, while du and v are derivatives and integrals of the respective.

5. Sep 13, 2011

### Staff: Mentor

Yes, I think KingBigness had his letters mixed around.

6. Sep 13, 2011

### 1MileCrash

Ok. For the first one I'm coming to:

(x/2) - (sin2x/4) + C

The homework marks this wrong, so I can only assume I'm messing up somewhere..

7. Sep 13, 2011

### KingBigness

Sorry I just had them back to front...

u=log^2(t) dv=dt
du=2log(t)/t dt v=t

do integration by parts to the above and tell me what you get.

8. Sep 13, 2011

### 1MileCrash

For the second one I seem to have found myself in an infinite loop..

$\int (lnt)^2 dt$

u = ln(t)^2
du = [(lnt)^2]/t dt

v = t
dv = dt

$t(lnt)^2 - \int\frac{t(lnt)^2}{t} dt$

$t(lnt)^2 - \int (lnt)^2 dt$

Which is just the same integral for the second term.

9. Sep 13, 2011

### KingBigness

[STRIKE]If I were you I would use my log laws to bring the ^2 out the front of the integral sign and again integrate the log(t) dt via integration by parts.[/STRIKE]

Last edited: Sep 13, 2011
10. Sep 13, 2011

### 1MileCrash

Right, that changes everything. Thanks!

Why couldn't I have done that from the get-go, though?

11. Sep 13, 2011

### KingBigness

Because I lied to you. Ignore my last comment and lets go back to your working.

If you look at your du you have [(lnt)^2]/t. This is not correct.

The derivative of (lnt)^2 = 2[lnt]/t NOT [(lnt)^2]/t

therefore, you get ...... t(lnt)2−∫2(lnt)dt

factor out the 2 because it is a constant and you get... t(lnt)2−2∫(lnt)dt

now integrate that by part.

12. Sep 13, 2011

### KingBigness

What I got mixed up and I think you got mixed up is thinking you can bring the power out the front. This is only the case if the variable is squared (ln(t^2)) and not the whole (lnt)^2

13. Sep 13, 2011

### KingBigness

Sorry for my confusing rant but I hope you got some help from it

14. Sep 13, 2011

### Rayquesto

use integration by parts and you should be able to get the right answer. I just worked out the first question doing so. Now, I'm gonna see what the second one will be using, but I think it's pretty much the same thing.

15. Sep 13, 2011