Some integrals I just don't know how to do

[1MileCrash]

Homework Statement

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

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

Homework Equations

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.

 

[Mark44]

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?

 

[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.

 

[1MileCrash]

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.

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.

 

[Mark44]

Yes, I think KingBigness had his letters mixed around.

 

[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..

 

[KingBigness]

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.

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.

 

[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.

 

[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]

 

[1MileCrash]

Right, that changes everything. Thanks!

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

 

[KingBigness]

Because I lied to you. Ignore my last comment and let's 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.

 

[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

 

[KingBigness]

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

 

[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 going to see what the second one will be using, but I think it's pretty much the same thing.

 

[Rayquesto]

oh for the second part, you should use this: http://wiki.answers.com/Q/Integral_of_sin_squared_x

where the sin^2theta= 1/2(1-cos(2X))
damn I got to review my trig identities too. I'm forgetting these practical things!