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Some integrals I just don't know how to do

  1. Sep 13, 2011 #1
    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. jcsd
  3. Sep 13, 2011 #2

    Mark44

    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?
     
  4. Sep 13, 2011 #3
    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.
     
  5. Sep 13, 2011 #4
    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.
     
  6. Sep 13, 2011 #5

    Mark44

    Staff: Mentor

    Yes, I think KingBigness had his letters mixed around.
     
  7. Sep 13, 2011 #6
    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..
     
  8. Sep 13, 2011 #7
    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.
     
  9. Sep 13, 2011 #8
    For the second one I seem to have found myself in an infinite loop..

    [itex]\int (lnt)^2 dt[/itex]

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

    v = t
    dv = dt

    [itex]t(lnt)^2 - \int\frac{t(lnt)^2}{t} dt[/itex]


    [itex]t(lnt)^2 - \int (lnt)^2 dt[/itex]

    Which is just the same integral for the second term.
     
  10. Sep 13, 2011 #9
    [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
  11. Sep 13, 2011 #10
    Right, that changes everything. Thanks!

    Why couldn't I have done that from the get-go, though?
     
  12. Sep 13, 2011 #11
    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.
     
  13. Sep 13, 2011 #12
    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
     
  14. Sep 13, 2011 #13
    Sorry for my confusing rant but I hope you got some help from it
     
  15. Sep 13, 2011 #14
    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.
     
  16. Sep 13, 2011 #15
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