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S=integral sine S(lnz)^2 integration

  1. Mar 15, 2006 #1
    can anyone do the following integration?! PLEASE!!!
    S=integral sine

    S(lnz)^2
     
  2. jcsd
  3. Mar 15, 2006 #2

    cepheid

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    1. Yes, we can do them, but we are not going to do your hw for you, as that will not help you in any way. Please read the forum guidelines on hw help here:

    https://www.physicsforums.com/showthread.php?t=94383

    The gist of it is that you must post your work and show that you have made a decent attempt, but got stuck somewhere. Then we can help point you in the right direction.

    2. Your question is not very clear, and your notation not very good. For the first integral, do you mean:

    [tex] \int{\sin x \ dx} [/tex]

    If so, are you serious??? You don't know this integral?

    As for this one:

    [tex] \int{(\ln z)^2 \ dz} [/tex]

    Did you attempt integration by parts?
     
  4. Mar 15, 2006 #3
    yes i dont no!!!!

    i tried it for like the whole half day

    i dont have too much time please HELP!!!!
     
  5. Mar 15, 2006 #4
    (zlnz-z)(z)-(zlnz-z)(1/z)????????
     
  6. Mar 15, 2006 #5

    cepheid

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    :rolleyes:

    Sorry, I don't believe you. If you attempted it for half a day, why not post your work as I suggested? Also, what exactly did you "try" for the first one? I can't imagine you getting very far putting pencil to paper if you don't already know the answer from beforehand. The integrals and derivatives of sine and cosine functions are something that you just look at the derivation/proof for once and then remember the result after that. Everybody knows them; there is no computation involved. This integral is given in any basic calculus textbook and so it makes no sense whatsoever that this integral would be given for hw.

    You never answered my question for the second one. Did you attempt integration by parts?
     
  7. Mar 15, 2006 #6
    I think by "S=integral sine" they meant that S will represent the integral sign. So they're trying to find [itex]\int (\ln{z})^2 \, dz[/itex], and as you suggested by parts would be a good idea.
     
  8. Mar 15, 2006 #7
    ................................................................................ i m dead
     
  9. Mar 15, 2006 #8

    cepheid

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    Thanks for clearing that up. That never even occurred to me (proper spelling always helps :tongue2: ). So I have maligned the OP somewhat needlessly, but not too much. :rofl:
     
  10. Mar 15, 2006 #9
    Notice that [itex] (\ln{z})^2 = (\ln{z})^2 * 1[/itex]. Can you see how integration by parts applies here?
     
  11. Mar 15, 2006 #10

    cepheid

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    I will ask one last time, (and hope you pay attention this time), DID YOU ATTEMPT INTEGRATION BY PARTS? If not, attempt it!!!
     
  12. Mar 15, 2006 #11
    thanx alot, but time out for me .... THANX ALOT THOUGH
     
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