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Integrating x^2 . sinx

  1. Dec 20, 2006 #1

    Mo

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    1. The problem statement, all variables and given/known data
    Find the integral of:


    2. Relevant equations
    x^2 . sinx dx (With upper and lower limits of pi and 0 respectively.)


    3. The attempt at a solution

    Integrating for the first time:
    -x^2 . cos x + 2 [integral of] x . cosx dx

    After integrating a part of my first integration:
    -x^2 . cosx + 2xsinx + 2cosx

    ----------------------------

    I used the method of integration by parts where u = x^2 and dv = sinx.

    I don't really need the final answer, i just want to see where i am going wrong with the actual integration.

    Anything with trig in it seems to completely throw me :grumpy:

    Thanks.

    ----------------------------
     
  2. jcsd
  3. Dec 20, 2006 #2
    Mo, your solution (-x^2 . cosx + 2xsinx + 2cosx) is CORRECT :: approve

    marlon
     
  4. Dec 20, 2006 #3

    Mo

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    Thanks for the help. The thing is, if i input the limits of pi and 0, i get a completely different answer to that in the book!

    The book gives the answer as ((pi^2) - 4)

    Either i am substituting the limits incorrectly (likely) or the book's answer is incorrect (unlikely..)

    Am I right in saying that first we substitute pi for x, get an answer, then substitute 0 for x and again get an answer.

    We then subtract the our the latter answer from the first.

    ie:

    (-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(0))

    which will give us the final answer.

    Regards.

    ps: thanks for spotting that, cristo
     
    Last edited: Dec 20, 2006
  5. Dec 20, 2006 #4

    cristo

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    Staff Emeritus
    Science Advisor

    Probably a typo, but this should read

    (-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(0))
     
  6. Dec 20, 2006 #5
    again correct...It must be a typo

    cos(pi) = -1
    sin(pi) = 0

    i get

    (-pi^2* (-1) + 2(pi)* 0 + 2* (-1)) - (2)
    marlon
     
    Last edited: Dec 20, 2006
  7. Dec 20, 2006 #6

    Mo

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    In that case the book was correct.

    I was doing the substitutions using my calculator which was not in radians mode. :cry:

    Thanks for your help though, much appreciated.
     
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