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Homework Help: Integral please check

  1. Jan 20, 2009 #1

    prime-factor

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

    ∫ -60x3tan(3x4)dx

    2. Relevant equations

    integration rules

    3. The attempt at a solution

    => ∫ -60x3tan(3x4)dx

    => u = (3x4)

    => du = 12x3dx

    => (-5)∫ (1/5)-60x3tan(3x4)dx

    => -5 ∫ tan u

    => -5 .-ln cos(u) , where u = (3x4)

    => 5 . ln cos (3x4)

    I am using this integral to check an original equation which I differentiated which was:
    ln (cos5(3x4) , so where have I gone wrong when integrating, because I'm not getting it back ? ( I have started just learning the basics of integration so be nice :) )
     
    Last edited: Jan 20, 2009
    prime-factor, Jan 20, 2009
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  3. Jan 21, 2009 #2

    Mark44

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    Staff: Mentor

    I don't see anything wrong with what you have, short of a missing constant of integration and missing absolute values around cos(u) and a few other minor things.

    I suspect that you made an error in your differentation and got the result you did.

    Assuming you started with ln (cos5(3x4)),
    d/dx[ ln (cos5(3x4)]
    = 1/cos5(3x4) * d/dx [cos5(3x4]
    = 1/cos5(3x4) * (5)*cos4(3x4) * d/dx( cos(3x4))
    = 1/cos5(3x4) * (5)*cos4(3x4) * -sin(3x4) * d/dx (3x4)
    = 1/cos5(3x4) * (5)*cos4(3x4) * -sin(3x4) * 12x3)
    [tex]= \frac{-60 x^3 sin(3x^4)}{cos(3x^4}[/tex]
    = -60x3 tan(3x4)

    One minor thing I referred to is your misuse of the "implies" symbol where you should be using "equals." Each of your integration steps involves an expression, not a statement, so from one step to the next you should use "=".

    You used ==> correctly in your substitution, namely
    u = (3x4)
    => du = 12x3dx​
    Here, the first line is an equation, a kind of statement, and the second equation follows from it.
     
    Mark44, Jan 21, 2009
  4. Jan 21, 2009 #3

    prime-factor

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    Thanks for the thorough response :).

    But I am still not sure why I can't get back: 'ln (cos5(3x4)'

    I differentiated it as you did and got the same answer, but
    I am not getting it back when integrating?
     
    prime-factor, Jan 21, 2009
  5. Jan 21, 2009 #4

    Mark44

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    When you say "get back" what are you starting with?
     
    Mark44, Jan 21, 2009
  6. Jan 21, 2009 #5

    prime-factor

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    Okay. Sorry I explained it so badly.

    I started with a function which was:

    y = ln cos5(3x4)

    differentiating this gives: -60x3tan(3x4)

    Now why is it that when I integrate '-60x3tan(3x4)' I don't get back the original function which was 'y = ln cos5(3x4)'.
     
    prime-factor, Jan 21, 2009
  7. Jan 21, 2009 #6

    Mark44

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    You sort of do. ln w5 = 5 ln w.
     
    Mark44, Jan 21, 2009
  8. Jan 21, 2009 #7

    prime-factor

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    Oh My God... I can't believe I did not see that before!!

    I forgot that cos5(3x4) = (cos(3x4)5

    and hence with the ln in front it is treated like a basic operation.

    You are very kind for helping me with this and being so pacient :).

    Thanks Again.
     
    prime-factor, Jan 21, 2009
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