1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral please check

  1. Jan 20, 2009 #1
    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
  2. jcsd
  3. Jan 21, 2009 #2


    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.
  4. Jan 21, 2009 #3
    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?
  5. Jan 21, 2009 #4


    Staff: Mentor

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


    Staff: Mentor

    You sort of do. ln w5 = 5 ln w.
  8. Jan 21, 2009 #7
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?