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Trig. Integratin Suggestion

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Can I get some help with:

    [tex] \int -\frac {sin^2(3x)}{3cos(x)} dx[/tex]

    2. Relevant equations



    3. The attempt at a solution

    It looks like a substitution would work but I'm striking out with:


    [tex] u = sin(3x)[/tex]
    [tex] du = 3cos(3x)[/tex]
    because this is now in the denominatior

    [tex] u = cos(3x)[/tex]
    [tex] du = -3sin(3x)[/tex]

    I'm left with a sin(3x) term.

    I'm guessing a trig identity is now going to be involved.

    I have found that

    [tex] \int sec(x) tan(x) dx = sec(x) [/tex]

    My problem can be re-written as

    [tex] \int - sin(3x) tan(3x) dx [/tex]

    ???

    Thanks for the help.

    -Sparky_
     
    Last edited: Jan 12, 2008
  2. jcsd
  3. Jan 12, 2008 #2

    mda

    User Avatar

    Just to clarify, is the argument of the cos correct?
    Its just that cos(3x) will be simpler...
     
  4. Jan 12, 2008 #3

    Gib Z

    User Avatar
    Homework Helper

    Assuming you meant the argument of the cosine to be 3x, its easy to see that you can solve your integral if you can solve:

    [tex]\int \frac{ \sin^2 x}{\cos x} dx[/tex]

    To do that, remember [itex]\sin^2 x = 1- \cos^2 x[/itex], so that the integral becomes

    [tex]\int \left( \sec x - \cos x \right) dx[/tex].

    Both of those are usually regarded as standard integrals, although sometimes the secant integral is not. To do that one;

    [tex]\int \sec x dx = \int \frac{\cos x}{\cos^2 x} dx = \int \frac{\cos x}{1- \sin^2 x} dx = \int \frac{1}{(1+u)(1-u)} du[/tex] when u = sin x. Now do partial fractions and your home free.
     
  5. Jan 13, 2008 #4
    Yes - I meant cos(3x).

    Thanks
    Sparky_
     
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