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Integral involving sec and tan

  1. Feb 24, 2004 #1
    How do I perform this integral?
    [tex]\int \sec^2{3x} \tan^5{3x} \,dx [/tex]
     
  2. jcsd
  3. Feb 24, 2004 #2

    NateTG

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    Re: Transcendentals

    Try [tex]u=\tan{3x}[/tex]
     
  4. Feb 24, 2004 #3

    ShawnD

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    Maple says the answer is
    [tex]\frac{tan^6(3x)}{18}[/tex]

    Now lets see if we can get that.
    First of all, the derivative of tan is sec^2. So that leads to me belive the process would be the substitution method.

    I'll assign "U" as tan(3x)

    [tex]\int sec^2(3x)tan^5(3x) dx[/tex]

    [tex]\int sec^2(3x)U^5 dx[/tex]

    Now take the derivative of U with respect to x

    [tex]U = tan(3x)[/tex]

    [tex]\frac{dU}{dx} = 3sec^2(3x)[/tex]

    [tex]dx = \frac{dU}{3sec^2(3x)}[/tex]

    Now fill that back into what we had above

    [tex]\int sec^2(3x)U^5 \frac{dU}{3sec^2(3x)}[/tex]

    [tex]\frac{1}{3} \frac{U^6}{6}[/tex]

    [tex]\frac{U^6}{18}[/tex]

    [tex]\frac{tan^6(3x)}{18}[/tex]


    Right on
     
  5. Feb 25, 2004 #4
    Awesome thats what I got. Thanks guys.
     
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