1. Limited time only! Sign up for a free 30min personal 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 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


    User Avatar
    Science Advisor
    Homework Helper

    Re: Transcendentals

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


    User Avatar
    Science Advisor

    Maple says the answer is

    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]



    Right on
  5. Feb 25, 2004 #4
    Awesome thats what I got. Thanks guys.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Integral involving sec and tan