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Integtrating secant and tangent

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data

    (integrate)sec^3(2x)*tan(2x)dx

    2. Relevant equations



    3. The attempt at a solution

    Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

    I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.
     
  2. jcsd
  3. Feb 5, 2007 #2
    u = 2x
    du = 2 dx

    [itex]\frac 12 \int \sec ^3 u \tan u du[/itex]

    Now let [itex]v = \sec ^2 u[/itex] What is dv?
     
  4. Feb 6, 2007 #3

    dextercioby

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    There's the mistake.
     
  5. Feb 6, 2007 #4

    Gib Z

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    Slight change to IMDereks post. I dont know wat dv is, cant be bothered working it out, i think my methods quicker. let v= sec u, split the integrand into sec^2 u (sec u tan u), that way the things in the brackets it dv/du, the integral is now integral: v^2 dv. simpler i think
     
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