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Trig integral

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int tan^5(6x) sec^3(6x) dx[/tex]



    2. Relevant equations



    3. The attempt at a solution
    first off I set u=6x to get 1/6[tex]\int tan^5(u) sec^3(u) dx[/tex]
    then I used trig identities to put tangent in terms of secant and I came up with

    [tex]\int sec^9(u)-3sec^7(u)+3sec^5(u)-sec^3(u) dx[/tex]
    Not sure where to go from here, or if I'm doing this the right way
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2008 #2
    Watch your substitutions.

    [tex]\frac 1 6\int\tan^{5}u\sec^{3}udu[/tex]

    [tex]\frac 1 6\int\tan^{4}u\sec^{2}\sec u\tan udu[/tex]

    *[tex]\tan^{2}u+1=\sec^{2}u[/tex]

    Take it from here.
     
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