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Integrating a trig function divided by a trig function

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the arc length of the curve r=4/θ, for ∏/2 ≤ θ ≤ ∏

    2. Relevant equations

    L= ∫ ds = ∫ √(r^2 + (dr/dθ)^2) dθ

    3. The attempt at a solution

    After some calculations, and letting θ = tanx, I now have to find ∫ ((secx)^3/(tanx)^2). I am not sure how to do this, but i have found online that ∫ (secx)^3 = (1/2)sec(x)tan(x)+(1/2)ln|sec(x)+tan(x)|. Using integration by parts and letting u = 1/(tan(x))^2= (cot(x))^2 and du/dx = -2cot(x).(cosec(x))^2 is looking very tedious. How do I solve this problem correctly?
     
  2. jcsd
  3. Mar 18, 2012 #2

    HallsofIvy

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    I don't see how you got [itex]sec^3(x)[/itex]. I get just sec(x) in the numerator.

    Eventually, I reduce it to
    [tex]\int \frac{sin^3(x)}{cos^2(x)}dx[/tex]

    Factor out a sin(x) to use with the dx and convert the remaining [itex]sin^2(x)[/itex] to [itex]1- cos^2(x)[/itex].
     
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