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Find length of curve. Integral of sec?

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve y=ln(cosx) from 0 to pi/3


    2. Relevant equations
    integral of (1+(y')^2)^1/2


    3. The attempt at a solution
    First, find y'. y' is equal to -sin/cos, or simply -tan(x).

    [tex]\int[/tex][tex]\sqrt{1+(-tan(x)^{2}}[/tex]
    = [tex]\int[/tex][tex]\sqrt{sec(x}[/tex]
    1. The problem statement, all variables and given/known data

    Can this possibly be right so far? This is a horrendous integral. I've expended so much energy on this problem, but if someone can at least validate that I am on the right track, I'll post my work from here out.
     
  2. jcsd
  3. Jun 18, 2009 #2

    rock.freak667

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    [tex]1+tan^2x=sec^2x \Rightarrow \sqrt{1+tan^2x}= \sqrt{sec^2x}=secx[/tex]
     
  4. Jun 18, 2009 #3
    :redface::cry:

    Thank you so much.
     
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