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Arclength (Difficult?)

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve [itex]y=2ln(sin\frac{1}{2}x)[/itex], [itex]\frac{\pi}{3}\leq x\leq\pi[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Alright so I figured out the derivative of y is cot(1/2)x so I put it into the arclength formula to get:
    [tex]\int_{\frac{\pi}{3}}^{\pi} \sqrt{1+{cot^{2}(\frac{x}{2})}}dx[/tex]

    But I don't know how to solve that.. Looking at wolfram it seems like a really ugly integral so I don't want to put a lot of effort into solving it if I have the integral wrong.

    Thanks for any help.
  2. jcsd
  3. Apr 3, 2013 #2
    Never forget your Pythagorean identities. Divide [itex] sin^2 \theta +cos^2 \theta = 1 [/itex] through by [itex] sin^2 \theta [/itex] to get [itex] 1+cot^2\theta = csc^2\theta [/itex]. Just let [itex] \theta = \frac{x}{2} [/itex] and the integral simplifies greatly.
  4. Apr 3, 2013 #3
    Yeah I just kept going and realized that. So I did:
    u=x/2 du=1/2 dx
    [tex]2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{1+cot^{2}u}du=2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sqrt{csc^{2}u}du=2ln(cscu+cotu)|_{\frac{\pi}{6}}^{\frac{\pi}{2}}[/tex]
  5. Apr 3, 2013 #4
    Yeah I finished it and evaluated, [itex]-2ln(2+\sqrt{3})[/itex]

    Forgot that I knew what the integral of csc was :P. Thanks.
  6. Apr 3, 2013 #5
    Don't forget to change your bounds after making a substitution.
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