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Math Help - Arclength Integration Help

  1. #1
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    Arclength Integration Help

    Hey,

    Ok I've been having a bit of trouble on a homework question and I was just wondering if I am even going about this the right way.

    QUESTION:

    Find the Arclength of y = 1/x from x=1 to x=2.

    So I got the following integral:

    I = \int^2_1 \sqrt{1 + \big(\tfrac{-1}{x^2}\big)^2}dx

    Then I did:

    \tan\theta = \frac{1}{x^2} \implies \sec^2\theta d\theta = \frac{-2}{x^3} dx

    So I get:

    I = \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\sec^3\theta \cdot x^3 \cdot d\theta}{-2} = -\tfrac{1}{2} \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\sec^3\theta \cdot d\theta}{\tan\theta\sqrt{\tan\theta}}

    And after that I used integration by parts to get:

    I = -\bigg(\frac{\sqrt{\tan\theta}}{\sin\theta}\bigg|^{  \arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} + \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\csc\theta \cdot d\theta}{\sqrt{\tan\theta}}\bigg)

    And this is where I hit a wall. I don't know how to solve that last integral. And I might be going about this the entirely wrong way.

    Thanks,

    Trevor
    Last edited by TrevorP; January 27th 2008 at 03:11 PM.
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  2. #2
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    Krizalid's Avatar
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    When you make a substitution, we require to get the new integration limits.

    Since you made \tan \theta  = \frac{1}<br />
{{x^2 }} plug x=(1,2) to get integration limits for \theta.
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  3. #3
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    Yeah you're right, I do normally write that in on my papers I hand in. I'll add that to my post though.
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  4. #4
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    BUMP.

    Anyone got any ideas?
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  5. #5
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    Nevermind my prof just sent out a message saying that it should be y = ln(x). Which should make this question...well considerably easier.
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