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Arc length (cosecant-cubed)

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the arc length of [tex]y=\sqrt{x}[/tex] from x=0 to x=2.

    3. The attempt at a solution

    I don't know, this is a nastier integral than it looks. From the substitutions,

    [tex]s = \int_0^2 \sqrt{1 + \frac{1}{4x}} dx[/tex]. From doing this over and over again I already know the answer will have square roots and logarithms but I keep getting it wrong!

    I've tried two methods so far, both end with evaluating [tex]\int cosec^3 x dx[/tex], but with different limits.

    Heres one: [tex]u = \sqrt{1 + \frac{1}{4x}}[/tex]. Then, I get [tex]s = \int_1^{\sqrt{9/8}} \frac{u^2}{- 2(u^2-1)^2} du[/tex]

    Substitute [tex]u = sec \theta[/tex], [tex]s = \int_{0}^{sec^{-1} \sqrt{(9/8)}} - cosec^3 \theta d\theta = \frac{1}{8}(3\sqrt{2}... [/tex] which has the wrong form.

    ([tex]2 \int cosec^3 x dx = - cot\theta cosec\theta - ln|cot\theta + cosec\theta| + c[/tex] by integration by parts.)

    The answer is [tex]\frac{1}{2} (3\sqrt{2} + ln (1 + \sqrt{2}) [/tex] or about 2.56. I'm quite sure my general method is correct (the square roots and logarithms are consistent with the ansewr), but there's a problem with the limits.

    The other one is to substitute a trig expression directly (ie [tex]\frac{1}{4x} = tan^2 \theta[/tex], with different limits.
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2


    Staff: Mentor

    Here's something different you might try. Instead of working with y = sqrt(x), you could instead work with y = x^2. The arc length along your curve between x = 0 and x = 2 is exactly the same as the arc length along the curve between x = 0 and x = sqrt(2). If you draw the graphs of y = sqrt(x) and y = x^2, you should be able to convince yourself of what I'm saying.

    The integral becomes:
    [tex]\int_0^{\sqrt{2}} \sqrt{1 + 4x^2} dx[/tex]

    You can use a trig substitution here, with tan [itex]\theta[/itex] = 2x.
  4. Jan 26, 2009 #3
    Mark, thanks so much, that worked. You meant that since x^2 is the inverse function, it is exactly symmetrical about the y=x, so (2, root-2) maps to (root-2, 2), right?

    But I hope to be able to solve the original problem as-is. Just changing functions unfortunately won't reveal what was wrong with what I did above.
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