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Math Help - arc length calc

  1. #1
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    arc length calc

    find the arclength of the curve y=lnx [1,square root 3]

    this is all i have so far: L= integral form [1,square root 3] of square root of [(1+ (1/x)^2]dx.
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  2. #2
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    Hello, twilightstr!

    Find the arclength of the curve y\,=\,\ln x \text{ on }[1, \sqrt{3}]

    This is all i have so far: . L\:=\:\int_1^{\sqrt{3}} \sqrt{1 + \left(\tfrac{1}{x}\right)^2}\,dx . . . . Good!

    We have: . \int\sqrt{\frac{x^2+1}{x^2}}\,dx \;=\;\int\frac{\sqrt{x^2+1}}{x}\,dx


    Let: . x = \tan\theta \quad\Rightarrow\quad dx = \sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{x^2+1} = \sec\theta<br />


    Substitute: . \int\frac{\sec\theta\,(\sec^2\!\theta\,d\theta)}{\  tan\theta} \;=\;\int\frac{\sec^3\!\theta}{\tan\theta}\,d\thet  a


    We have: . \sec\theta\cdot\frac{\sec^2\!\theta}{\tan\theta} \;=\;\sec\theta\,\frac{\tan^2\!\theta+1}{\tan\thet  a} \;=\;\sec\theta\left(\tan\theta + \frac{1}{\tan\theta}\right)

    . . . . = \;\sec\theta\tan\theta + \frac{\sec\theta}{\tan\theta} \;=\;\sec\theta\tan\theta + \frac{1}{\sin\theta}


    So we have: . \int(\sec\theta\tan\theta + \csc\theta)\,d\theta \;=\;\sec\theta + \ln|\csc\theta - \cot\theta| + C

    . . and so on . . .

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