arc length calc

**twilightstr** · February 19th 2009, 07:18 PM

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.

**Soroban** · February 19th 2009, 08:52 PM

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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