# What is the arc length of f(x) = ln(x^2)  on x in [1,3] ?

Apr 2, 2016

$\sqrt{13} - \sqrt{5} + 2 \cdot \ln \left(\left(\sqrt{65} + 2 \cdot \sqrt{13} - 2 \cdot \sqrt{5} - 4\right)\right) \cong 3.006$

#### Explanation:

To do this we need to apply the formula for the length of the curve
mentioned in:
How do you find the length of a curve using integration?

We start from
$L = {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$
$f \left(x\right) = \ln {x}^{2} = 2 \ln x$ => $f ' \left(x\right) = \frac{2}{x}$
Then
$L = {\int}_{1}^{3} \sqrt{1 + \frac{4}{x} ^ 2} \mathrm{dx} = {\int}_{1}^{3} \frac{\sqrt{{x}^{2} + 4}}{x} \mathrm{dx} = F \left(x = 3\right) - F \left(x = 1\right)$

Making
$x = 2 \tan y$
$\mathrm{dx} = 2 {\sec}^{2} y \cdot \mathrm{dy}$
we get
$F \left(x\right) = \int \frac{\sqrt{{x}^{2} + 4}}{x} \mathrm{dx} = \int \frac{2 \sec y \cdot \cancel{2} {\sec}^{2} y}{\cancel{2} \tan y} \mathrm{dy} = 2 \int \left(\frac{1}{\cos} ^ 3 y\right) \left(\cos \frac{y}{\sin} y\right) \mathrm{dy} = 2 \int \frac{\mathrm{dy}}{{\cos}^{2} y \cdot \sin y}$

But
$\frac{1}{{\cos}^{2} y \cdot \sin y} = \frac{\sin y}{{\cos}^{2} y} + \frac{1}{\sin} y$

So
$F \left(x\right) = 2 \int \sin \frac{y}{\cos} ^ 2 y \mathrm{dy} + 2 \int \frac{\mathrm{dy}}{\sin} y$

Making $\cos y = z$ => $\sin y \cdot \mathrm{dy} = - \mathrm{dz}$
The first part becomes
$- 2 \int \frac{\mathrm{dz}}{z} ^ 2 = \frac{2}{z} = \frac{2}{\cos} y$

Therefore
$F \left(x\right) = \frac{2}{\cos} y + 2 \ln | \csc y - \cot y | + c o n s t .$
But
$x = 2 \tan y$ => $\sin y = \frac{x}{2} \cos y$
${\sin}^{2} y + {\cos}^{2} y = 1$ => $\left({x}^{2} / 4 + 1\right) {\cos}^{2} y = 1$ => $\cos y = \frac{2}{\sqrt{{x}^{2} + 4}}$
$\to \sin y = \frac{x}{2.} \left(\frac{2}{\sqrt{{x}^{2} + 4}}\right)$ => $\sin y = \frac{x}{\sqrt{{x}^{2} + 4}}$
So
$F \left(x\right) = \sqrt{{x}^{2} + 4} + 2 \ln | \frac{\sqrt{{x}^{2} + 4}}{x} - \frac{2}{x} | + c o n s t .$
$F \left(x\right) = \sqrt{{x}^{2} + 4} + 2 \ln | \sqrt{{x}^{2} + 4} - 2 | - 2 \ln | x | + c o n s t .$

Finally
$L = F \left(x = 3\right) - F \left(x = 1\right) = \sqrt{13} + 2 \ln \left(\sqrt{13} - 2\right) - 2 \ln 3 - \left(\sqrt{5} + 2 \ln \left(\sqrt{5} - 2\right) - 2 \ln 1\right)$
$L = \sqrt{13} - \sqrt{5} + \ln \left(\frac{\sqrt{13} - 2}{3 \left(\sqrt{5} - 2\right)}\right)$
But
$\frac{\sqrt{13} - 2}{\sqrt{5} - 2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \sqrt{65} + 2 \sqrt{13} - 2 \sqrt{5} - 4$
So

$L = \sqrt{13} - \sqrt{5} + \ln \left(\frac{\sqrt{65} + 2 \sqrt{13} - 2 \sqrt{5} - 4}{3}\right) \cong 3.006$