# How do you find the arc length of y=ln(cos(x)) on the interval [pi/6,pi/4]?

Aug 21, 2014

You can find the Arc Length of a function by first finding its derivative and plugging into the known formula:

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Process:

With our function of $\ln \left(\cos \left(x\right)\right)$, we must first find its derivative. The shortcut for $\ln \left(u\right)$ -type functions is to just take the derivative of the inside and place it over the original of what's inside. Since the derivative of $\cos \left(x\right)$ is ($- \sin x$), we end up with:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \frac{x}{\cos} x$,

which is equal to:

$- \tan x$.

Plugging into our Arc Length formula, we have:

$L = {\int}_{a}^{b} \sqrt{1 + {\left(- \tan x\right)}^{2}} \mathrm{dx}$.

If we square the $- \tan x$ term, we get:

$L = {\int}_{a}^{b} \sqrt{1 + {\tan}^{2} \left(x\right)} \mathrm{dx}$

Since $1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right)$ is one of our known trig identities, we can change our equation into:

$L = {\int}_{a}^{b} \sqrt{{\sec}^{2} \left(x\right)} \mathrm{dx}$, which simplifies to $L = {\int}_{a}^{b} \sec x \mathrm{dx}$

Now we must remember that $\int \sec x$ = $\ln \left(\sec x + \tan x\right) + C$, so for our equation we must solve:

$\ln \left(\sec x + \tan x\right)$ from $\frac{\pi}{6}$ to $\frac{\pi}{4}$, giving us:

$L = \ln \left(\frac{2}{\sqrt{2}} + 1\right) - \ln \left(\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right)$

$\frac{2}{\sqrt{2}}$ can be simplified to $\sqrt{2}$, and the right term has a common denominator, which lets you add them together to become $\frac{3}{\sqrt{3}}$, which is simplified$\sqrt{3}$, giving us:

$L = \ln \left(\sqrt{2} + 1\right) - \ln \left(\sqrt{3}\right)$

If you remember that $\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$, we can simplify our answer to get:

$L = \ln \left(\frac{\sqrt{2} + 1}{\sqrt{3}}\right)$

We can evaluate this for a decimal answer:

$L \approx 0.332067 \ldots$