# Find the arc length of the function below?

## $y = \setminus \ln \left(\setminus \sec x\right)$, with parameters $0 \setminus \le x \setminus \le \setminus \frac{\pi}{4}$?

Apr 24, 2018

The arc has a length of $\ln \left(\sqrt{2} + 1\right) .$

#### Explanation:

For a function in the form $y = f \left(x\right) ,$ the arc length from $\left[a , b\right]$ is given by

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

The given function $y = \ln \sec x$ is in the form $y = f \left(x\right)$, so we'll use the above formula. Furthermore, $\left[a , b\right] = \left[0 , \frac{\pi}{4}\right]$, so these are our integral's bounds.

Differentiate using the Chain Rule:

$y = \ln \sec x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} x \cdot \frac{d}{\mathrm{dx}} \sec x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cancel{\sec} x \tan x}{\cancel{\sec} x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tan x$

Then,

$L = {\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + {\tan}^{2} x} \mathrm{dx}$

Recall the identity $1 + {\tan}^{2} x = {\sec}^{2} x$. Then we get

$L = {\int}_{0}^{\frac{\pi}{4}} \sqrt{{\sec}^{2} x} \mathrm{dx}$

$L = {\int}_{0}^{\frac{\pi}{4}} \sec x \mathrm{dx}$

This is a common integral and worth memorizing. In general,

$\int \sec x \mathrm{dx} = \ln | \sec x + \tan x | + C$

Then,

${\int}_{0}^{\frac{\pi}{4}} \sec x \mathrm{dx} = \ln \left(\sec x + \tan x\right) {|}_{0}^{\frac{\pi}{4}}$

Absolute value bars are dropped because secant and tangent are positive on $\left[0 , \frac{\pi}{4}\right]$.

$= \ln \left(\sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right)\right) - \ln \left(\sec \left(0\right) + \tan \left(0\right)\right)$

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

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

The arc has a length of $\ln \left(\sqrt{2} + 1\right) .$