# How do you find the length of the curve y=lnabs(secx) from 0<=x<=pi/4?

Feb 2, 2017

$\ln \left(\sqrt{2} + 1\right) = 0.8814$ length units, nearly..

#### Explanation:

In the given interval,$| \sec x | = \sec x$.

$y ' = \frac{1}{\sec} x \left(\sec x\right) ' = \sec x \tan \frac{x}{\sec} x = \tan x$1

Length = $\int \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$,

with $y = \ln | \sec x |$ and x from $0 \to \frac{\pi}{4}$

$= \int \sqrt{1 + {\tan}^{2} x} \mathrm{dx}$, for the limits

$= \int \sec x \mathrm{dx}$, for the limita

$= \left[\ln \left(\sec x + \tan x\right)\right] ,$ between $x = 0 \mathmr{and} \frac{\pi}{4}$

$= \ln \left(\sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right)\right) = \ln \left(\sqrt{2} + 1\right)$