Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 15

$ \ln(\sqrt{2} + 1)$

Given: $y = \ln{(\sec x)}$ $dy/dx = \frac{\sec x \tan x}{\sec x} $ $= \tan x$ and $1+(dy/dx)^{2} $ $= 1+\tan^{2} x = \sec^{2} x$ $L = \int^{\frac{\pi}{4}}_{0} |\sec x| dx $ $= \int^{\frac{\pi}{4}}_{0}\sec x dx $ $= [\ln{(\sec x + \tan x)}]^{\frac{\pi}{4}}_{0} $ $= \ln(\sqrt{2} + 1) - \ln(1+0)$ $ = \ln(\sqrt{2} + 1)$