Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 15

Answer

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

Work Step by Step

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