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: 14



Work Step by Step

$y = \ln{\cos x}$ then $dy/dx = -\tan x$ and $1+(dy/dx)^{2} = 1+\tan^{2} x = \sec^{2} x$ So $L = \int^{\frac{\pi}{3}}_{0} \sqrt{\sec^{2} x}dx = \int^{\frac{\pi}{3}}_{0} \sec x dx = [\ln|\sec x +\tan x|]^{\frac{\pi}{3}}_{0} = \ln(2+\sqrt{3}) - \ln(1+0) = \ln(2+\sqrt{3})$
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