Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 59


ln(1+$\sqrt 2$)

Work Step by Step

The arc length L=$\int$ √[1+$(dy/dx)^{2}$] dx y=ln(cosx) dy/dx=$\frac{-sinx}{cosx}$=-tanx L=$\int$ √[1+$(-tanx)^{2}$] dx = $\int$ √($sec^{2}x$) dx=$\int$secx dx (secx>0 when x is between 0 and $\frac{\pi}{4}$) Now $\int$secx dx=ln|tanx + secx| When x∈[0, $\frac{\pi}{4}$], L=ln|tan$\frac{\pi}{4}$ + sec$\frac{\pi}{4}$| -ln|tan0+sec0|=ln(1+$\sqrt 2$)
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