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: 3

Answer

$L = \int^\pi_0\sqrt{1+\cos^2x}dx$

Work Step by Step

$y = \sin x$ then $dy/dx = \cos x$ and $ 1+(dy/dx)^2 = 1+\cos^2x$ $L = \int^\pi_0\sqrt{1+\cos^2x}dx$
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