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$
$L = \int^\pi_0\sqrt{1+\cos^2x}dx$
Calculus 8th Edition
by Stewart, James
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 - Page 589: 3
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