University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 278: 110

Answer

$3 \sec (\dfrac{\theta}{3})+C$

Work Step by Step

Calculate the anti-derivative: Since, we know $\int \sec x \tan x =\sec x+C$ Substitute $k=\dfrac{\theta}{3} \implies dk=(1/3) d \theta$ $=3 \int \sec k \tan k+C$ or, $=3 \sec k+C$ Back substitution: $k=\dfrac{\theta}{3} $ $=3 \sec (\dfrac{\theta}{3})+C$
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