University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 48


$- \dfrac{1}{3} \tan x+C $

Work Step by Step

Given: $\int \dfrac{-1}{3} \sec^2 x dx$ Thus, $\int ( \dfrac{-1}{3} \sec^2 x ) dx= -\dfrac{1}{3} \int \sec^2 x dx=- \dfrac{1}{3} \tan x+C $
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