Answer
$- \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 $
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 48
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