Answer
$- \dfrac{1}{3} \tan x+C $
Work Step by Step
Here, $\int \dfrac{-1}{3} \sec^2 x dx$
Thus, we have $\int ( \dfrac{-1}{3} \sec^2 x ) dx= (-\dfrac{1}{3} )\int \sec^2 x dx=(- \dfrac{1}{3}) \tan x+C $
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 40
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
