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: 60


$\dfrac{1}{2} t -\dfrac{1}{12} \sin 6t+C$

Work Step by Step

Given: $ \dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt$ Thus, $\dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt= \dfrac{1}{2} t -(\dfrac{1}{2})(\dfrac{1}{6}) (\sin 6t) +C=\dfrac{1}{2} t -\dfrac{1}{12} \sin 6t+C$
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