Thomas' Calculus 13th Edition

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: 48


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

Work Step by Step

Solve. $ \dfrac{1}{2} \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt$ Thus, we have $(\dfrac{1}{2}) \int 1 dt -\dfrac{1}{2} \int (\cos 6t) dt= \dfrac{1}{2} t -(\dfrac{1}{2})(\dfrac{1}{6}) (\sin 6t) +C$ or, $=(\dfrac{1}{2}) t +\dfrac{1}{12} \sin 6t+C$
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