Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 32


$$\frac{1}{2}\theta^2+\sin(-\theta ) +c.$$

Work Step by Step

Since $(\sin(-\theta ))'= -\cos (-\theta)$, we have $$\int (\theta -\cos(-\theta))d\theta =\frac{1}{2}\theta^2+\sin(-\theta ) +c.$$
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