Answer
$$\frac{1}{2}\theta^2+\sin(-\theta ) +c.$$
Work Step by Step
Since $(\sin(-\theta ))'= -\cos (-\theta)$, we have
$$\int (\theta -\cos(-\theta))d\theta =\frac{1}{2}\theta^2+\sin(-\theta ) +c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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