Answer
$$\int \cos \theta -\theta d\theta =\sin \theta -\frac{1}{2} \theta^2+c.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
$$\int \cos \theta -\theta d\theta =\sin \theta -\frac{1}{2} \theta^2+c.$$
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 - Chapter Review Exercises - Page 278: 27
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