Calculus (3rd Edition)

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


$$\int \cos \theta -\theta d\theta =\sin \theta -\frac{1}{2} \theta^2+c.$$

Work Step by Step

Recall that $(\sin x)'=\cos x$. $$\int \cos \theta -\theta d\theta =\sin \theta -\frac{1}{2} \theta^2+c.$$
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