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: 39


$$\frac{1}{3}\sin(3 \theta)-2\tan\left(\frac{\theta}{4}\right)+c. $$

Work Step by Step

Since $(\sin(3 \theta))'= 3\cos (3 \theta)$ and $(\tan\left(\frac{\theta}{4}\right))'=\frac{1}{4}\sec ^{2}\left(\frac{\theta}{4}\right)$ , we have $$ \int\left(\cos (3 \theta)-\frac{1}{2} \sec ^{2}\left(\frac{\theta}{4}\right)\right) d \theta\\ =\frac{1}{3}\sin(3 \theta)-\frac{4}{2}\tan\left(\frac{\theta}{4}\right)+c\\ =\frac{1}{3}\sin(3 \theta)-2\tan\left(\frac{\theta}{4}\right)+c. $$
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