University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 41


$$\int\sec^4\theta d\theta=\frac{\tan^3\theta}{3}+\tan\theta+C$$

Work Step by Step

$$A=\int\sec^4\theta d\theta$$ $$A=\int\sec^2\theta(\sec^2\theta d\theta)$$ $$A=\int(\tan^2\theta+1)d(\tan \theta)$$ We set $u=\tan\theta$, which means $$A=\int(u^2+1)du$$ $$A=\frac{u^3}{3}+u+C$$ $$A=\frac{\tan^3\theta}{3}+\tan\theta+C$$
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