Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 23


$$\int\sec^2\theta\tan^3\theta d\theta=\frac{\tan^4\theta}{4}+C$$

Work Step by Step

$$A=\int\sec^2\theta\tan^3\theta d\theta$$ Let $u=\tan\theta$. Then $du=\sec^2\theta d\theta$. Substitute into $A$, we have $$A=\int u^3du$$ $$A=\frac{u^4}{4}+C$$ $$A=\frac{\tan^4\theta}{4}+C$$
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