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: 12


$$\int\sec^22\theta d\theta=\frac{1}{2}\tan(2\theta)+C$$

Work Step by Step

$$A=\int\sec^22\theta d\theta$$ Let $u=2\theta$ Then $du=2d\theta$. So $d\theta=\frac{1}{2}du$ Substitute into $A$, we have $$A=\int\sec^2u(\frac{1}{2})du$$ $$A=\frac{1}{2}\int \sec^2udu$$ $$A=\frac{1}{2}(\tan u)+C$$ $$A=\frac{1}{2}\tan(2\theta)+C$$
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