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: 35


$$\int\sec^3x\tan xdx=\frac{\sec^3x}{3}+C$$

Work Step by Step

$$A=\int\sec^3x\tan xdx$$ $$A=\int\sec^2x(\sec x\tan xdx)$$ $$A=\int\sec^2xd(\sec x)$$ We set $a=\sec x$. Therefore, $$A=\int a^2da$$ $$A=\frac{a^3}{3}+C$$ $$A=\frac{\sec^3x}{3}+C$$
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