Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises: 27

Answer

$\frac{\sec^64x}{24}$

Work Step by Step

Find the indefinite integral $\int sec^64x tan4x dx$ $\int sec^54x sec4xtan4xdx$ Use u-substitution, let $u=sec4x$, $du=4sec4xtan4x$ $\frac{1}{4}\int u^5 du$ $\frac{1}{24}u^6+C$ $\frac{\sec^64x}{24} +C$
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