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: 20

Answer

$\frac{1}{6}\tan^32x + \frac{1}{2}\tan2x +C$

Work Step by Step

Find the indefinite integral $\int sec^42x dx$ Split into two factors of $sec^22x$ $\int sec^22x sec^22x dx$, Convert secants to tangents $\int (tan^22x +1) sec^22x dx$, Divide into two integrations $\int (tan^22x sec^22x dx + \int sec^22x dx$ Use u-substitution for both integrals For the first, let $u=tan2x$, and $du=2sec^22x$ For the second, let $u=2x$, and $du=2dx$ $\frac{1}{2}\int u^2 du + \frac{1}{2}\int \sec^2u du$, Integrate $\frac{1}{6}u^3 + \frac{1}{2}\tan u +C$ $\frac{1}{6}\tan^32x + \frac{1}{2}\tan2x +C$, Resubstitute
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