Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 35

Answer

$$ - \frac{1}{4}\ln \left| {\cos 4x} \right| - \frac{{{{\tan }^2}4x}}{8} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\tan }^3}4xdx} \cr & {\text{substitute }}u = 4x,{\text{ }}du = 4dx \cr & = \frac{1}{4}\int {{{\tan }^3}udu} \cr & {\text{split off }}\tan \theta \cr & = \frac{1}{4}\int {{{\tan }^2}u\tan udu} \cr & {\text{pythagorean identity}} \cr & = \frac{1}{4}\int {\left( {1 - {{\sec }^2}u} \right)} \tan udu \cr & {\text{Split }} \cr & = \frac{1}{4}\int {\tan u} du - \frac{1}{4}\int {{{\sec }^2}u} \tan udu \cr & {\text{evaluate the integral}} \cr & = - \frac{1}{4}\ln \left| {\cos u} \right| - \frac{{{{\tan }^2}u}}{8} + C \cr & {\text{replace }}u = 4x \cr & = - \frac{1}{4}\ln \left| {\cos 4x} \right| - \frac{{{{\tan }^2}4x}}{8} + C \cr} $$
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