Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 514: 18

Answer

$$\frac{{{{\tan }^4}x}}{4} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\sin }^3}x}}{{{{\cos }^5}x}}dx} \cr & {\text{rewrite the integrand}} \cr & = \int {\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}\left( {\frac{1}{{{{\cos }^2}x}}} \right)dx} \cr & {\text{using identities}} \cr & = \int {{{\tan }^3}x{{\sec }^2}xdx} \cr & {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr & = \int {{{\tan }^3}x{{\sec }^2}xdx} \cr & = \int {{u^3}} du \cr & {\text{find the antiderivative}} \cr & = \frac{{{u^4}}}{4} + C \cr & {\text{ with}}\,\,\,u = \tan x \cr & = \frac{{{{\tan }^4}x}}{4} + C \cr} $$
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