Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 499: 29

Answer

$\frac{1}{4}{\tan ^4}x + \frac{1}{3}{\tan ^6}x + \frac{1}{8}{\tan ^8}x + C$

Work Step by Step

$$\eqalign{ &\text{Let }I= \int {{{\tan }^3}x{{\sec }^6}x} dx \cr & {\text{Split }}{\sec ^6}x{\text{ as se}}{{\text{c}}^4}x{\sec ^2}x \cr & I = \int {{{\tan }^3}x{{\sec }^4}x} {\sec ^2}xdx \cr & {\text{Use the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr & I= \int {{{\tan }^3}x{{\left( {{{\sec }^2}x} \right)}^2}} {\sec ^2}xdx \cr & {\text{Rewrite }}{\sec ^2}x{\text{ using }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr & I= \int {{{\tan }^3}x{{\left( {1 + {{\tan }^2}x} \right)}^2}} {\sec ^2}xdx \cr & {\text{Expand the binomial}} \cr & I= \int {{{\tan }^3}x\left( {1 + 2{{\tan }^2}x + {{\tan }^4}x} \right)} {\sec ^2}xdx \cr & = \int {\left( {{{\tan }^3}x + 2{{\tan }^5}x + {{\tan }^7}x} \right)} {\sec ^2}xdx \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr & {\text{Substituting}} \cr & I= \int {\left( {{u^3} + 2{{\tan }^5}x + {u^7}} \right)du} \cr & {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr & I= \frac{1}{4}{u^4} + \frac{1}{3}{u^6} + \frac{1}{8}{u^8} + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}\tan x{\text{ for }}u \cr & I= \frac{1}{4}{\tan ^4}x + \frac{1}{3}{\tan ^6}x + \frac{1}{8}{\tan ^8}x + C \cr} $$
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