Calculus, 10th Edition (Anton)

Published by Wiley

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 33

Answer

$$\frac{{{{\sec }^7}x}}{7} - \frac{{{{\sec }^5}x}}{5} + C$$

Work Step by Step

\eqalign{ & \int {{{\sec }^5}x{{\tan }^3}x} dx \cr & {\text{split }}{\sec ^5}x{\text{ and }}{\tan ^3}x \cr & = \int {{{\sec }^4}x{{\tan }^2}x\sec x\tan x} dx \cr & {\text{identity se}}{{\text{c}}^2}\theta = {\tan ^2}\theta + 1 \cr & = \int {{{\sec }^4}x\left( {{{\sec }^2}x - 1} \right)\sec x\tan x} dx \cr & = \int {\left( {{{\sec }^6}x - {{\sec }^4}x} \right)\sec x\tan x} dx \cr & {\text{substitute }}u = \sec x,{\text{ }}du = \sec x\tan xdx \cr & = \int {\left( {{u^6} - {u^4}} \right)} du \cr & {\text{find the antiderivative by the power rule}} \cr & = \frac{{{u^7}}}{7} - \frac{{{u^5}}}{5} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \sec x \cr & = \frac{{{{\sec }^7}x}}{7} - \frac{{{{\sec }^5}x}}{5} + C \cr}

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