Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 39

Answer

$$\tan x + \frac{{{{\tan }^3}x}}{3} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sec }^4}x} dx \cr & {\text{split exponent se}}{{\text{c}}^4}x \cr & = \int {{{\sec }^2}x} {\sec ^2}xdx \cr & {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}x = {\sec ^2}x \cr & = \int {{{\sec }^2}x} \left( {{\text{1 + ta}}{{\text{n}}^2}x} \right)dx \cr & = \int {{{\sec }^2}x} \left( {{\text{1 + ta}}{{\text{n}}^2}x} \right)dx \cr & {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr & = \int {\left( {{\text{1 + }}{u^2}} \right)} du \cr & {\text{find the antiderivative by the power rule}} \cr & = u + \frac{{{u^3}}}{3} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr & = \tan x + \frac{{{{\tan }^3}x}}{3} + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.