Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 24

Answer

$$2\sec x + \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {{{\left( {1 + \tan x} \right)}^2}\sec x} dx \cr & {\text{Expand the binomial}} \cr & = \int {\left( {1 + 2\tan x + {{\tan }^2}x} \right)\sec x} dx \cr & {\text{Multiply}} \cr & = \int {\left( {\sec x + 2\sec x\tan x + {{\tan }^2}x\sec x} \right)} dx \cr & {\text{Distribute}} \cr & = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {{{\tan }^2}x\sec x} dx \cr & {\text{Use pythagorean identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr & = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {\left( {{{\sec }^2}x - 1} \right)\sec x} dx \cr & = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {{{\sec }^3}x} dx - \int {\sec x} dx \cr & = 2\int {\sec x\tan x} dx + \int {{{\sec }^3}x} dx \cr & {\text{Integrating}} \cr & = 2\sec x + \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr} $$
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