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: 42

Answer

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

Work Step by Step

$$\eqalign{ & \int {{{\tan }^4}x} dx \cr & {\text{split exponent of }}{\tan ^4}x \cr & = \int {{{\tan }^2}x} {\tan ^2}xdx \cr & {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta \cr & = \int {\left( {{{\sec }^2}x - 1} \right)} {\tan ^2}xdx \cr & = \int {\left( {{{\sec }^2}x{{\tan }^2}x - {{\tan }^2}x} \right)} dx \cr & = \int {\left( {{{\sec }^2}x{{\tan }^2}x - {{\sec }^2}x + 1} \right)} dx \cr & {\text{sum rule}} \cr & = \int {{{\tan }^2}x} {\sec ^2}xdx - \int {{{\sec }^2}x} dx + \int {dx} \cr & {\text{find antiderivatives}} \cr & = \frac{{{{\tan }^3}x}}{3} - \tan x + x + C \cr} $$
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