## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 41

#### Answer

$$\frac{1}{8}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {{{\tan }^3}4x} dx \cr & {\text{split exponent of }}{\tan ^3}4x \cr & = \int {{{\tan }^2}4x} \tan 4xdx \cr & {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta \cr & = \int {\left( {{{\sec }^2}4x - 1} \right)} \tan 4xdx \cr & = \int {\left( {\tan 4x{{\sec }^2}4x - \tan 4x} \right)} dx \cr & {\text{sum rule}} \cr & = \int {\tan 4x{{\sec }^2}4x} dx - \int {\tan 4x} dx \cr & {\text{find the antiderivatives}} \cr & = \frac{1}{{4\left( 2 \right)}}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C \cr & = \frac{1}{8}{\tan ^2}4x + \frac{1}{4}\ln \left| {\cos 4x} \right| + C \cr}

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