## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 34

#### Answer

$$\frac{{{{\sec }^5}\theta }}{5} - \frac{{2{{\sec }^3}\theta }}{3} + \sec \theta + C$$

#### Work Step by Step

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

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