## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 24

#### Answer

$$\frac{1}{5}\sec {x^5} + C$$

#### Work Step by Step

\eqalign{ & \int {{x^4}\sec {x^5}\tan {x^5}} dx \cr & {\text{set }}u = {x^5}{\text{ then }}\frac{{du}}{{dx}} = 5{x^4},\,\,\,\,\,\,\,\,\frac{{du}}{{5{x^4}}} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {{x^4}\sec {x^5}\tan {x^5}} dx = \int {{x^4}\sec u\tan u} \left( {\frac{{du}}{{5{x^4}}}} \right) \cr & {\text{cancel the common factor }}{x^4} \cr & = \int {\sec u\tan u} \left( {\frac{{du}}{5}} \right) \cr & = \frac{1}{5}\int {\sec u\tan u} du \cr & {\text{integrate by using the Basic Trigonometric integral }}\int {\sec x\tan x} dx = \sec x + C \cr & = \frac{1}{5}\sec u + C \cr & {\text{write in terms of }}x \cr & = \frac{1}{5}\sec {x^5} + C \cr}

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