## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 2

#### Answer

$$- \frac{1}{5}\cos \left( {5x} \right) + C$$

#### Work Step by Step

\eqalign{ & \int {\sin 5x} dx \cr & {\text{set }}u = 5x{\text{ then }}\frac{{du}}{{dx}} = 5,\,\,\,\,\,\,\,\,\frac{{du}}{5} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {\sin 5x} dx = \int {\sin u} \left( {\frac{{du}}{5}} \right) \cr & {\text{use multiple constant rule}} \cr & = \frac{1}{5}\int {\sin u} du \cr & {\text{integrate by using the Basic Trigonometric integral }}\int {\sin x} dx = - \cos x + C \cr & = \frac{1}{5}\left( { - \cos u} \right) + C \cr & = - \frac{1}{5}\cos u + C \cr & {\text{write in terms of }}x{\text{ }} \cr & = - \frac{1}{5}\cos \left( {5x} \right) + C \cr}

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