## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 25

#### Answer

$$- \frac{{6x}}{5}\sin 5x - \frac{6}{{25}}\cos 5x + C$$

#### Work Step by Step

\eqalign{ & \int { - 6x\cos 5x} dx \cr & {\text{setting }}\,\,\,\,\,\,u = - 6x{\text{ then }}du = - 6dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \cos 5xdx{\text{ then }}v = \frac{1}{5}\sin 5x \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int { - 6x\cos 5x} dx = \left( { - 6x} \right)\left( {\frac{1}{5}\sin 5x} \right) - \int {\left( {\frac{1}{5}\sin 5x} \right)\left( { - 6dx} \right)} \cr & {\text{simplifying}} \cr & \int { - 6x\cos 5x} dx = - \frac{{6x}}{5}\sin 5x + \frac{6}{5}\int {\sin 5x} dx\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{solving}}\,\,\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 & = \frac{1}{5}\int {\sin u} du \cr & {\text{integrate }} \cr & = \frac{1}{5}\left( { - \cos u} \right) + C \cr & = - \frac{1}{5}\cos u + C \cr & {\text{write in terms of }}x \cr & = - \frac{1}{5}\cos \left( {5x} \right) + C \cr & \cr & {\text{substitute the result of }}\int {\sin 5x} dx{\text{ into }}\left( {\bf{1}} \right) \cr & \int { - 6x\cos 5x} dx = - \frac{{6x}}{5}\sin 5x + \frac{6}{5}\left( { - \frac{1}{5}\cos \left( {5x} \right)} \right) + C \cr & \int { - 6x\cos 5x} dx = - \frac{{6x}}{5}\sin 5x - \frac{6}{{25}}\cos 5x + C \cr}

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