# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 26

$$- \frac{{7x}}{5}\cos 5x + \frac{7}{{25}}\sin 5x + C$$

#### Work Step by Step

\eqalign{ & \int {7x\sin 5x} dx \cr & {\text{setting }}\,\,\,\,\,\,u = 7x{\text{ then }}du = 7dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \sin 5xdx{\text{ then }}v = - \frac{1}{5}\cos 5x \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {7x\sin 5x} dx = \left( {7x} \right)\left( { - \frac{1}{5}\cos 5x} \right) - \int {\left( { - \frac{1}{5}\cos 5x} \right)\left( {7dx} \right)} \cr & {\text{simplifying}} \cr & \int {7x\sin 5x} dx = - \frac{{7x}}{5}\cos 5x + \frac{7}{5}\int {\cos 5x} dx\,\left( {\bf{1}} \right) \cr & \cr & {\text{solving}}\,\,\int {\cos 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 {\cos 5x} dx = \int {\cos u} \left( {\frac{{du}}{5}} \right) \cr & = \frac{1}{5}\int {\cos } du \cr & {\text{integrate}} \cr & = \frac{1}{5}\sin u + C \cr & {\text{write in terms of }}x \cr & = \frac{1}{5}\sin 5x + C \cr & \cr & {\text{substitute the result of }}\int {\cos 5x} dx{\text{ into }}\left( {\bf{1}} \right) \cr & \int {7x\sin 5x} dx = - \frac{{7x}}{5}\cos 5x + \frac{7}{5}\left( {\frac{1}{5}\sin 5x} \right) + C \cr & \int {7x\sin 5x} dx = - \frac{{7x}}{5}\cos 5x + \frac{7}{{25}}\sin 5x + C \cr}

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