Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 490: 11

Answer

$\left( {{x^2} + 2x} \right)\sin x + \left( {2x + 2} \right)\cos x - 2\sin x + C$

Work Step by Step

$$\eqalign{ & \int {\left( {{x^2} + 2x} \right)\cos x} dx \cr & {\text{Integrating by parts}}{\text{, }} \cr & {\text{Let }}u = {x^2} + 2x,{\text{ }}du = \left( {2x + 2} \right)dx \cr & {\text{ }}dv = \cos xdx,{\text{ }}v = \sin x \cr & {\text{Using integration by parts formula}} \cr & \int u dv = uv - \int {vdu} \cr & \int {\left( {{x^2} + 2x} \right)\cos x} dx = \left( {{x^2} + 2x} \right)\left( {\sin x} \right) - \int {\left( {\sin x} \right)\left( {2x + 2} \right)dx} {\text{ }}\left( {\bf{1}} \right) \cr & \cr & {\text{Integrate by parts }}\int {\left( {\sin x} \right)\left( {2x + 2} \right)dx} \cr & {\text{Let }}u = 2x + 2,{\text{ }}du = 2dx \cr & {\text{ }}dv = \sin xdx,{\text{ }}v = - \cos x \cr & \int {\left( {\sin x} \right)\left( {2x + 2} \right)dx} = \left( {2x + 2} \right)\left( { - \cos x} \right) + \int {2\cos x} dx \cr & \int {\left( {\sin x} \right)\left( {2x + 2} \right)dx} = - \left( {2x + 2} \right)\cos x + 2\sin x + C \cr & {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr & = \left( {{x^2} + 2x} \right)\left( {\sin x} \right) - \left[ { - \left( {2x + 2} \right)\cos x + 2\sin x} \right] + C \cr & {\text{Simplifying}} \cr & = \left( {{x^2} + 2x} \right)\sin x + \left( {2x + 2} \right)\cos x - 2\sin x + C \cr} $$
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