Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 38

Answer

$$\frac{1}{2}{y^2} = - x\cos x + \sin x + 8$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{x}{y}\sin x,{\text{ }}\left( {0,4} \right) \cr & {\text{Separate the variables}} \cr & ydy = x\sin xdx \cr & {\text{Integrate both sides}} \cr & \frac{1}{2}{y^2} = \int {x\sin x} dx{\text{ }}\left( {\bf{1}} \right) \cr & {\text{*Integrate }}\int {x\sin x} dx{\text{ by parts}} \cr & {\text{Integrate by parts}} \cr & {\text{Let }}u = x \Rightarrow du = dx \cr & dv = \sin xdx \Rightarrow v = - \cos x \cr & {\text{Integration by parts formula }} \cr & \int {x\sin x} dx = - x\cos x + \int {\cos x} dx \cr & \int {x\sin x} dx = - x\cos x + \sin x + C \cr & {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr & \frac{1}{2}{y^2} = - x\cos x + \sin x + C{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Use the intial condition }}\left( {0,4} \right) \cr & \frac{1}{2}{\left( 4 \right)^2} = - 0\cos \left( 0 \right) + \sin \left( 0 \right) + C \cr & C = 8 \cr & {\text{Substituting }}C{\text{ into }}\left( {\bf{2}} \right) \cr & \frac{1}{2}{y^2} = - x\cos x + \sin x + 8 \cr & \cr & {\text{Graph}} \cr} $$
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