Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 21

Answer

$$\frac{1}{2}x\sin \left( {\ln x} \right) - \frac{1}{2}x\cos \left( {\ln x} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\sin \left( {\ln x} \right)} dx \cr & {\text{substitute }}u = \sin \left( {\ln x} \right),{\text{ }} \cr & du = \cos \left( {\ln x} \right)\left( {\frac{1}{x}} \right)dx \cr & du = \frac{{\cos \left( {\ln x} \right)}}{x}dx \cr & dv = dx \cr & v = x \cr & \cr & {\text{use integration by parts }} \cr & uv - \int v du \cr & = x\sin \left( {\ln x} \right) - \int x \left( {\frac{{\cos \left( {\ln x} \right)}}{x}} \right)dx \cr & = x\sin \left( {\ln x} \right) - \int {\cos \left( {\ln x} \right)} dx \cr & \cr & {\text{substitute }}u = \cos \left( {\ln x} \right),{\text{ }} \cr & du = - \sin \left( {\ln x} \right)\left( {\frac{1}{x}} \right)dx \cr & du = - \frac{{\sin \left( {\ln x} \right)}}{x}dx \cr & dv = dx \cr & v = x \cr & \cr & {\text{use integration by parts }} \cr & = x\sin \left( {\ln x} \right) - \left( {x\cos \left( {\ln x} \right) + \int {\frac{{x\sin \left( {\ln x} \right)}}{x}dx} } \right) \cr & = x\sin \left( {\ln x} \right) - \left( {x\cos \left( {\ln x} \right) + \int {\sin \left( {\ln x} \right)dx} } \right) \cr & \int {\sin \left( {\ln x} \right)} dx = x\sin \left( {\ln x} \right) - x\cos \left( {\ln x} \right) - \int {\sin \left( {\ln x} \right)dx} \cr & {\text{solving for }}\int {\sin \left( {\ln x} \right)} dx \cr & 2\int {\sin \left( {\ln x} \right)} dx = x\sin \left( {\ln x} \right) - x\cos \left( {\ln x} \right) \cr & \int {\sin \left( {\ln x} \right)} dx = \frac{1}{2}x\sin \left( {\ln x} \right) - \frac{1}{2}x\cos \left( {\ln x} \right) + C \cr} $$
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