Calculus: Early Transcendentals 9th Edition

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

Chapter 7 - Review - Exercises - Page 553: 12

Answer

$ - \frac{1}{2}{\cos ^2}x\ln \left( {\cos x} \right) + \frac{1}{4}{\cos ^2}x + C$

Work Step by Step

$$\eqalign{ & \int {\sin x\cos x\ln \left( {\cos x} \right)} dx \cr & = \int {\cos x\ln \left( {\cos x} \right)\left( {\sin x} \right)} dx \cr & {\text{Make the substitution }}z = \cos x,{\text{ }}dz = - \sin xdx,\,{\text{ then}} \cr & \int {\cos x\ln \left( {\cos x} \right)\left( {\sin x} \right)} dx = \int {z\ln z\left( { - 1} \right)} dz \cr & = - \int {z\ln z} dz \cr & {\text{Integrating by parts}}{\text{, }} \cr & {\text{Let }}u = \ln z,{\text{ }}du = \frac{1}{z}dz, \cr & {\text{ }}dv = zdz,{\text{ }}v = \frac{1}{2}{z^2} \cr & {\text{Using integration by parts formula}} \cr & \int u dv = uv - \int {vdu} \cr & - \int {z\ln z} dz = - \left( {\ln z} \right)\left( {\frac{1}{2}{z^2}} \right) + \int {\left( {\frac{1}{2}{z^2}} \right)} \left( {\frac{1}{z}dz} \right) \cr & = - \frac{1}{2}{z^2}\ln z + \frac{1}{2}\int z dz \cr & = - \frac{1}{2}{z^2}\ln z + \frac{1}{2}\left( {\frac{{{z^2}}}{2}} \right) + C \cr & = - \frac{1}{2}{z^2}\ln z + \frac{1}{4}{z^2} + C \cr & {\text{Write in terms of }}x,{\text{ substitute cos }}x{\text{ for }}z \cr & = - \frac{1}{2}{\cos ^2}x\ln \left( {\cos x} \right) + \frac{1}{4}{\cos ^2}x + C \cr} $$
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