## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 48

#### Answer

$$- \sin \left( {1 - \ln v} \right) + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\cos \left( {1 - \ln v} \right)}}{v}} dv \cr & {\text{integrate by the substitution method}} \cr & {\text{set }}u = 1 - \ln v{\text{ then }}\frac{{du}}{{dv}} = - \frac{1}{v},\,\,\,\,dv = - vdu \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{\cos \left( {1 - \ln v} \right)}}{v}} dv = \int {\frac{{\cos u}}{v}} \left( { - vdu} \right) \cr & {\text{cancel common terms}} \cr & = - \int {\cos u} du \cr & {\text{integrating}}{\text{,}} \cr & = - \sin u + C \cr & {\text{replace }}1 - \ln v{\text{ for }}u \cr & = - \sin \left( {1 - \ln v} \right) + C \cr}

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