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