Answer
$$\sin x\left[ {\ln \left( {\sin x} \right) - 1} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\cos x\ln \left( {\sin x} \right)} dx \cr
& {\text{Let }}t = \sin x,{\text{ }}dt = \cos xdx \cr
& \int {\cos x\ln \left( {\sin x} \right)} dx = \int {\ln t} dt \cr
& {\text{Integrate by tables, formula 87: }}\int {\ln u} du = u\left( { - 1 + \ln u} \right) + C,{\text{ }} \cr
& \int {\ln t} dt = t\left( { - 1 + \ln t} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sin x{\text{ for }}t \cr
& = \sin x\left[ { - 1 + \ln \left( {\sin x} \right)} \right] + C \cr
& = \sin x\left[ {\ln \left( {\sin x} \right) - 1} \right] + C \cr} $$
