Answer
$$\sin x\left[ {\ln \left( {\sin x} \right) - 1} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\cos x} \right)\ln \left( {\sin x} \right)} dx \cr
& \int {\ln \left( {\sin x} \right)\left( {\cos x} \right)} dx \cr
& {\text{Let }}t = \sin x,{\text{ }}dt = \cos xdx \cr
& \cr
& {\text{Apply the substitution}} \cr
& \int {\ln \left( {\sin x} \right)\left( {\cos x} \right)} dx = \int {\ln t} dt \cr
& \cr
& {\text{Use the Integral of }}\ln x\left( {{\text{See page 519}}} \right):{\text{ }} \cr
& \int {\ln xdx} = x\ln x - x + C,{\text{ then}} \cr
& \int {\ln t} dt = t\ln t - t + C \cr
& {\text{Back - substitute }}t = \sin x \cr
& \int {\left( {\cos x} \right)\ln \left( {\sin x} \right)} dx = \sin x\ln \left( {\sin x} \right) - \sin x + C \cr
& {\text{Factoring}} \cr
& \int {\left( {\cos x} \right)\ln \left( {\sin x} \right)} dx = \sin x\left[ {\ln \left( {\sin x} \right) - 1} \right] + C \cr} $$
