Answer
$$ - \frac{1}{4}\cos \left( {\ln x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin \left( {\ln x} \right)}}{{4x}}} dx \cr
& = \frac{1}{4}\int {\frac{{\sin \left( {\ln x} \right)}}{x}} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& = \frac{1}{4}\int {\sin udu} \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{4}\cos u + C \cr
& {\text{with }}u = \ln x \cr
& = - \frac{1}{4}\cos \left( {\ln x} \right) + C \cr} $$
