Answer
$$\frac{1}{2}x\sin \left( {\ln x} \right) - \frac{1}{2}x\cos \ln \left( x \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\sin \left( {\ln x} \right)} dx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = \sin \left( {\ln x} \right),\,\,\,\,du = \frac{{\cos \left( {\ln x} \right)}}{x}dx,\,\,\,\,dv = dx,\,\,\,\,v = x \cr
& {\text{Integration by parts then gives}} \cr
& = x\sin \left( {\ln x} \right) - \int {x\left( {\frac{{\cos \left( {\ln x} \right)}}{x}} \right)dx} \cr
& = x\sin \left( {\ln x} \right) - \int {\cos \left( {\ln x} \right)dx} \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = \cos \left( {\ln x} \right)dx,\,\,\,\,du = - \frac{{\sin \left( {\ln x} \right)}}{x}dx,\,\,\,\,dv = dx,\,\,\,\,v = x \cr
& = x\sin \left( {\ln x} \right) - \left( {x\cos \ln \left( x \right) - \int {x\left( { - \frac{{\sin \left( {\ln x} \right)}}{x}} \right)} dx} \right) \cr
& \int {\sin \left( {\ln x} \right)} dx = x\sin \left( {\ln x} \right) - x\cos \ln \left( x \right) - \int {\sin \left( {\ln x} \right)} dx \cr
& {\text{combine like integrals and add a constant of integration}} \cr
& 2\int {\sin \left( {\ln x} \right)} dx = x\sin \left( {\ln x} \right) - x\cos \ln \left( x \right) + C \cr
& {\text{Divide both sides by 2}} \cr
& \int {\sin \left( {\ln x} \right)} dx = \frac{1}{2}x\sin \left( {\ln x} \right) - \frac{1}{2}x\cos \ln \left( x \right) + C \cr} $$
