Answer
$$\ln \sqrt x - \frac{1}{4}\sin \left( {\ln {x^2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sin }^2}\left( {\ln x} \right)}}{x}} dx,\,\,\,\,u = \ln x \cr
& {\text{Using the given substitution}} \cr
& u = \ln x,\,\,\,\,\,\,\,du = \frac{1}{x}dx,\, \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{{{\sin }^2}\left( {\ln x} \right)}}{x}} dx = \int {{{\sin }^2}u} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 26}} \cr
& \left( {26} \right):\,\,\,\,\,\,\int {{{\sin }^2}udu = \frac{1}{2}u - \frac{1}{4}\sin 2u} + C \cr
& = \frac{1}{2}u - \frac{1}{4}\sin 2u + C \cr
& {\text{write in terms of }}x{\text{, and replace }}\ln x{\text{ for }}u \cr
& = \frac{1}{2}\ln x - \frac{1}{4}\sin \left( {2\ln x} \right) + C \cr
& {\text{using logarithmic properties}} \cr
& = \ln \sqrt x - \frac{1}{4}\sin \left( {\ln {x^2}} \right) + C \cr} $$
