\(
\eqalign{
& \int {{{\tan (\ln (\sqrt x ))} \over x}} \,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \cdot {1 \over x}\,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \,d\left( {\ln (x)} \right) = \cr
& \downarrow u = \ln (x) \cr
& \int {\tan \left( {{1 \over 2}u} \right)} \,du = \cr
& \int {{{\sin \left( {{1 \over 2}u} \right)} \over {\cos \left( {{1 \over 2}u} \right)}}} \,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}}
\cdot - {1 \over 2}\sin \left( {{1 \over 2}u} \right)\,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}} \cdot
\,d\left( {\cos \left( {{1 \over 2}u} \right)} \right) = \cr
& \downarrow v = \cos \left( {{1 \over 2}u} \right) \cr
& \int {{{ - 2} \over v}} \cdot \,dv = \cr
& - 2\ln (v) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}u} \right)} \right) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}\ln (x)} \right)} \right) + C \cr
& - 2\ln \left( {\cos \left( {\ln (\sqrt x )} \right)} \right) + C \cr}
\)
