Hoe moeilijk kan dat zijn?:-)

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