This integral can be evaluated using integration by parts .
Firstly, integrating by parts where u=xanddv=sin(x)dx, we get
∫xsin(x)dx=−xcos(x)+sin(x)+c
In the same way, for u=xanddv=cos(x)dx, we have
∫xcos(x)dx=xsin(x)−sin(x)+c
Now, using integration by parts again with u=tan(x)anddv=xsin(x)dx and the first integral we have
∫xsin(x)tan(x)dx=
=−xtan(x)cos(x)+sin(x)tan(x)+∫xcos(x)sec2(x)dx−∫sin(x)sec2(x)dx
=−xsin(x)+sin(x)tan(x)+∫xcos(x)dx−∫sin(x)cos2(x)dx
=−xsin(x)+sin(x)tan(x)+xsin(x)−sin(x)−∫sin(x)cos2(x)dx
=sin(x)tan(x)−sin(x)−∫sin(x)cos2(x)dx
Finally, using substitution u=cos(x),du=−sin(x)dx in the last integral we have
∫xsin(x)tan(x)dx=sin(x)tan(x)−sin(x)+∫1u2du
=sin(x)tan(x)−sin(x)−13cos3(x)+c.