This integral can be evaluated using integration by parts .
Firstly, integrating by parts where u=x and dv=sin(x)dxu=xanddv=sin(x)dx, we get
int xsin(x)dx=-x cos(x)+sin(x)+c∫xsin(x)dx=−xcos(x)+sin(x)+c
In the same way, for u=x and dv=cos(x)dxu=xanddv=cos(x)dx, we have
int xcos (x)dx=x sin (x)-sin (x)+c∫xcos(x)dx=xsin(x)−sin(x)+c
Now, using integration by parts again with u=tan(x) and dv=x sin (x)dxu=tan(x)anddv=xsin(x)dx and the first integral we have
int x sin(x) tan(x)dx=∫xsin(x)tan(x)dx=
=-x tan(x)cos(x)+sin(x)tan(x)+int xcos(x)sec^2(x)dx-int sin(x)sec^2(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)+int xcos(x)dx-int sin(x)/cos^2(x)dx=−xsin(x)+sin(x)tan(x)+∫xcos(x)dx−∫sin(x)cos2(x)dx
=-xsin(x)+sin(x)tan(x)+x sin (x)-sin (x)-int sin(x)/cos^2(x)dx=−xsin(x)+sin(x)tan(x)+xsin(x)−sin(x)−∫sin(x)cos2(x)dx
=sin(x)tan(x)-sin (x)-int sin(x)/(cos^2(x))dx=sin(x)tan(x)−sin(x)−∫sin(x)cos2(x)dx
Finally, using substitution u=cos(x), du=-sin(x) dxu=cos(x),du=−sin(x)dx in the last integral we have
int x sin(x) tan(x)dx=sin(x)tan(x)-sin (x)+int 1/u^2du ∫xsin(x)tan(x)dx=sin(x)tan(x)−sin(x)+∫1u2du
=sin(x)tan(x)-sin (x)-1/3cos^3(x)+c.=sin(x)tan(x)−sin(x)−13cos3(x)+c.
