This integral can be evaluated using integration by parts .
Firstly, integrating by parts where #u=x and dv=sin(x)dx#, we get
#int xsin(x)dx=-x cos(x)+sin(x)+c#
In the same way, for #u=x and dv=cos(x)dx#, we have
#int xcos (x)dx=x sin (x)-sin (x)+c#
Now, using integration by parts again with #u=tan(x) and dv=x sin (x)dx# and the first integral we have
#int x sin(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#
#=-xsin(x)+sin(x)tan(x)+int xcos(x)dx-int sin(x)/cos^2(x)dx#
#=-xsin(x)+sin(x)tan(x)+x sin (x)-sin (x)-int sin(x)/cos^2(x)dx#
#=sin(x)tan(x)-sin (x)-int sin(x)/(cos^2(x))dx#
Finally, using substitution #u=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 #
#=sin(x)tan(x)-sin (x)-1/3cos^3(x)+c.#
