∫xcos(2x)dx

Question

What I did wrong
u=x
du=dx
dv=cos(2x)dx
v=-sin(2x)/2
-xsin(2x)/2 +∫sin(2x)/2dx--> -xsin(2x)/2 -cos(2x)/4+C

Answer 1

oooup, this one why my answer wrong need steps and explain.
u=x
du=dx
dv=cos(2x)dx
v=sin(2x)/2
xsin(2x)/2 -∫sin(2x)/2dx--> = xsin(2x)/2 -cos(2x)/4+C

Answer 2

-∫sin(2x)/2dx--> = +cos(2x)/4