Explain the method of integration of function y=x*sin2x?

We'll use the formula sin ax = -(cos ax)'/a and we'll integrate
by parts:

Int x*sin2x dx = Int x*[-(cos 2x)]'
dx/2

We'll note u = x => u'du =
dx

v'dv = -(cos 2x)'dx/2 => v = -(cos
2x)/2

Int udv = uv - Int vdu

Int
x*sin2x dx =-x*(cos 2x)/2 - Int-(cos 2x)dx/2

Int x*sin2x dx =
-x*(cos 2x)/2 + (sin 2x)/4 + C

Therefore, using
integration by parts, we've get the result: Int x*sin2x dx = -x*(cos 2x)/2 + (sin 2x)/4 +
C.