I need a hand for solving the integration part of the differential equation $y''+4y=x^2sin2x$ . $(D-2i)(D+2i)y=x^2sin2x$ , $t= \dfrac{x{^2}sin2x}{D+2i}$
$t'+2it=x^2sin2x$, $t=uv$
$v=e^{-2ix}$
$du=(e^{2ix}) x^2sin2xdx$
I am stuck at this part. Can someone help me to solve it? Thanks in advance.