Answer
$-\frac{1}{4}(2x^{2}cos2x-2xsin2x-cos2x)+C$
Work Step by Step
Use integration by parts twice.
$\int x^{2}sin2xdx$
$u=x^{2},dv=\int sin2xdx$
$u'=2xdx,v=-\frac{1}{2}cos2x$
$=-\frac{1}{2}x^{2}cos2x-\int (2x)(-\frac{1}{2}cos2x)dx$
$=-\frac{1}{2}x^{2}cos2x+\int xcos2xdx$
$u=x,dv=\int cos2x$
$u'=dx,v=\frac{1}{2}sin2x$
$=-\frac{1}{2}x^{2}cos2x+\frac{1}{2}xsin2x-\int\frac{1}{2}sin2xdx$
$=-\frac{1}{2}x^{2}cos2x+\frac{1}{2}xsin2x+\frac{1}{4}cos2x+C$
Take out $-\frac{1}{4}$ to get:
$-\frac{1}{4}(2x^{2}cos2x-2xsin2x-cos2x)+C$