Before going into the calculus, consider the curve itself. Note that if f(x)=x^2sin2x, f(x)=-f(-x), because x^2 is always positive while sin2x=-sin(-2x). Thus, where f(x) for positive x is above the x axis, the point for the corresponding negative value of x is below the x axis by the same amount. And where positive x is below the x axis, the corresponding negative value is above the x axis by the same amount. When we integrate between limits -n and n to find the area under the graph we will always get zero. So we can expect the result between -1 and 1 to be zero.
On with the calculus.
Let u=x^2 then du/dx=2x; let dv=sin2xdx, so dv/dx=sin2x, then v=-cos2x/2.
d(uv)/dx=vdu/dx+udv/dx, so udv/dx=d(uv)/dx-vdu/dx; thus x^2sin2xdx=d(-x^2cos2x/2)/dx-int(-cos2x/2.2xdx).
Integrating: int(x^2sin2xdx)=-x^2cos2x/2+int(xcos2xdx).
Now, let u=x, dv=cos2xdx; du/dx=1; v=sin2x/2.
Applying the same formula and integrating:
int(xcos2xdx)=xsin2x/2-int(sin2xdx)/2=xsin2x/2+cos2x/4.
Substitute for int(xcos2xdx):
int(x^2sin2xdx)=-x^2cos2x/2+xsin2x/2+cos2x/4.
Consider the limits -1 and 1. Rather than angles in degrees, it would seem more appropriate to consider angles in radians. Apply the limits: cos2=-0.416147, sin2=0.909297. (The expression becomes cos2/2+sin2/2+cos2/4-cos(-2)/2+sin(-2)/2-cos(-2)/4. Cosz=cos(-z) and sinz=-sin(z). The terms therefore cancel out.) The definite integral is 0.5587-0.5587=0, as expected.