Thanks

Is this it?

Integrate by parts;

$\displaystyle x^2.sin2x $

$\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

$\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

$\displaystyle v= -\frac{cos2x}{2}+C $

$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substituting

$\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

$\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

Need to solve the integral of 2x.Cos2x by parts:

$\displaystyle \int 2x.cos2x $

$\displaystyle let \ u=2x, then\ \frac {du}{dx}=2.dx $

$\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx $

$\displaystyle v= \frac{sin2x}{2}+C $

Good so far.
$\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

Substitute back in;

$\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$

Yep
$\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

$\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

$\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

$\displaystyle =x.sin2x+\frac{cos2x}{2}+C$

Yes, I think that's fine!