$$\int x^2\sin{x}dx = -\int x^2d(\cos{x})=-x^2\cos{x} +\int \cos{x}d(x^2)=-x^2\cos{x}+2\int x\cos{x}dx$$

The remaining integral can be evaluated using integration by parts again:

$$\int x\cos{x}dx = \int xd(\sin{x}) = x\sin{x}-\int\sin{x}dx = x\sin{x}+\cos{x}+C$$

Combining both finds the final result as

$$\int x^2\sin{x}dx=\boxed{2x\sin{x}+(2-x^2)\cos{x}+C} $$