Evaluate the integral integral x2 (x) dx.

Choose $u$ and $dv$:
Let $u = x^2$
Then $du = 2x\,dx$

Let $dv = \cos(x)\,dx$
Then $v = \int \cos(x)\,dx = \sin(x)$

Apply the integration by parts formula:

$$\int x^2 \cos(x)\,dx = x^2 \sin(x) - \int \sin(x) (2x)\,dx$$ $$\int x^2 \cos(x)\,dx = x^2 \sin(x) - 2 \int x \sin(x)\,dx$$

Choose new $u$ and $dv$:
Let $u_1 = x$
Then $du_1 = dx$

Let $dv_1 = \sin(x)\,dx$
Then $v_1 = \int \sin(x)\,dx = -\cos(x)$

Apply the integration by parts formula to $\int x \sin(x)\,dx$:

$$\int x \sin(x)\,dx = x(-\cos(x)) - \int (-\cos(x))\,dx$$ $$\int x \sin(x)\,dx = -x\cos(x) + \int \cos(x)\,dx$$ $$\int x \sin(x)\,dx = -x\cos(x) + \sin(x)$$

$$\int x^2 \cos(x)\,dx = x^2 \sin(x) - 2 (-x\cos(x) + \sin(x)) + C$$ $$\int x^2 \cos(x)\,dx = x^2 \sin(x) + 2x\cos(x) - 2\sin(x) + C$$

The final answer is $x^2 \sin(x) + 2x\cos(x) - 2\sin(x) + C$.