The value of the integral given below is $$\int_0^\pi x^2 \cos x \, dx$$

To evaluate the definite integral:

Use integration by parts:

Step 1: Let $u = x^2$ and $v^{\prime} = \cos x \implies u^{\prime} = 2x$ and $v = \sin x$.

Since $\sin \pi = 0$ and $\sin 0 = 0$, the boundary term is:

Thus,

Step 2: Integrate $\int_0^\pi x \sin x \, dx$ by parts again with $u = x$ and $v^{\prime} = \sin x \implies u^{\prime} = 1$ and $v = -\cos x$:

Step 3: Substitute back into $I$:

Hence, the correct option is (A).