We need to find the correct solution for the integral $\int x^2 \sin x \, dx$.

1. Problem Description

We need to find the correct solution for the integral $\int x^2 \sin x \, dx$.

2. Solution Steps

We will use integration by parts twice. The integration by parts formula is:

$\int u \, dv = uv - \int v \, du$

First, let $u = x^2$ and $dv = \sin x \, dx$. Then $du = 2x \, dx$ and $v = -\cos x$. Applying integration by parts:

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

Now, we need to evaluate $\int x \cos x \, dx$. We'll use integration by parts again.

Let $u = x$ and $dv = \cos x \, dx$. Then $du = dx$ and $v = \sin x$. Applying integration by parts:

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

Substituting this back into the original integral:

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

3. Final Answer

The correct solution is $-x^2 \cos x + 2x \sin x + 2 \cos x + C$. This matches option B.

Final Answer: Option B