To integrate the function $x^2 \cos(x)$, we will use integration by parts. Recall the formula for integration by parts:

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

Here, we choose:

• $u = x^2$ (because differentiating $x^2$ will simplify it),
• $dv = \cos(x) \, dx$.

Now, let's compute the necessary derivatives and integrals:

1. $du = 2x \, dx$
2. $v = \int \cos(x) \, dx = \sin(x)$

Now, apply the integration by parts formula:

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

Next, we need to solve $\int 2x \sin(x) \, dx$, which again requires integration by parts.

For this, choose:

• $u = 2x$,
• $dv = \sin(x) \, dx$.

Then:

• $du = 2 \, dx$,
• $v = \int \sin(x) \, dx = -\cos(x)$.

Now, apply the integration by parts formula again:

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

The remaining integral is straightforward:

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

Thus, the integral $\int 2x \sin(x) \, dx$ becomes:

$-2x \cos(x) + 2 \sin(x)$

Now, substitute this back into the original equation:

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

Simplify:

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

So the final answer is:

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

Would you like any further explanation or have any questions about this process? Here are some related questions to consider:

1. How does the choice of $u$ and ( dv \