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The integral of the function $x^3 \cos(x)$ can be expressed as:

where $C$ is the constant of integration.

To evaluate this integral, we will employ the method of integration by parts. We start by letting:

• $u = x^3$
• $dv = \cos(x) \, dx$

From this, we differentiate and integrate to find $du$ and $v$:

• $du = 3x^2 \, dx$
• $v = \sin(x)$

Applying the integration by parts formula, which is given by:

we have:

Next, we need to evaluate the integral $\int 3x^2 \sin(x) \, dx$. We will use integration by parts again, this time setting:

• $u = 3x^2$
• $dv = \sin(x) \, dx$

Differentiating and integrating gives us:

• $du = 6x \, dx$
• $v = -\cos(x)$

Substituting into the integration by parts formula results in:

Now, we turn our attention to the integral $\int 6x \cos(x) \, dx$. We apply integration by parts once more, with:

• $u = 6x$
• $dv = \cos(x) \, dx$

This leads to:

• $du = 6 \, dx$
• $v = \sin(x)$

Using the integration by parts formula again, we obtain:

The integral $\int 6 \sin(x) \, dx$ is straightforward:

Now, we can combine all the parts together. Substituting back, we have:

Putting everything together, we find:

Thus, the final result is:

This concludes our evaluation of the integral.