To find the integral $\int 2x^2 \cos(4x) \, dx$, we can use the method of integration by parts. This method states:

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

For this problem, we can set $u = 2x^2$ and $dv = \cos(4x) \, dx$.

First, we need to compute $du$ and $v$: $u = 2x^2 \quad \Rightarrow \quad du = 4x \, dx$ $dv = \cos(4x) \, dx \quad \Rightarrow \quad v = \frac{\sin(4x)}{4}$

Now we can apply the integration by parts formula: $\int 2x^2 \cos(4x) \, dx = 2x^2 \cdot \frac{\sin(4x)}{4} - \int \frac{\sin(4x)}{4} \cdot 4x \, dx$ $= \frac{2x^2 \sin(4x)}{4} - \int x \sin(4x) \, dx$ $= \frac{x^2 \sin(4x)}{2} - \int x \sin(4x) \, dx$

We now need to integrate $\int x \sin(4x) \, dx$ using integration by parts again. Let $u = x$ and $dv = \sin(4x) \, dx$: $u = x \quad \Rightarrow \quad du = dx$ $dv = \sin(4x) \, dx \quad \Rightarrow \quad v = -\frac{\cos(4x)}{4}$

Applying the integration by parts formula again: $\int x \sin(4x) \, dx = x \left( -\frac{\cos(4x)}{4} \right) - \int \left( -\frac{\cos(4x)}{4} \right) \, dx$ $= -\frac{x \cos(4x)}{4} + \frac{1}{4} \int \cos(4x) \, dx$ $= -\frac{x \cos(4x)}{4} + \frac{1}{4} \cdot \frac{\sin(4x)}{4}$ $= -\frac{x \cos(4x)}{4} + \frac{\sin(4x)}{16}$

Combining both parts, we have: $\int 2x^2 \cos(4x) \, dx = \frac{x^2 \sin(4x)}{2} - \left( -\frac{x \cos(4x)}{4} + \frac{\sin(4x)}{16} \right)$ $= \frac{x^2 \sin(4x)}{2} + \frac{x \cos(4x)}{4} - \frac{\sin(4x)}{16} + C$

Thus, the integral $\int 2x^2 \cos(4x) \, dx$ is: $\int 2x^2 \cos(4x) \, dx = \frac{x^2 \sin(4x)}{2} + \frac{x \cos(4x)}{4} - \frac{\sin(4x)}{16} + C$

where $C$ is the constant of integration.

Would you like more details or have any questions about this solution?

Here are 5 numbered questions you might ask:

1. How does the method of integration by parts work?
2. How do you choose $u$ and $dv$ in integration by parts?
3. Can you explain the steps to integrate $\cos(4x)$ and $\sin(4x)$?
4. What is the general strategy for integrating products of polynomials and trigonometric functions?
5. How do you verify the result of an integral?

Tip: For complex integrals involving products of polynomial and trigonometric functions, consider using integration by parts multiple times as needed.