We are asked to evaluate the integral $\int 4x \cos(2x) \, dx$ using integration by parts. The formula for integration by parts is:

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

Step 1: Choose $u$ and $dv$

We need to choose $u$ and $dv$ from the given integral. A good choice is:

• $u = 4x$ (because the derivative of $x$ is simpler than its integral),
• $dv = \cos(2x) \, dx$.

Step 2: Find $du$ and $v$

Now we differentiate $u$ and integrate $dv$:

• $du = 4 \, dx$,
• $v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x)$ (since the integral of $\cos(kx)$ is $\frac{1}{k} \sin(kx)$).

Step 3: Apply the integration by parts formula

Now, applying the integration by parts formula:

$\int 4x \cos(2x) \, dx = uv - \int v \, du$

Substitute $u = 4x$, $du = 4 \, dx$, and $v = \frac{1}{2} \sin(2x)$:

$= 4x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot 4 \, dx$

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

Step 4: Evaluate the remaining integral

We now evaluate $\int \sin(2x) \, dx$:

$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)$

Step 5: Substitute back

Now substitute this result into the expression:

$= 2x \sin(2x) - 2 \left( -\frac{1}{2} \cos(2x) \right)$

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

Final Answer:

Thus, the integral is:

$\int 4x \cos(2x) \, dx = 2x \sin(2x) + \cos(2x) + C$

where $C$ is the constant of integration.

Would you like further clarification on any part? Here are five related questions you might find useful:

1. How do you determine the best choice for $u$ and $dv$ in integration by parts?
2. What happens when you reverse the choices of $u$ and $dv$?
3. Can integration by parts be used repeatedly on more complex integrals?
4. What is the general strategy for integrating trigonometric functions combined with polynomials?
5. How does the integration by parts formula derive from the product rule in differentiation?

Tip: When integrating trigonometric functions like sine and cosine, pay attention to their derivatives and antiderivatives, as they alternate between each other!