To find the integral $\int (4x - 3) \sin(3x) \, dx$, we can use integration by parts. Let's proceed step by step.

Step 1: Identify $u$ and $dv$

We can choose: $u = 4x - 3 \quad \text{and} \quad dv = \sin(3x) \, dx$

Step 2: Differentiate $u$ and integrate $dv$

Find $du$ by differentiating $u$ with respect to $x$: $du = 4 \, dx$ Now, integrate $dv$ to find $v$: $v = \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x)$

Step 3: Apply the integration by parts formula

The integration by parts formula is: $\int u \, dv = uv - \int v \, du$ Substituting the values we found: $\int (4x - 3) \sin(3x) \, dx = (4x - 3) \left(-\frac{1}{3} \cos(3x)\right) - \int \left(-\frac{1}{3} \cos(3x)\right) \cdot 4 \, dx$

Step 4: Simplify and solve the remaining integral

Simplify the expression: $\int (4x - 3) \sin(3x) \, dx = -\frac{4x - 3}{3} \cos(3x) + \frac{4}{3} \int \cos(3x) \, dx$ Now, integrate $\cos(3x)$: $\int \cos(3x) \, dx = \frac{1}{3} \sin(3x)$ Substitute back: $\int (4x - 3) \sin(3x) \, dx = -\frac{4x - 3}{3} \cos(3x) + \frac{4}{9} \sin(3x) + C$ where $C$ is the constant of integration.

Final Answer

$\boxed{\int (4x - 3) \sin(3x) \, dx = -\frac{4x - 3}{3} \cos(3x) + \frac{4}{9} \sin(3x) + C}$

Would you like to see any details or have any questions?

Here are 5 related questions to consider:

1. How do you apply integration by parts in different scenarios?
2. What happens if you choose different functions for $u$ and $dv$?
3. How would the integration process change if it were $(4x^2 - 3x) \sin(3x)$?
4. Can you perform the integral using a different method, like substitution?
5. How does the choice of $u$ and $dv$ affect the complexity of the integral?

Tip: When choosing $u$ and $dv$ in integration by parts, pick $u$ to be the function that becomes simpler when differentiated, and $dv$ to be the part that can be easily integrated.