The problem involves finding the constants $A$ and $B$ in the integral equation:

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

Approach:

We can use integration by parts to evaluate the integral and determine $A$ and $B$. Let's solve step by step.

Integration by Parts Formula:

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

Let: $u = x \quad \text{and} \quad dv = \cos(4x) \, dx$

1. Differentiate $u$ to find $du$: $du = dx$

2. Integrate $dv$ to find $v$: $v = \int \cos(4x) \, dx = \frac{\sin(4x)}{4}$

Substitution in the formula:

$\int x \cos(4x) \, dx = u v - \int v \, du$ Substituting $u$, $v$, and $du$: $\int x \cos(4x) \, dx = x \cdot \frac{\sin(4x)}{4} - \int \frac{\sin(4x)}{4} \, dx$

Simplify:

1. First term: $x \cdot \frac{\sin(4x)}{4} = \frac{x \sin(4x)}{4}$

2. Second term: $\int \frac{\sin(4x)}{4} \, dx = \frac{-\cos(4x)}{16}$

Combining these terms: $\int x \cos(4x) \, dx = \frac{x \sin(4x)}{4} + \frac{\cos(4x)}{16} + C$

Match to the given form:

The solution is given as: $\int x \cos(4x) \, dx = A x \sin(4x) + B \cos(4x) + C$

Compare coefficients:

1. The coefficient of $x \sin(4x)$ gives: $A = \frac{1}{4}$

2. The coefficient of $\cos(4x)$ gives: $B = \frac{1}{16}$

Final Answer:

$A = \frac{1}{4}, \quad B = \frac{1}{16}$

Would you like further details on integration by parts or related examples? Here are 5 related questions for further exploration:

1. How do you approach integration by parts for trigonometric functions?
2. What are alternative methods to solve this integral, such as using reduction formulas?
3. How does the choice of $u$ and $dv$ affect the integration process?
4. What are common applications of integration by parts in physics or engineering?
5. Can this method be extended to solve definite integrals involving similar terms?

Tip: When solving integrals, always verify results by differentiating the final expression to see if it matches the original integrand.