The integral provided is:

$\int 5x \cos(19x) \, dx$

To solve this, we will use integration by parts. The integration by parts formula is:

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

Step 1: Choose $u$ and $dv$

Let:

• $u = 5x$ (so that $du = 5 \, dx$)
• $dv = \cos(19x) \, dx$ (so that $v = \frac{\sin(19x)}{19}$, as the integral of $\cos(19x)$ is $\frac{1}{19} \sin(19x)$)

Step 2: Apply the integration by parts formula

$\int 5x \cos(19x) \, dx = 5x \cdot \frac{\sin(19x)}{19} - \int 5 \cdot \frac{\sin(19x)}{19} \, dx$

Step 3: Solve the remaining integral

Now we need to solve the integral:

$\int \frac{5 \sin(19x)}{19} \, dx$

The integral of $\sin(19x)$ is $-\frac{1}{19} \cos(19x)$, so:

$\int \frac{5 \sin(19x)}{19} \, dx = -\frac{5}{19^2} \cos(19x)$

Step 4: Combine the results

Now, combine the results from the previous steps:

$\int 5x \cos(19x) \, dx = \frac{5x \sin(19x)}{19} + \frac{5}{19^2} \cos(19x) + C$

Thus, the exact answer is:

$\frac{5x \sin(19x)}{19} + \frac{5}{361} \cos(19x) + C$

Would you like further explanation or additional details? Here are some related questions:

1. What is the purpose of using integration by parts?
2. Can you explain why we choose $u$ and $dv$ in this method?
3. What are other situations where integration by parts is useful?
4. How would the result change if the argument of the cosine function were different?
5. How does integration by parts simplify more complex integrals?

Tip: Integration by parts is particularly helpful when you have the product of a polynomial and a trigonometric or exponential function.