The problem asks us to find the antiderivative of $x \cos(5x)$ given that $f(x) = x \sin(5x)$ and $f'(x) = 5x \cos(5x) + \sin(5x)$.

1. Problem Description

The problem asks us to find the antiderivative of $x \cos(5x)$ given that $f(x) = x \sin(5x)$ and $f'(x) = 5x \cos(5x) + \sin(5x)$.

2. Solution Steps

We are given $f(x) = x \sin(5x)$ and $f'(x) = 5x \cos(5x) + \sin(5x)$.

We want to find the antiderivative of $x \cos(5x)$, which is $\int x \cos(5x) \, dx$.

From $f'(x) = 5x \cos(5x) + \sin(5x)$, we can write

$5x \cos(5x) = f'(x) - \sin(5x)$.

Dividing by 5, we get

$x \cos(5x) = \frac{f'(x)}{5} - \frac{\sin(5x)}{5}$.

Now, we integrate both sides with respect to $x$:

$\int x \cos(5x) \, dx = \int \left( \frac{f'(x)}{5} - \frac{\sin(5x)}{5} \right) \, dx$

$\int x \cos(5x) \, dx = \frac{1}{5} \int f'(x) \, dx - \frac{1}{5} \int \sin(5x) \, dx$.

We know that $\int f'(x) \, dx = f(x)$, so

$\int x \cos(5x) \, dx = \frac{1}{5} f(x) - \frac{1}{5} \int \sin(5x) \, dx$.

To evaluate $\int \sin(5x) \, dx$, we use substitution. Let $u = 5x$, so $du = 5 \, dx$, and $dx = \frac{1}{5} \, du$.

$\int \sin(5x) \, dx = \int \sin(u) \cdot \frac{1}{5} \, du = \frac{1}{5} \int \sin(u) \, du = \frac{1}{5} (-\cos(u)) + C = -\frac{1}{5} \cos(5x) + C$.

Therefore,

$\int x \cos(5x) \, dx = \frac{1}{5} f(x) - \frac{1}{5} \left( -\frac{1}{5} \cos(5x) \right) + C$.

Since $f(x) = x \sin(5x)$,

$\int x \cos(5x) \, dx = \frac{1}{5} (x \sin(5x)) + \frac{1}{25} \cos(5x) + C$.

$\int x \cos(5x) \, dx = \frac{x \sin(5x)}{5} + \frac{\cos(5x)}{25} + C$.

3. Final Answer

$\frac{x \sin(5x)}{5} + \frac{\cos(5x)}{25} + C$