To solve the integral \( I = \int_0^{4\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx \) and find the value of \( k \) such that \( I = k \pi \ln 7 \), we will proceed step by step.
### Step 1: Rewrite the integral
Let
\[
I = \int_0^{4\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx.
\]
### Step 2: Use the properties of the sine and cosine functions
Since \( \sin x \) and \( \cos x \) are periodic functions with a period of \( 2\pi \), we can break the integral into two parts:
\[
I = \int_0^{2\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx + \int_{2\pi}^{4\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx.
\]
Using the substitution \( x = u + 2\pi \) for the second integral, we have:
\[
\int_{2\pi}^{4\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx = \int_0^{2\pi} \ln |13 \sin(u + 2\pi) + 3\sqrt{3} \cos(u + 2\pi)| \, du = \int_0^{2\pi} \ln |13 \sin u + 3\sqrt{3} \cos u| \, du.
\]
Thus,
\[
I = 2 \int_0^{2\pi} \ln |13 \sin x + 3\sqrt{3} \cos x| \, dx.
\]
### Step 3: Simplify the expression inside the logarithm
We can express \( 13 \sin x + 3\sqrt{3} \cos x \) in a different form. Let \( r = \sqrt{13^2 + (3\sqrt{3})^2} \):
\[
r = \sqrt{169 + 27} = \sqrt{196} = 14.
\]
Now, we can find \( \theta \) such that:
\[
\cos \theta = \frac{13}{14}, \quad \sin \theta = \frac{3\sqrt{3}}{14}.
\]
Thus, we can rewrite:
\[
13 \sin x + 3\sqrt{3} \cos x = 14 \left( \sin x \cos \theta + \cos x \sin \theta \right) = 14 \sin(x + \theta).
\]
### Step 4: Substitute back into the integral
Now, we can rewrite the integral:
\[
I = 2 \int_0^{2\pi} \ln |14 \sin(x + \theta)| \, dx.
\]
This can be split into two parts:
\[
I = 2 \int_0^{2\pi} \ln 14 \, dx + 2 \int_0^{2\pi} \ln |\sin(x + \theta)| \, dx.
\]
The first integral evaluates to:
\[
2 \int_0^{2\pi} \ln 14 \, dx = 2 \cdot 2\pi \ln 14 = 4\pi \ln 14.
\]
### Step 5: Evaluate the second integral
The integral \( \int_0^{2\pi} \ln |\sin(x + \theta)| \, dx \) can be evaluated using the known result:
\[
\int_0^{2\pi} \ln |\sin x| \, dx = -2\pi \ln 2.
\]
Thus,
\[
\int_0^{2\pi} \ln |\sin(x + \theta)| \, dx = -2\pi \ln 2.
\]
So,
\[
I = 4\pi \ln 14 + 2(-2\pi \ln 2) = 4\pi \ln 14 - 4\pi \ln 2 = 4\pi (\ln 14 - \ln 2) = 4\pi \ln \frac{14}{2} = 4\pi \ln 7.
\]
### Step 6: Relate to the original equation
We have found that:
\[
I = 4\pi \ln 7.
\]
Comparing this with the given \( I = k \pi \ln 7 \), we find that:
\[
k = 4.
\]
### Final Answer
Thus, the value of \( k \) is \( \boxed{4} \).
