Let's solve the given integral step-by-step. The integral is:

$u = -\int \frac{\sin(3x)}{3} \cdot \frac{1}{1 + \sin(3x)} \, dx$

Step 1: Simplify the integrand

The integrand simplifies to:

$-\frac{1}{3} \int \frac{\sin(3x)}{1 + \sin(3x)} \, dx$

Step 2: Substitution

Let: $t = 1 + \sin(3x)$ Then: $\frac{dt}{dx} = 3\cos(3x)$ or equivalently: $dt = 3\cos(3x) \, dx$

Rewriting the integral: $\sin(3x) = t - 1$

The integral becomes: $-\frac{1}{3} \int \frac{t - 1}{t} \cdot \frac{1}{3\cos(3x)} \, 3\cos(3x) \, dt$

This simplifies to: $-\frac{1}{3} \int \frac{t - 1}{t} \, dt$

Step 3: Split the fraction

Split the fraction: $\frac{t - 1}{t} = 1 - \frac{1}{t}$

The integral becomes: $-\frac{1}{3} \int \left( 1 - \frac{1}{t} \right) \, dt$

Split into two integrals: $-\frac{1}{3} \left( \int 1 \, dt - \int \frac{1}{t} \, dt \right)$

Step 4: Evaluate each term

1. The integral of $1$ with respect to $t$ is: $\int 1 \, dt = t$
2. The integral of $\frac{1}{t}$ is: $\int \frac{1}{t} \, dt = \ln|t|$

Thus, the integral becomes: $-\frac{1}{3} \left( t - \ln|t| \right)$

Step 5: Substitute back $t = 1 + \sin(3x)$

Substitute $t = 1 + \sin(3x)$ back into the result: $-\frac{1}{3} \left( 1 + \sin(3x) - \ln|1 + \sin(3x)| \right)$

Simplify: $u = -\frac{1}{3} \left( 1 + \sin(3x) - \ln|1 + \sin(3x)| \right)$

Final Answer:

$u = -\frac{1}{3} - \frac{\sin(3x)}{3} + \frac{\ln|1 + \sin(3x)|}{3}$

Let me know if you'd like further clarification or details!