`I= int sin^(2) 3x dx`.

Answer

To solve the integral \( I = \int \sin^2(3x) \, dx \), we can use the trigonometric identity for \(\sin^2\) which states:

\[ \sin^2(a) = \frac{1 - \cos(2a)}{2} \]

### Step-by-Step Solution:

1. **Apply the Identity**:
We apply the identity to \(\sin^2(3x)\):
\[ \sin^2(3x) = \frac{1 - \cos(6x)}{2} \]

2. **Rewrite the Integral**:
Substitute the identity into the integral:
\[ I = \int \sin^2(3x) \, dx = \int \frac{1 - \cos(6x)}{2} \, dx \]

3. **Separate the Integral**:
We can separate the integral:
\[ I = \frac{1}{2} \int (1 - \cos(6x)) \, dx = \frac{1}{2} \left( \int 1 \, dx - \int \cos(6x) \, dx \right) \]

4. **Integrate Each Term**:
- The integral of \(1\) with respect to \(x\) is:
\[ \int 1 \, dx = x \]
- The integral of \(\cos(6x)\) is:
\[ \int \cos(6x) \, dx = \frac{1}{6} \sin(6x) \]
Therefore, we have:
\[ I = \frac{1}{2} \left( x - \frac{1}{6} \sin(6x) \right) \]

5. **Combine the Results**:
Now, combine the results:
\[ I = \frac{x}{2} - \frac{1}{12} \sin(6x) + C \]
where \(C\) is the constant of integration.

### Final Answer:
\[ I = \frac{x}{2} - \frac{1}{12} \sin(6x) + C \]