1. The problem asks for the integral of $\sin(3x + 2)$ with respect to $x$. 2. Recall the formula for integrating sine of a linear function: $$\int \sin(ax + b) \, dx = -\frac{1}{a} \cos(ax + b) + C$$ where $a$ and $b$ are constants, and $C$ is the constant of integration. 3. In this problem, $a = 3$ and $b = 2$. 4. Applying the formula, we get: $$\int \sin(3x + 2) \, dx = -\frac{1}{3} \cos(3x + 2) + C$$ 5. This matches option A. Therefore, the answer is: **A.** $-\frac{1}{3} \cos(3x + 2) + C$