1. **State the problem:** We need to evaluate the integral $$\int \frac{3}{\sin^2 3x} \, dx$$ and match it with one of the given options. 2. **Rewrite the integrand:** Recall that $$\csc x = \frac{1}{\sin x}$$, so $$\frac{1}{\sin^2 3x} = \csc^2 3x$$. Thus, the integral becomes: $$\int 3 \csc^2 3x \, dx$$ 3. **Use substitution:** Let $$u = 3x$$, then $$du = 3 dx$$ or $$dx = \frac{du}{3}$$. 4. **Substitute into the integral:** $$\int 3 \csc^2 u \cdot \frac{du}{3} = \int \csc^2 u \, du$$ 5. **Integrate:** The integral of $$\csc^2 u$$ is $$-\cot u + C$$, so: $$\int \csc^2 u \, du = -\cot u + C$$ 6. **Back-substitute:** Replace $$u$$ with $$3x$$: $$-\cot 3x + C$$ 7. **Compare with options:** The answer matches option (c) $$-\cot 3x$$. **Final answer:** $$\int \frac{3}{\sin^2 3x} \, dx = -\cot 3x + C$$