1. **State the problem:** We need to find the integral $$\int \frac{\cos 2x}{\sin^3 2x} \, dx$$. 2. **Rewrite the integral:** Notice that the integral can be written as $$\int \cos 2x \cdot \sin^{-3} 2x \, dx$$. 3. **Use substitution:** Let $$u = \sin 2x$$. Then, $$\frac{du}{dx} = 2 \cos 2x$$, so $$du = 2 \cos 2x \, dx$$, which implies $$\cos 2x \, dx = \frac{du}{2}$$. 4. **Substitute into the integral:** $$\int \cos 2x \cdot \sin^{-3} 2x \, dx = \int u^{-3} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-3} \, du$$. 5. **Integrate:** Recall that $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. Here, $$n = -3$$, so $$\frac{1}{2} \int u^{-3} \, du = \frac{1}{2} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{4} u^{-2} + C$$. 6. **Back-substitute:** Replace $$u$$ with $$\sin 2x$$: $$-\frac{1}{4} \sin^{-2} 2x + C = -\frac{1}{4 \sin^2 2x} + C$$. **Final answer:** $$\int \frac{\cos 2x}{\sin^3 2x} \, dx = -\frac{1}{4 \sin^2 2x} + C$$.