1. **State the problem:** We want to find the integral $$\int \sin^2(3x) \cos^2(2x) \, dx$$. 2. **Recall formulas and identities:** Use the power-reduction formulas: $$\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$$ $$\cos^2(\phi) = \frac{1 + \cos(2\phi)}{2}$$ 3. **Apply these to the integral:** $$\sin^2(3x) = \frac{1 - \cos(6x)}{2}$$ $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$ 4. **Rewrite the integral:** $$\int \sin^2(3x) \cos^2(2x) \, dx = \int \left(\frac{1 - \cos(6x)}{2}\right) \left(\frac{1 + \cos(4x)}{2}\right) dx = \int \frac{(1 - \cos(6x))(1 + \cos(4x))}{4} \, dx$$ 5. **Expand the numerator:** $$ (1 - \cos(6x))(1 + \cos(4x)) = 1 + \cos(4x) - \cos(6x) - \cos(6x)\cos(4x) $$ 6. **Use product-to-sum for the last term:** $$ \cos(6x)\cos(4x) = \frac{\cos(10x) + \cos(2x)}{2} $$ 7. **Substitute back:** $$ 1 + \cos(4x) - \cos(6x) - \frac{\cos(10x) + \cos(2x)}{2} $$ 8. **Rewrite the integral:** $$ \int \frac{1 + \cos(4x) - \cos(6x) - \frac{\cos(10x) + \cos(2x)}{2}}{4} \, dx = \int \left( \frac{1}{4} + \frac{\cos(4x)}{4} - \frac{\cos(6x)}{4} - \frac{\cos(10x)}{8} - \frac{\cos(2x)}{8} \right) dx $$ 9. **Integrate term-by-term:** $$ \int \frac{1}{4} dx = \frac{x}{4} $$ $$ \int \frac{\cos(4x)}{4} dx = \frac{\sin(4x)}{16} $$ $$ \int -\frac{\cos(6x)}{4} dx = -\frac{\sin(6x)}{24} $$ $$ \int -\frac{\cos(10x)}{8} dx = -\frac{\sin(10x)}{80} $$ $$ \int -\frac{\cos(2x)}{8} dx = -\frac{\sin(2x)}{16} $$ 10. **Combine all results:** $$ \int \sin^2(3x) \cos^2(2x) \, dx = \frac{x}{4} + \frac{\sin(4x)}{16} - \frac{\sin(6x)}{24} - \frac{\sin(10x)}{80} - \frac{\sin(2x)}{16} + C $$ **Final answer:** $$ \boxed{\frac{x}{4} + \frac{\sin(4x)}{16} - \frac{\sin(6x)}{24} - \frac{\sin(10x)}{80} - \frac{\sin(2x)}{16} + C} $$