1. **State the problem:** Evaluate the integral $$\int 20 \sin^2 x \cos^2 x \, dx$$. 2. **Use a trigonometric identity:** Recall that $$\sin^2 x \cos^2 x = \left(\sin x \cos x\right)^2 = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x$$. 3. **Rewrite the integral:** $$\int 20 \sin^2 x \cos^2 x \, dx = \int 20 \cdot \frac{1}{4} \sin^2 2x \, dx = \int 5 \sin^2 2x \, dx$$. 4. **Use the power-reduction formula:** $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$. Apply this with $$\theta = 2x$$: $$\sin^2 2x = \frac{1 - \cos 4x}{2}$$. 5. **Substitute into the integral:** $$\int 5 \sin^2 2x \, dx = \int 5 \cdot \frac{1 - \cos 4x}{2} \, dx = \frac{5}{2} \int (1 - \cos 4x) \, dx$$. 6. **Integrate term-by-term:** $$\int 1 \, dx = x$$ $$\int \cos 4x \, dx = \frac{\sin 4x}{4}$$ So, $$\frac{5}{2} \int (1 - \cos 4x) \, dx = \frac{5}{2} \left(x - \frac{\sin 4x}{4}\right) + C = \frac{5}{2} x - \frac{5}{8} \sin 4x + C$$. 7. **Final answer:** $$\int 20 \sin^2 x \cos^2 x \, dx = \frac{5}{2} x - \frac{5}{8} \sin 4x + C$$.