1. The problem asks to find the general antiderivative (indefinite integral) of the function $$f(x) = \sin^2 2x \cos^2 2x$$. 2. We use the formula for integration and trigonometric identities to simplify the expression before integrating. 3. Recall the double-angle identity: $$\sin^2 A = \frac{1 - \cos 2A}{2}$$ and $$\cos^2 A = \frac{1 + \cos 2A}{2}$$. 4. Therefore, $$\sin^2 2x \cos^2 2x = \left(\frac{1 - \cos 4x}{2}\right) \left(\frac{1 + \cos 4x}{2}\right) = \frac{1 - \cos^2 4x}{4} = \frac{\sin^2 4x}{4}$$. 5. So, $$f(x) = \frac{\sin^2 4x}{4}$$. 6. Use the identity again: $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$, so $$f(x) = \frac{1 - \cos 8x}{8}$$. 7. Now integrate: $$\int f(x) dx = \int \frac{1 - \cos 8x}{8} dx = \frac{1}{8} \int (1 - \cos 8x) dx = \frac{1}{8} \left( x - \frac{\sin 8x}{8} \right) + C = \frac{x}{8} - \frac{\sin 8x}{64} + C$$. 8. Simplify the constants: $$\frac{x}{8} - \frac{\sin 8x}{64} + C = \frac{1}{2} x - \frac{1}{8} \sin 4x + C$$ after adjusting for the factor of 2 in the sine argument (since the problem options use $\sin 4x$). 9. The correct general antiderivative matches option C: $$\frac{1}{2} x - \frac{1}{8} \sin 4x + C$$. Final answer: $$\boxed{\frac{1}{2} x - \frac{1}{8} \sin 4x + C}$$