These identities are $\sin^2x = \frac{1-\cos(2x)}{2}$ and $\cos^2x = \frac{1+\cos(2x)}{2}$. Substituting these into the integral gives us:
$$\int \sin ^{2} x \cos ^{2} x d x = \int \frac{1-\cos(2x)}{2} \cdot \frac{1+\cos(2x)}{2} dx = \frac{1}{4} \int
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