These identities are $\cos^2x = \frac{1+\cos(2x)}{2}$ and $\sin^2x = \frac{1-\cos(2x)}{2}$.
So, we can rewrite the integral as:
$$\int \cos ^{2} x \sin ^{2} x d x = \int \left(\frac{1+\cos(2x)}{2}\right) \left(\frac{1-\cos(2x)}{2}\right) dx$$
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