\(\sin^2x=\dfrac{1-\cos2x}2\)
\(\cos^2x=\dfrac{1+\cos2x}2\)
From the identities above, you have
\(\sin^22x\cos^22x=\dfrac{(1-\cos4x)(1+\cos4x)}4=\dfrac{1-\cos^24x}4\)
Applying once more, you have
\(\dfrac{1-\cos^24x}4=\dfrac{1-\dfrac{1+\cos8x}2}4=\dfrac{1-\cos8x}8\)
So,
\(\displaystyle\int\sin^22x\cos^22x\,\mathrm dx=\frac18\int(1-\cos8x)\,\mathrm dx=\frac x8-\frac1{64}\sin8x+C\)
