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integrate sin^2xcos^2x dx
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\[\int\limits_{?}^{?}\sin^2xcos^2xdx\]
sin^2x= 1-cos^2x/2
well express everything as sine or cosine using \(cos^2x+sin^2x=1\)
You could also use the fact that \(\sin{2x} = 2\sin{x}\cos{x}\) and then your integral will reduce to \(\displaystyle \int \left (\frac{\sin{2x}}{2} \right)^{2} dx\) Now you can use the fact that \( \cos{2\theta} = 1 - 2\sin^{2}\theta\) or alternatively \(2\sin^{2}\theta = 1 - \cos{2\theta}\) to give \(\displaystyle \int \frac{\sin^{2}{2x}}{4} dx = \int \frac{(1-\cos{4x})}{8} dx\)
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