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\int \sin ^{2}x\cos ^{2}x\,dx\,

When both \sin x and \cos x have even powers, turn both into \cos(2x) by using the power reducing formulas

\sin ^{2}u={\frac  {1-\cos(2u)}{2}}\,

\cos ^{2}u={\frac  {1+\cos(2u)}{2}}\,

\int \sin ^{2}x\cos ^{2}x\,dx=\int {\frac  {1-\cos(2x)}{2}}{\frac  {1+\cos(2x)}{2}}={\frac  {1-\cos ^{2}(2x)}{4}}\,dx\,

Now, if there is an odd number of \cos x, simply turn all but one into \sin x and we know the procedure. If there is still an even number of \cos x (and there is always an even number of \sin x, that is 0), continue to reduce them until you get to an odd power of \cos x. First, for to lessen the mess, I will change the \cos to \sin .

\int {\frac  {1-\cos ^{2}(2x)}{4}}\,dx=\int {\frac  {sin^{2}(2x)}{4}}\,dx={\frac  {1}{4}}\int {\frac  {1-\cos(4x)}{2}}\,dx\,


\int \sin ^{2}x\cos ^{2}x\,dx={\frac  {1}{8}}x-{\frac  {1}{32}}\sin(4x)+C\,

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