How do you find the antiderivative of #sin^2 (x)cos^2 (x) dx#?

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Answer 1 · 2016-10-22T14:05:56

#intsin^2xcos^2xdx=x/8-(sin4x)/32#

As #sin2x=2sinxcosx#, #sinxcosx=1/2sin2x# and

#sin^2xcos^2x=1/4sin^2 2x=1/4(1/2(1-cos4x))=(1-cos4x)/8#

Hence #intsin^2xcos^2xdx=int(1-cos4x)/8 dx#

= #x/8-1/8intcos4xdx#

Let #u=4x#, then #du=4dx#

and #intcos4xdx=1/4intcosudu=1/4sinu=1/4sin4x#

Hence #intsin^2xcos^2xdx=x/8-1/8xx1/4sin4x=x/8-(sin4x)/32#