How do you find the antiderivative of #int sin^2xcos^2x dx#?

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Answer 1 · 2017-01-11T15:45:46

The answer is #=x/8-1/32sin4x+C#

#sin2x=2sinxcosx#

So,

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

#cos4x=1-2sin^2(2x)#

#sin^2(2x)=1/2(1-cos(4x))#

So,

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

#=1/8(x-1/4sin4x)+C#

#=x/8-1/32sin4x+C#