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,

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

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

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

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