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∫sin²x cos²x dx

= ∫(1-cos²x)cos²x dx

=∫cos²xdx-⌠cos²xcos²xdx

=1/2⌠1+cos2x dx-1/2⌠[(1+cos2x)(1+cos2x)] Do the operations, distributions, arrange common numbers, and try to sort out the factors as a polynomiom. Then,

=1/2x+1/4sin2x-1/2x-1/2sin2x-1/4x-1/16sin4x

=-1/4x-1/16sin4x

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Q: What is the integral of sin squared x times cos squared x?
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