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(sin^4x-cos^4x)(sin^4x+cos^4x)/[1-2sin^2xcos^2x]
=(sin^2x+cos^2x)(sin^2x-cos^2x)[(sin^2x+cos^2x)^2-2sin^2xcos^2x]/[1-2sin^2xcos^2x]
=(-cos2x)(1-2sin^2xcos^2x)/[1-2sin^2xcos^2x]
=integral(-cos2x)
-sin2x/2+c
all the best for your exams.
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