sin^2xcos^2x=1/4（2sinxcosx）^2=1/4sin2x
原式=∫cos2x/sin^2xcos^2xdx=∫cos2x/（1/4sin2x）dx
=∫2/sin2xd(sin2x)
=2lnsin2x+C
题目是cos2x/ (sin^2 x cos^2 x) dx ？
= (cos^2x - sin^2x)/ (sin^2 x cos^2 x) dx
答：由cos2x=1-2(sinx)^2得：(sinx)^2=1/2-cos2x/2 ∫(sinx)^2dx =∫ 1/2-cos2x/2 dx =x/2-sin2x/4 C
∫sin^2xcos^2xdx=1/4 ∫sin^2 (2x)dx
=1/4 ∫(1-cos4x)/2 dx
=1/8 x-1/32 sin4x +C
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