∫1/(sin^4xcos^4x)dx =∫16/sin^4(2x)dx =∫16csc^4(2x)
（sin∧4xcos∧2x）的原函数是 ∫（sin∧4xcos∧2x）dx =∫[sin^2x(
不是d(cos2x)，二是d(cos²x) 因为(cos²x)
=(cos^2x+sin^2x)^2-2sin^2xcos^2x =1-sin2x/2
化简一步 y=(sinx+cosx)-2sinxcosx =1-(1/2)(2sinxcosx) =
f(x)=sin^4x-sin^2x+1 =(sin^2x-1/2)^2+3/4 0<=s
y=(cosx)^4+2sinxcosx-(sinx)^4 =[(cosx)^2+(sinx)^2
f(x)=cos^4x-2sinxcosx-sin^4x =(cos^2x+sin^2x)(cos