dx/[1/16(sin2x)^3(cos2x+1)]=16sin2xdx/[(sin2x)^4(cos2x+1)]=-8dcos2x/[(1-cos2x)^2(1+cos2x)]
设y=cos2x
dy/(1-y)^2(1+y)=dy/4(1+y)+dy/2(1-y)^2+dy/4(y-1)=1/4ln|y+1||y-1|-1/2(y-1)+c
原题=2ln(sin^2(2x))+4/(cos2x-1)+c
dx/[1/16(sin2x)^3(cos2x+1)]=16sin2xdx/[(sin2x)^4(cos2x+1)]=-8dcos2x/[(1-cos2x)^2(1+cos2x)]
设y=cos2x
dy/(1-y)^2(1+y)=dy/4(1+y)+dy/2(1-y)^2+dy/4(y-1)=1/4ln|y+1||y-1|-1/2(y-1)+c
原题=2ln(sin^2(2x))+4/(cos2x-1)+c