求解∫1/(cos^4(x)sin^2(x))dx

∫1/[(cosx)^4(sinx)]dx=∫[(sinx)^2+(cosx)^2]/[(cosx)^4(sinx)]dx
=∫(secx)^4dx+4∫(csc2x)^2dx
∫(secx)^4dx=∫(secx)^2[(tanx)^2+1]dx=∫[(tanx)^2+1]dtanx=(tanx)^3/3+tanx
∫(csc2x)^2dx=-1/2*cot2x
所以∫1/[(cosx)^4(sinx)]dx=(tanx)^3/3+tanx-2cot2x+C

sin^2(x)cos^4(x)
=1/4*sin²2xcos²x
=1/4*(1-cos4x)/2*(1+cos2x)/2
=1/16*(1+cos2x-cos4x-cos2xcos4x)
=1/16*(1+cos2x-cos4x-cos2xcos4x)
=1/16*[1+cos2x-cos4x-1/2*(cos3x+cos6x)]