求不定积分∫cos2x/(cosx+sinx)^4 dx

∫cos2x/(cosx+sinx)^4 dx
=∫（(cosx)^2-(sinx)^2）/(cosx+sinx)^4 dx
=∫（cosx-sinx）/(cosx+sinx)^3 dx
=∫1/(cosx+sinx)^3 d(cosx+sinx)
=-1/2(cosx+sinx)^(-2) + C

∫ cos2x / (sin²x * cos²x) dx
= ∫ cos2x / (1/2 * sin2x)² dx
= 4∫ cos2x / (sin²2x) dx
= 4∫ csc2x * cot2x dx
= -2∫ csc2x * cot2x d(2x)
= -2csc2x + C
= -2/(sin2x) + C
= -secx*cscx + C本回答被网友采纳