∫ sin³x/(2 + cosx) dx
= ∫ (cos²x - 1)/(2 + cosx) dcosx
= ∫ [cosx(cosx + 2 - 2) - 1]/(2 + cosx) dcosx
= ∫ [cosx(cosx + 2) - 2(cosx + 2 - 2) - 1]/(2 + cosx) dcosx
= ∫ [cosx(cosx + 2) - 2(cosx + 2) + 3]/(2 + cosx) dcosx
= ∫ [cosx - 2 + 3/(2 + cosx)] dcosx
= (1/2)cos²x - 2cosx + 3ln(2 + cosx) + C,这样对吧?