第1个回答 推荐于2016-01-07
∫dx/[(2+cosx)sinx]
=∫sinxdx/[(2+cosx)sin²x]
=∫-d(cosx)/[(2+cosx)(1-cos²x)]
=∫{1/[(cosx+1)(cosx-1)]-1/[(cosx-1)(cosx+2)]}d(cosx)
=∫{1/[2(cosx-1)]-1/[2(cosx+1)-1/[3(cosx-1)]+1/[3(cosx+2)]}d(cosx)
=(1/2)∫d(cosx)/(cosx-1)-(1/2)∫d(cosx)/(cosx+1)-(1/3)∫d(cosx)/(cosx-1)+(1/3)∫d(cosx)/(cosx+2)
=(1/2)ln｜cosx-1｜-(1/2)ln｜cosx+1｜-(1/3)ln｜cosx-1｜+(1/3)ln｜cosx+2｜
=(1/6)ln｜cosx-1｜+(1/3)ln｜cosx+2｜-(1/2)ln｜cosx+1｜
本回答被提问者和网友采纳第2个回答 2014-12-27
上下同乘sinx
第3个回答 2014-12-27
化简