sinx⼀(sinx+1)的原函数怎么求啊? 倒数第二步不对啊?

2025-12-25 08:00:18
推荐回答(1个)
回答1:

∫sinx/(sinx+1) dx
=∫(sinx+1-1)/(sinx+1) dx
=∫[1-1/(sinx+1)] dx
=x-∫1/(sinx+1) dx
=x-∫[1+tan^2(x/2)]/[1+tan^2(x/2)+2tan(x/2) ]dx
=x-∫1/[1+tan(x/2)]^2dtan(x/2)
=x+1/[1+tan(x/2)]+C