求不定积分∫1/(3+cosx)dx,麻烦大家帮帮忙哈,谢谢啦^^

Question

求不定积分∫1/(3+cosx)dx,麻烦大家帮帮忙哈,谢谢啦^_^

Answer 1

就是万能代换.令t=tanx/2,x=2arctant,dx=2/(1+t^2)dt,cosx=(1-t^2)/(1+t^2),代入得：
∫1/(3+cosx)dx=∫1/(3+(1-t^2)/(1+t^2)）*2/(1+t^2)dt=∫1/(2+t^2)dt=(1/√2)arctan(t/√2)+C
=(1/√2)arctan(tan(x/2)/√2)+C