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

Answer 2

令u = tan(x/2)，cosx = (1 - u²)/(1 + u²)，dx = 2du/(1 + u²)
∫ 1/(3 + cosx) dx
= ∫ 1/[3 + (1 - u²)/(1 + u²)] · 2/(1 + u²) · du
= ∫ (1 + u²)/(3 + 3u² + 1 - u²) · 2/(1 + u²) · du
= 2∫ 1/(4 + 2u²) · du
= ∫ 1/(2 + u²) · du
= (1/√2)arctan(u/√2) + C
= (1/√2)arctan[(1/√2)tan(x/2)] + C