三角函数换元积分法,
∫3/(3-cosx) dx
令p=tan(x/2),dx=2dp/(1+p²),cosx=(1-p²)/(1+p²)
=3∫1/{3-[(1-p²)/(1+p²)]}*2/(1+p²) dp
=6∫1/[3(1+p²)-(1-p²)] dp
=3∫1/(1+2p²) dp
令p=1/√2*tanθ,dp=1/√2*sec²θdθ
=(3/√2)∫sec²θ/(1+2*1/2*tan²θ) dθ
=(3/√2)∫dθ
=(3/√2)θ+C
=(3/√2)arctan(√2*p)+C
=(3/√2)arctan[√2*tan(x/2)]+C