令u=tan（x/2)
cosx=(1-u^2)/(1+u^2) dx=2/(1+u^2)du
1/(3+cosx)=1/{2+[2/(1+u^2)]}
所以原始变为：∫[1/(3+cosx)]dx=∫1/(u^2+2)du=√2/2*arctan[√2/2*tan(x/2)]+c