积分∫dx/3+cosx=?请尽可能详细，谢谢。

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

∫1/(3-cosx) dx
=∫1/[2+2(sin0.5x)^2]dx
=∫1/[1+(sin0.5x)^2]d(0.5x)
=∫(csc0.5x)^2/[(csc0.5x)^2+1]d(0.5x)
=-∫1/[(cot0.5x)^2+2]d(cot0.5x)
=-(1/√2)arctan(cot0.5x/√2)+C