令u=tan(x/2) => dx=2du/(1+u²),cosx=(1-u²)/(1+u²)当x=0,u=0 // 当x=π,u=+∞原式= ∫[0,+∞] 1/[2+(1-u²)/(1+u²)] * 2/(1+u²) du= ∫[0,+∞] (1+u²)/(u²+3) * 2/(1+u...