原式=∫[∫[-π/2,π/2][cosxdx/(1+(cosx)^2]+∫[-π/2,π/2][xcosxdx/(1+(cosx)^2]前一项是偶函数,后一项是奇函数,积分为0,原式=2∫[0,π/2][cosxdx/(1+(cosx)^2]=2∫[0,π/2]dsinx/[2-(sinx)^2]设t=sinx,原式=2∫[...