令 x = sintI = ∫<0, π/2>sintcostdt/{[2-(sint)^2]cost} = ∫<0, π/2>sintdt/[2-(sint)^2] = -∫<0, π/2>dcost/[1+(cost)^2]= -[arctan(cost)]<0, π/2> = π/4
可以尝试三角代换。
