Question

1. ∫(-π/2到π/2) [（sinx/1+x^2)+(cosx)^2]dx 2. ∫(-π/2到π/2) [（绝对值x）+sinx)^2]dx

怎么解到答案
1. π/2
2. π/2+(π^3/12)
谢谢！！！！！

Answer 1

∫(-π/2到π/2) [（sinx/1+x^2)+(cosx)^2]dx
奇函数的积分为0
故∫(-π/2到π/2) [（sinx/1+x^2)+(cosx)^2]dx
＝∫(-π/2到π/2) (cosx)^2dx
＝2∫(0到π/2) (cosx)^2dx
＝∫(0到π/2) (1＋cos2x）dx
＝
∫(-π/2到π/2) [（绝对值x）+sinx)^2]dx
＝∫(-π/2到π/2) [（绝对值x）＾2+（sinx)^2＋2绝对值x*sinx]dx
绝对值x*sinx是奇函数
∫(-π/2到π/2) [（绝对值x）+sinx)^2]dx
＝∫(-π/2到π/2) [（绝对值x）＾2+（sinx)^2]dx
＝2∫(0到π/2) [x＾2+（sinx)^2]dx