急急急急急，这道高数题怎么做，求大佬

Answer 1

x^3(cosx)^2属于奇函数，同时积分区间是对称区间，所以积分为0.
则原积分=∫(-π/2,π/2)(sinxcosx)^2dx
=1/4∫(-π/2,π/2)(sin2x)^2dx
=1/8∫(-π/2,π/2)(1-cos4x)dx
=1/8*(x-1/4sin4x)|(-π/2,π/2)
=π/8