学帮网求定积分,
来源：学生作业学帮网 编辑：学帮网 时间：2024/05/26 20:49:25
求定积分,
因为x^3(cosx)^2是个奇函数,所以∫(-π/2->π/2) x^3(cosx)^2dx=0
所以原积分=∫(-π/2->π/2) (sinx)^2(cosx)^2dx=(1/4)∫(-π/2->π/2) (sin2x)^2dx
=(1/4)∫(0->π/2) (sin2x)^2d(2x)
=(1/4)∫(0->π) (sinu)^2du
=(1/2)∫(0->π/2) (sinu)^2du
=π/8