求定积分,

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求定积分,

求定积分,
求定积分,

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