求定积分∫(-π/2→π/2)(x|x|+cosx)dx/[1+(sinx)^2]

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∫(-π/2→昌悉π/2)(x|x|+cosx)dx/[1+(sinx)^2]
=∫(-π/2→弊友π/2)x|x|*dx/[1+(sinx)^2]+∫(-π/耐卜乎2→π/2)cosx*dx/[1+(sinx)^2]
由于x|x|*dx/[1+(sinx)^2]是奇函数，故∫(-π/2→π/2)x|x|*dx/[1+(sinx)^2]=0
原式=∫(-π/2→π/2)cosx*dx/[1+(sinx)^2]
=∫(-π/2→π/2)d(sinx)/[1+(sinx)^2]
=[arctan(sinx)]|(-π/2,π/2)
=arctan[sin(π/2)]-arctan[sin(-π/2)]
=arctan1-arctan(-1)
=2arctan1
=2*π/4
=π/2

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