let
f(x) = x/[1+(sinx)^2]
f(-x)=-f(x)
=>∫(-π/2->π/2) x/[1+(sinx)^2] dx =0
∫(-π/2->π/2) (x+cosx)/[1+(sinx)^2] dx
=∫(-π/2->π/2) cosx/[1+(sinx)^2] dx
=2∫(0->π/2) cosx/[1+(sinx)^2] dx
=2∫(0->π/2) dsinx/[1+(sinx)^2]
=2[arctan(sinx)]|(0->π/2)
=π/2
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