∫(-π/2→π/2)(x|x|+cosx)dx/[1+(sinx)^2]=∫(-π/2→π/2)x|x|*dx/[1+(sinx)^2]+∫(-π/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→...