解:∵A=[0,π]*[0,π]
∴0≤x+y≤2π
∵当0≤x+y≤π/2时,cos(x+y)≥0
当π/2≤x+y≤3π/2时,cos(x+y)≤0
当3π/2≤x+y≤2π时,cos(x+y)≥0
∴∫∫|cos(x+y)|dxdy=∫<0,π/2>dx[∫<0,π/2-x>cos(x+y)dy-∫<π/2-x,π>cos(x+y)dy]
+∫<π/2,π>dx[-∫<0,3π/2-x>cos(x+y)dy+∫<3π/2-x,π>cos(x+y)dy]
=∫<0,π/2>[(sin(π/2)-sinx)-(sin(π+x)-sin(π/2))]dx
+∫<π/2,π>[(sin(π+x)-sin(3π/2))-(sin(3π/2)-sinx)]dx
=2∫<0,π/2>dx+2∫<π/2,π>dx
=2*(π/2)+2*(π-π/2)
=2π。
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