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(3) I = ∫∫<D>xcos(x+y)dxdy = ∫<0, π>xdx∫<0, x>cos(x+y)dy
= ∫<0, π>xdx∫<0, x>cos(x+y)d(x+y) = ∫<0, π>xdx[sin(x+y)]<y=0, y=x>
= ∫<0, π>x(sin2x-sinx)dx = ∫<0, π>xsin2xdx - ∫<0, π>xsinxdx
= -(1/2)∫<0, π>xdcos2x + ∫<0, π>xdcosx
= -(1/2)[xcos2x]<0, π> + (1/2)∫<0, π>cos2xdx + [xcosx]<0, π> - ∫<0, π>cosxdx
= -π/2 + (1/4)[sin2x]<0, π> - π - [sinx]<0, π> = -3π/2;
I = ∫∫<D>xcos(x+y)dxdy = ∫<0, π>dy∫<y, π>xcos(x+y)dx
= ∫<0, π>dy∫<y, π>xdsin(x+y)
= ∫<0, π>dy{[xsin(x+y)]<y, π> - ∫<y, π>sin(x+y)dx}
= ∫<0, π>dy{-πsiny - ysin2y +[cos(x+y)]<y, π>}
= ∫<0, π>[-πsiny - ysin2y - cosy + cos2y]dy
= [πcosy + (1/2)ycos2y - (1/4)sin2y - siny + (1/2)sin2y]<0, π>
= -π + (1/2)π - π = -3π/2.
= ∫<0, π>xdx∫<0, x>cos(x+y)d(x+y) = ∫<0, π>xdx[sin(x+y)]<y=0, y=x>
= ∫<0, π>x(sin2x-sinx)dx = ∫<0, π>xsin2xdx - ∫<0, π>xsinxdx
= -(1/2)∫<0, π>xdcos2x + ∫<0, π>xdcosx
= -(1/2)[xcos2x]<0, π> + (1/2)∫<0, π>cos2xdx + [xcosx]<0, π> - ∫<0, π>cosxdx
= -π/2 + (1/4)[sin2x]<0, π> - π - [sinx]<0, π> = -3π/2;
I = ∫∫<D>xcos(x+y)dxdy = ∫<0, π>dy∫<y, π>xcos(x+y)dx
= ∫<0, π>dy∫<y, π>xdsin(x+y)
= ∫<0, π>dy{[xsin(x+y)]<y, π> - ∫<y, π>sin(x+y)dx}
= ∫<0, π>dy{-πsiny - ysin2y +[cos(x+y)]<y, π>}
= ∫<0, π>[-πsiny - ysin2y - cosy + cos2y]dy
= [πcosy + (1/2)ycos2y - (1/4)sin2y - siny + (1/2)sin2y]<0, π>
= -π + (1/2)π - π = -3π/2.
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