Question

求∫∫丨cos(x+y)丨dxdy 区域D为0<=x<=π/2 0<=y<=π/2 求高手指点

Answer 1

x和y可互换，因此原式
= 2∫∫丨 cos(x+y) 丨dxdy 区域为D0： 0<=x<=π/2 0<=y<=x
= 2[∫∫<积分区域D1> cos(x+y) dxdy + ∫∫<积分区域D2> -cos(x+y) dxdy ]
D1： 0<=y<=π/4， y<=x<=π/2 -y
D2： 0<=x<=π/4， x<=y<=π/2 -x
D0=D1∪D2
则原式= 2[∫∫<积分区域D1> cos(x+y) dxdy + ∫∫<积分区域D2> -cos(x+y) dxdy ]
= 2∫∫<积分区域D1> cos(x+y) dxdy - 2∫∫<积分区域D2> cos(x+y) dxdy
= 2∫<y从0到π/4> dy ∫<x从y到π/2 -y> cos(x+y) dx - 2∫<x从π/4到π/2> dx ∫<y从π/2 -x到x> cos(x+y) dy
= 2∫<y从0到π/4> sin(x+y)|<x从y到π/2 -y> dy - 2∫<x从π/4到π/2> sin(x+y)|<y从π/2 -x到x>dx
= 2∫<y从0到π/4> [sin(π/2) -sin2y] dy - 2∫<x从π/4到π/2> [sin2x - sin(π/2)] dx
= 2∫<y从0到π/4> (1-sin2y) dy - 2∫<x从π/4到π/2> (sin2x - 1) dx
= 2[y+(1/2)cos2y]|<y从0到π/4> - 2[-(1/2)cos2x -x]|<x从π/4到π/2>
= [2y+cos2y]|<y从0到π/4> + [cos2x + 2x]|<x从π/4到π/2>
= [π/2 + (cos(π/2) - cos0)] + [(cosπ - cos(π/2) ) + π/2]
= [π/2 +(0-1)] + [(-1 - 0 ) + π/2]
= π - 2