>>477 ∬e^(x-y)cos(x+y)dxdy I={(x,y):-パイ/2<=x<=パイ/2,0<=y<=パイ/4}

=∬[I] e^x e^(-y) (cos x cos y -sin x sin y) dxdy
=∬[I] e^x cos x e^(-y) cos y dx dy -∬[I] e^x sin x e^(-y) sin y dx dy
= (∫[-π/2 , π/2] e^x cos x dx ) ( ∫[0,π/4] e^(-y) cos y dy )
- (∫[-π/2 , π/2] e^x sin x dx ) ( ∫[0,π/4] e^(-y) sin y dy )

１変数の積分になったからあとはご自分でどうぞ