设I=∫∫D [sin(x^2)+cos(y^2)]dxdy原式=∫(0,1)[sinx2+∫(0,1)cosy2dy]dx=∫(0,1)sin(x^2)dx+∫(0,1)cos(y^2)dy=∫(0,1)[sin(x^2)+cos(x^2)]dx=∫(0,1)2^(1/2)*sin(x^2+pi/4)dx∵pi/4≤x^2+pi/4≤1+pi/4...