（3）原式=∫(1,2)dy∫(y,y^2)sin(πx/2y)dx =∫(1,2)dy*[-(2y/π)*cos(πx/2y)|(y,y^2)] =∫(1,2) (-2y/π)*cos(πy/2)dy =∫(1,2) (-4y/π^2)*d[sin(πy/2)] =(-4y/π^2)*sin(πy/2)|(1,2)+∫(1,2) (4/π^2)*sin(πy/2)dy =4/π^2-(8/π^3)*cos(πy/2)|(1,2) =4/π^2+8/π^3
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024