∫(1,2)dx∫(√x,x)sin(πx/2y)dy+∫(2,4)dx+∫(√x,2)sin(πx/2y)dy其中(1,2)1是下限,2是上限.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 00:09:03
![∫(1,2)dx∫(√x,x)sin(πx/2y)dy+∫(2,4)dx+∫(√x,2)sin(πx/2y)dy其中(1,2)1是下限,2是上限.](/uploads/image/z/5573859-51-9.jpg?t=%E2%88%AB%281%2C2%29dx%E2%88%AB%28%E2%88%9Ax%2Cx%29sin%28%CF%80x%2F2y%29dy%2B%E2%88%AB%282%2C4%29dx%2B%E2%88%AB%28%E2%88%9Ax%2C2%29sin%28%CF%80x%2F2y%29dy%E5%85%B6%E4%B8%AD%EF%BC%881%2C2%EF%BC%891%E6%98%AF%E4%B8%8B%E9%99%90%2C2%E6%98%AF%E4%B8%8A%E9%99%90.)
∫(1,2)dx∫(√x,x)sin(πx/2y)dy+∫(2,4)dx+∫(√x,2)sin(πx/2y)dy其中(1,2)1是下限,2是上限.
∫(1,2)dx∫(√x,x)sin(πx/2y)dy+∫(2,4)dx+∫(√x,2)sin(πx/2y)dy
其中(1,2)1是下限,2是上限.
∫(1,2)dx∫(√x,x)sin(πx/2y)dy+∫(2,4)dx+∫(√x,2)sin(πx/2y)dy其中(1,2)1是下限,2是上限.
你得先把积分区域画出来,然后看图改变积分顺序.积分区域是y=x,y=√x,和y=2围成的区域.所以原式=∫(1,2)dy∫(y,y∧2)sin(πx/2y)dx=(4π 8)/π∧3
∫sin^2x/(1+sin^2x )dx求解,
∫sin(1/x)dx
∫sin(√2x)dx
∫(1-sin^2( x/2))dx
∫(cosx/1+sin^2x)dx
∫sin^2x(1+tanx)dx
∫1/sin(x/2)dx
∫ x sin(x+1) dx
∫x/sin^2(x) dx
∫sin(x) cos^2(x)dx
∫1+sin^2x/1-cos2x dx∫1+sin^2x/1-cos2x dx
∫√(cotx+1)/(sin^2x)dx的不定积分
∫1/√x*sin√x dx
∫1/ √X sin √x dx.
∫1/x^2 * sin 1/x dx
∫(1+sin^2x)/(cos^2x)dx
∫e^(-2x)sin(1/2x)dx
求∫1/tan^2x+sin^2x dx