d/du ln((sin u^2)^2) 他舅~
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
∫du/(u^2-1)^(1/2)=ln[u+(u^2-1)^(1/2)]+C1
x=ln(u^2-1),dx={2u/(u^2-1)}du
du/(u^2-1)^(1/2)=dx/x 如何得到ln(u+(u^2-1))=lnx
设u=ln√(x^2+y^2+z^2) 求du
求函数u=ln(2x+3y+4z^2)的全微分du
设u=ln(x+y^2+z^3),求du
x=e^-t y=∫(0到t)ln(1+u^2)du
matlab du/dt=d(du)/dx^2 x属于(0,1),t属于(0,T]u(0,t)=u(1,t)=0u(x,
原式=∫du/(1+u^2)(2u-1) =(-1/5)∫d(1+u^2)/(1+u^2)-(1/5)∫du(1+u^2
设u=ln(sinx/y^0.5),其中x=3t^2,y=(1+t^2)^0.5,求du/dt
u=ln(xy+z)求du=