Substitute term with variable

Dear all,

I'm trying to substitute a rather long term

    (L/T*z-U(i-1)*z-(Ui(i)-Ui(i-1))/(di(i-1)-di(i))*(z^2/2+di(i-1)*z)+ ...
    (Xi(i)*sinh(2*pi*z/L)+Yi(i)*cosh(2*pi*z/L))*cos(2*pi*x/L)

with one variable, Psi(i). I tried on Mupad using subsex, but this doesn't work. I know, however, if you re-arrange the equation below, that the term given above exists. There is probably no explicit matlab/mupad function that could handle the issue (is there??) but maybe you came across with a procedure that solves the problem. Also maybe in more general terms.

The idea behind all this is, that I want to differentiate the equation below with respect to Psi. Therefore, I thought re-arranging would be the most applicable approach, but, again, I'd also appreciate other ideas.

Thanks a lot in advance, Cheers,

&nbsp

    (int(((int(((-Ui(4.0)+L/T+cos((pi*x*2.0)/L)*((pi*Xi(5.0)*...

    cosh((H*pi*cos((pi*x*2.0)/L))/L)*2.0)/L+(pi*Yi(5.0)*...

    sinh((H*pi*cos((pi*x*2.0)/L))/L)*2.0)/L)+((Ui(4.0)-Ui(5.0))*(di(4.0)+...

    H*cos((pi*x*2.0)/L)*(1.0/2.0)))/(di(4.0)-di(5.0)))^2*(1.0/2.0)+...

    1.0/L^2*pi^2*sin((pi*x*2.0)/L)^2*(Yi(5.0)*...

    cosh((H*pi*cos((pi*x*2.0)/L))/L)+Xi(5.0)*...

    sinh((H*pi*cos((pi*x*2.0)/L))/L))^2*2.0)/g+H*cos((pi*x*2.0)/L)*...

    (1.0/2.0),x,0.0,L*(1.0/2.0))*-2.0)/L+((-Ui(4.0)+L/T+cos((pi*x*2.0)/L)*...

    ((pi*Xi(5.0)*cosh((H*pi*cos((pi*x*2.0)/L))/L)*2.0)/L+(pi*Yi(5.0)*...

    sinh((H*pi*cos((pi*x*2.0)/L))/L)*2.0)/L)+((Ui(4.0)-Ui(5.0))*...

    (di(4.0)+H*cos((pi*x*2.0)/L)*(1.0/2.0)))/(di(4.0)-di(5.0)))^2*...

    (1.0/2.0)+1.0/L^2*pi^2*sin((pi*x*2.0)/L)^2*(Yi(5.0)*...

    cosh((H*pi*cos((pi*x*2.0)/L))/L)+Xi(5.0)*sinh((H*pi*...

    cos((pi*x*2.0)/L))/L))^2*2.0)/g+H*cos((pi*x*2.0)/L)*...

    (1.0/2.0))^2,x,0.0,L*(1.0/2.0))*2.0)/L;

[José-Luis]

I don't understand the problem, sorry. Why don't you make the "term" a function of i and call it when needed.

[Magnus]

Because Psi is differentiated before. The problem is something like this: diff(diff(Psi,x),Psi). In order to do the d/dPsi I would like to re-collect the definition of Psi, see above, and then substitute this expression with Psi again.