Maple procedure issue

I know some of you are into maple so I figured that perhaps someone might be able to help me with the issue that I am currently having.

Code:

---

restart;
calcDistributionFunc := proc (f)
    cdf := unapply(int(int(f(x)*sin(x), x = 0 .. x2), phi = 0 .. phi2), x2);
    phi2 := 2*Pi;
   
    cte := cdf((1/2)*Pi);

    if 1 < evalf(cte) then
        cte := 1/cte
    end if;

    cdf := apply(cdf, x);
    cdf := cte*cdf;
    cdf := unapply(cdf, x);

    s := solve(y = cdf(x), x);

    if 1 < nops([s]) then
        b := unapply(s[1], y)
    else
        b := unapply(s, y)
    end if
end proc

calcDistributionFunc(cos(x));

---

The above code works fine as long as you do not put in into a procedure, from the moment I try to put the code into a procedure and call it using calcDistributionFunc(cos(x)); I get the error:

Error, (in solve) cannot solve expressions with int((cos(x))(x)*sin(x), x = 0 .. (1/2)*Pi) for x

For example a function f(x) = cos(x) will yield a distribution function: arccos(sqrt(-y+1))

In case you are wondering what it is that I am trying to do, I have a userdefined distribution function and I want to generate random numbers using this distribution by generating uniform random numbers between 0 and 1 and evaluating this random number using the calculated distribution function. This in turn is used for Monte Carlo based simulations...

Probably I am just missing some basic tips/pointers on how to use procedures in maple. I have to admit I am more a matlab person then a maple one, but I need an analytical solution and not a numerical one. I know that using Matlab I can also perform analytical math using the symbolic link toolbox, but that would be like driving around in a Porsche 959 trying to get to the finishline of Paris-Dakar.

Thanks in advance for any help :).

Just looking at that error message, I would guess that

Code:

---

cdf := unapply(int(int(f(x)*sin(x), x = 0 .. x2), phi = 0 .. phi2), x2);

---

should really be

Code:

---

cdf := unapply(int(int(f*sin(x), x = 0 .. x2), phi = 0 .. phi2), x2);

---

since it seems to be dumping cos(x) in place of "f", not "f(x)"

Wow, that I have been looking over this so many times... Thanks!

Anyhow, I will have to move away from the analytical solution as finding the inverse of a function is not that straightforward anymore as soon as the function to be inverted gets quite complicated (which is the case as I am trying to fit functions to measured BSDF data).

Seems I'm back where I started by performing my monte carlo simulations using distribution functions that are being sampled instead of using an analytical solution.