plot an obtained function after several calculus

Good morning!

I want to plot the folloing function which is obtained after several calculus on Matlab,

syms s t a
Y=(sin(int(cos((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/9))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/18)))^2/100)^2*(exp(-2*s^(1/3)) + cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + 0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s^(2/3)))^2/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/6)))^2/100)^2)^2, s, 0, 1))/100 + sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/18))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/36)))^2/100)*(exp(-2*t^(1/6)) + cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t^(1/3)))^2/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100))^2/100)*(exp(-2*t) + cos(int(sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(s^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/12)))^2/100)*(exp(-2*s^(1/2)) + sin(0.00008414709848078965066525023216303*cos(s^(1/2))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/4)))^2/100 + cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + 0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s))^2/100)) + 1, s, 0, t^2))^2/100 + sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100)*(exp(-2*t^(1/2)) + cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t))^2/100 + sin(0.00008414709848078965066525023216303*cos(t^(1/2))^2 + 0.008414709848078965066525023216303*exp(-2*t^(1/4)))^2/100))^2/100)

Unrecognized function or variable 'a'.

in this function another variable appeared (denoted by ' a' ) so i add it in symbolique variable

To plot this function i used the folowing code:

t_vals=linspace(0, 1, 10);
figure(1)
hold on
y_eval = double(eval(subs(Y(i+1),t, t_vals)))
plot(t_vals, y_eval;
hold off

[L.GREEN]

I tried also the following code , i got an error message:

Error using arrayfun

sym output type is not supported. Set 'UniformOutput' to false.

F=@(t) Y
figure(2)
hold on
Fnum = arrayfun(@(t)F(t),t_vals);
plot(t_vals,Fnum)
grid on
hold off

[Torsten]

You did not yet tell us what the problem is with your code. The evaluation for one value for t already lasts too long to be executed here. Is the problem the long time it takes on your PC ?

syms s t a
Y=(sin(int(cos((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
    sin(0.00008414709848078965066525023216303*cos(s^(1/9))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/18)))^2/100)^2*(exp(-2*s^(1/3)) + ...
    cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + ...
    0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s^(2/3)))^2/100 + ...
    sin(0.00008414709848078965066525023216303*cos(s^(1/3))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/6)))^2/100)^2)^2, s, 0, 1))/100 + ...
    sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
    sin(0.00008414709848078965066525023216303*cos(t^(1/18))^2 + ...
    0.008414709848078965066525023216303*exp(-2*t^(1/36)))^2/100)*(exp(-2*t^(1/6)) + ...
    cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t^(1/3)))^2/100 + ...
    sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + ...
    0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100))^2/100)*(exp(-2*t) + ...
    cos(int(sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
    sin(0.00008414709848078965066525023216303*cos(s^(1/6))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/12)))^2/100)*(exp(-2*s^(1/2)) + ...
    sin(0.00008414709848078965066525023216303*cos(s^(1/2))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/4)))^2/100 + ...
    cos(int(sin(0.00008414709848078965066525023216303*cos(a)^2 + ...
    0.008414709848078965066525023216303*exp(-2*a^(1/2))) + 1, a, 0, s))^2/100)) + 1, s, 0, t^2))^2/100 + ...
    sin((sin(int(cos((0.00008414709848078965066525023216303*cos(s^(2/3))^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/3)))^2)^2, s, 0, 1))/100 + ...
    sin(0.00008414709848078965066525023216303*cos(t^(1/6))^2 + ...
    0.008414709848078965066525023216303*exp(-2*t^(1/12)))^2/100)*(exp(-2*t^(1/2)) + ...
    cos(int(sin(0.00008414709848078965066525023216303*cos(s)^2 + ...
    0.008414709848078965066525023216303*exp(-2*s^(1/2))) + 1, s, 0, t))^2/100 + ...
    sin(0.00008414709848078965066525023216303*cos(t^(1/2))^2 + ...
    0.008414709848078965066525023216303*exp(-2*t^(1/4)))^2/100))^2/100);
double(subs(Y,t,1))

[L.GREEN]

I let my PC run, it's been 15 minutes and i still don't have the result.

[Torsten]

I think you must change your strategy ...

Cut the expression into pieces. Use numerical instead of symbolic integration whereever possible.

[L.GREEN]

Ok, that you for your help.

[Walter Roberson]

y_eval = double(eval(subs(Y(i+1),t, t_vals)))

You should never eval() a symbolic result. eval() of a symbolic result has no documented meaning.

In practice, eval() of a symbolic result is the same as eval() of char() of the symbolic result. The problem with that is that char() of symbolic results produces something that is not in MATLAB expression format and is not in MuPAD internal code: char() of symbolic results is intended to be human readable, and is not intended for machine processing.

What you need to do instead is perform all necessary subs() on the expression, and then double() the result without using eval()