Evaluate indefinite integral numerically in matlab/mathematica that it cannot do symbolically

Question

I am trying to calculate the integral of a function in Matlab and Mathematica that the software cannot do symbolically.

Here is my MatLab code so far, but I understand it may not be very helpful as is.

f = @(t) asin(0.5*sin(t));
a = @(t) sin(t);
F = int(f,t)   % Matlab can't do this
F = 
int(asin(sin(t)/2), t)
A = int(a,t)   % This works
A =
-cos(t)

dt = 1/(N-1); % some small number
for i=1:N
    F(i) = integral(f,(i-1)*dt,i*dt);
    A(i) = integral(a,(i-1)*dt,i*dt);
end

Both of the calculations in the for loop give a rough approximation of f or a not their integrals after multiplying by dt.

On the math stack-exchange I found a question that derives a finite difference like method for the integral at a point. However, when I did the calculation in Matlab it output a scaled down version of f which was evident after plotting (see above for what I mean by scaled down). I think that's because for smaller intervals the integral basically approximates the function to varying degrees of accuracy (again see above).

I am trying to get either a symbolic equation for the integral, or an approximation of the integral of the function at each location.

So my question is then if I have a function f that MatLab and Mathematica cannot easily take the integral of

1. can I approximate the integral directly with an integral calculator besides the default ones? (int,integral,trapz)

or

2. can I approximate the function with finite differences first and then evaluate the integral symbolically?

Answer 1

Your code is nearly fine it's just that

for i=1:N
    F(i) = integral(f,0,i*dt);
end

You could also do

F(1)=integral(f,0,dt)
for i=2:N
    F(i) = F(i-1)+integral(f,(i-1)*dt,i*dt);
end

Second option is surely more efficient

Because the primitive is really F(x)=int(f(x), 0, x) (0 defines a certain constant ) and for sufficiently small dx you have shown that f(x)=int(f(x), x,x+dx)/dx i. You have proven that MATLAB intégral function does its job.

For example let's take = the function above will compute if you wish to compute just replace 0 above by the constant a you like.

now and so you should get F containing a discretization of