Find the value from the integral equation

I have the function A(e) as:

A=(cos(4*w)*cos(4*w)+1)/(1+exp(w-m0);

where the w is the variable and and m0 is the unknown value which should to be obtained by:

int(A,e,-20,20)-2=0.

I need to obtain the m0 parameter by solving the above equation.

[Sargondjani]

I'm just going to give hints:

1. make a function, or function handle of A
2. define an objective function (the equation that should be zero. You can use Matlabs "integral" function.
3. you need solve solve the objective function. You can use fsolve, or alternatively look an algorithm using Newton's method on file exhange (for example this link might work: https://uk.mathworks.com/matlabcentral/fileexchange/28227-newton-s-method)

This should get you started.

Answer 1

[Torsten]

fun = @(w,m0) (cos(4*w).*cos(4*w)+1)./(1+exp(w-m0));
m0sol = fsolve(@(m0)integral(@(w)fun(w,m0),-20,20)-2,1)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

m0sol = -18.9804

integral(@(w)fun(w,m0sol),-20,20)-2

ans = 2.6910e-11

[Arad]

Thank you very much.

Answer 2

[Dyuman Joshi]

You can solve it symbollically (This requires symbolic toolbox)

syms w m0
A=(cos(4*w)*cos(4*w)+1)/(1+exp(w-m0));
eq=int(A,w,-20,20)-2==0;
val=vpasolve(eq)

val = 

%Verification
syms y(x)
y(x)=(cos(4*x)*cos(4*x)+1)/(1+exp(x-val));
inty=int(y,x,-20,20)

inty = 

%As we can see here, we don't get the numeric value of the integral
%so we use vpa to approximate the value
vpa(inty)

ans = 

2.0

[Arad]

Thank you very much.