"which makes no sense since I'm multiplying scalars."
Nope, you aren't.
"Why is this?"
Because you are multiplying vectors. The INTEGRAL documentation states "For scalar-valued problems, the function y = fun(x) must accept a vector argument, x, and return a vector result, y. This generally means that fun must use array operators instead of matrix operators. For example, use .* (times) rather than * (mtimes)." The same documentation also states one solution to this: "If you set the 'ArrayValued' option to true, then fun must accept a scalar and return an array of fixed size."
Lets try following what the documentation advises:
[a,b,k,I0,t,cte] = deal(1);
fun=@(x) sqrt(exp((-(k*a*I0*2*(1+cos(cte*x))/(b+a*2*I0*(1+cos(cte*x)))))*(1-exp(-t*(i)*(b+2*a*I0*(1+cos(cte*x)))))));
fun2=@(x) cos(cte*x); % check this ^
fun3=@(x) fun(x)*fun2(x);
integral(fun3,-1e-7,1e-7, 'ArrayValued',true)
Alternatively you could write the function to accept a vector, just as the documentation describes.
