I am trying to get a numerical answer of following improper integral but not getting the correct value

syms x;
f =(exp(1/cos(x)));
A=vpa(int(f,-11/7,11/7));
A

Please correct me where I am going wrong and how to evaluate this value using some approximations or other functions.

Answer 1

[Chunru]

Your integration interval [-11/7 11/7] include two points with infinity.

Try the following:

syms x;
f =(exp(1/cos(x)));
A=vpa(int(f, x, -10/7,10/7));
A

A = 

78.865629794567528914520336604775

[Walter Roberson]

In particular, 11/7 is an approximation of pi/2, and is slightly larger than pi/2 . cos(pi/2) is 0, so 1/cos(pi/2) would pass through infinity.

The situation would have been different if it had been exp(-1/cos(x))

[John D'Errico]

To expand on the answer here...

syms x
f =(exp(1/cos(x)));
fplot(f,[-11/7,-10/7])

fplot(f,[10/7,11/7])

The point being that cos(pi/2) == 0, and that pi/2 is just slightly less than 11/7.

pi/2 < 11/7

ans = logical
   1

But just slightly so. That means your integrand has a singularity at +/- pi/2. And that begs the question of whether the integral is well posed, having a finite value at all.