The curve in this problem is `y = ln\ cos\ x`.

We need its derivative:

`(dy)/(dx)=-(sin\ x)/(cos\ x)=-tan\ x`

For this problem, we'll make use of an earlier result,

`1+tan^2 x=sec^2 x`

Applying the formula gives:

`s=int_a^bsqrt(1+((dy)/dx)^2)dx`

`=int_0^(pi//3)sqrt(1+(-tan x)^2)dx`

`=int_0^(pi//3)sqrt(1+tan^2x) dx`

`=int_0^(pi//3)sqrt(sec^2x) dx`

`=int_0^(pi//3)sec\ x\ dx`

`=[ln\ |sec\ x+tan\ x|]_0^(pi//3)`

`=[ln\ |sec\ (pi/3)+tan\ (pi/3)|` `{:-|ln(1)+ln(0)|]`

`=1.317`