Answer
$$\kappa \left( x \right) = \cos x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \cos x \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}f''\left( x \right) \cr
& f'\left( x \right) = \frac{{ - \sin x}}{{\cos x}} = - \tan x \cr
& f''\left( x \right) = - {\sec ^2}x \cr
& {\text{Use the formula }}\kappa \left( x \right) = \frac{{\left| {f''\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{{\left| { - {{\sec }^2}x} \right|}}{{{{\left( {1 + {{\left( { - \tan x} \right)}^2}} \right)}^{3/2}}}} \cr
& {\text{Simplifying}} \cr
& \kappa \left( x \right) = \frac{{{{\sec }^2}x}}{{{{\left( {1 + {{\tan }^2}x} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{{{{\sec }^2}x}}{{{{\left( {{{\sec }^2}x} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{1}{{\sec x}} \cr
& \kappa \left( x \right) = \cos x \cr} $$
