Answer
$ \ln(2+\sqrt{3})$
Work Step by Step
Given: $y = \ln{\cos x}$
$dy/dx = -\tan x$
and $1+(dy/dx)^{2} = 1+\tan^{2} x = \sec^{2} x$
Therefore,
$L = \int^{\frac{\pi}{3}}_{0} \sqrt{\sec^{2} x}dx $
$= \int^{\frac{\pi}{3}}_{0} \sec x dx$
$ = [\ln|\sec x +\tan x|]^{\frac{\pi}{3}}_{0}$
$ = \ln(2+\sqrt{3}) - \ln(1+0) $
$= \ln(2+\sqrt{3})$
