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