Answer
The arc length is
$$l=\ln|2+\sqrt{3}|.$$
Work Step by Step
The arc length is given by the formula
$$l=\int_{x_1}^{x_2}\sqrt{1+(y')^2}dx.$$
In this problem $x_1=0$ and $x_2=\pi/3$ and $y=\ln(\cos x)$ so we have
$$l=\int_0^{\pi/3}\sqrt{1+((\ln(\cos x))')^2}dx.$$
The derivative using chain rule is
$$(\ln(\cos x))'=\frac{1}{\cos x}(\cos x)'=\frac{1}{\cos x}(-\sin x) = -\tan x.$$
This gives
$$l=\int_0^{\pi/3}\sqrt{1+\tan^2x}dx =\int_0^{\pi/3}\sqrt{\frac{\cos^2x}{\cos^2 x}+\frac{\sin^2x}{\cos^2x}}dx=\int_0^{\pi/3}\sqrt{\frac{\sin^2x+\cos^2x}{\cos^2x}}dx.$$
We use $\sin^2x+\cos^2x=1$ to get
$$l=\int_0^{\pi/3}\sqrt{\frac{1}{\cos^2x}}dx.$$
In the region of integration the cosine is positive so $\sqrt{\cos^2 x} = \cos x$ and we have
$$l=\int_0^{\pi/3}\frac{1}{\cos x}dx=\int_0^{\pi/3}\sec xdx= \ln|\sec x+\tan x|\left.\right|_0^{\pi/3}.$$
Substituting the bounds of integration
$$l=\ln|\sec(\pi/3)+\tan(\pi/3)|-\ln|\sec0+\tan0|=\\ \ln|2+\sqrt{3}|-\ln1=\ln|2+\sqrt3|$$
