[tex]y=\ln\cos x\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\sin x}{\cos x}=-\tan x[/tex]
[tex]\displaystyle\int_{\ell}\mathrm dS=\int_{x=0}^{x=\pi/3}\sqrt{1+(-\tan x)^2}\,\mathrm dx=\int_0^{\pi/3}\sqrt{\sec^2x}\,\mathrm dx[/tex]
Recall that [tex]\sqrt{x^2}=|x|[/tex], but since [tex]\sec x>0[/tex] over the integration domain, we can reduce [tex]|\sec x|=\sec x[/tex]. So we have
[tex]\displaystyle\int_0^{\pi/3}\sec x\,\mathrm dx=\ln|\sec x+\tan x|\bigg|_{x=0}^{x=\pi/3}=\ln\left|\sec\frac\pi3+\tan\frac\pi3\right|=\ln(2+\sqrt3)[/tex]
