## Calculus: Early Transcendentals 8th Edition

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