Answer
$\approx{1.317}$
Work Step by Step
The formula for the arc length is
$s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Differentiate $y = \ln{\cos{x}}$ with respect to x.
$y' = \frac{-\sin{x}}{\cos{x}}$
$=-\tan{x}$
Substitute the value of y' in $1 + (y')^2$
$1+(y')^2=1+{\tan{^2}{x}}$
$=1+{\tan{^2}{x}}$
$=\sec{^2}{x}$
Substitute the value of $1 + (y')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Solve for the distance s
$s=\int_{0}^\frac{\pi}{3}\sqrt{\sec{^2}{x}} dx$
$=[\ln\lvert \sec{x}+\tan{x} \rvert]_2^5\ $
$\approx{1.317}$
