Answer
$$\left. a \right)L = \int_0^\pi {\sqrt {1 + 4{{\sin }^2}2x} } dx,\,\,\,\,\left. b \right)\,\,L \approx 5.27$$
Work Step by Step
$$\eqalign{
& y = \cos 2x{\text{ on the interval }}\left[ {0,\pi } \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ }} \cr
& {\text{The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = \ln x{\text{ and }} \cr
& \left[ {0,\pi } \right] \to a = 0{\text{ and }}b = \pi .{\text{ Then}} \cr
& \cr
& \left. a \right)\,\,f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x} \right] \cr
& f'\left( x \right) = - 2\sin 2x \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_0^\pi {\sqrt {1 + {{\left( { - 2\sin 2x} \right)}^2}} } dx \cr
& L = \int_0^\pi {\sqrt {1 + 4{{\sin }^2}2x} } dx \cr
& \cr
& \left. b \right){\text{ Use technology to evaluate or approximate the integral}} \cr
& L \approx 1.08 \cr} $$
