Answer
$${\sqrt {1 + 36{{\sin }^2}3x} dx}$$
Work Step by Step
$$\eqalign{
& y = 2\cos 3x{\text{ on }}\left[ { - \pi ,\pi } \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = 2\cos 3x{\text{ and }}\left[ { - \pi ,\pi } \right] \to a = - \pi {\text{ and }}b = \pi .{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2\cos 3x} \right] \cr
& f'\left( x \right) = 2\left( { - 3\sin 3x} \right) \cr
& f'\left( x \right) = - 6\sin 3x \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_{ - \pi }^\pi {\sqrt {1 + {{\left( { - 6\sin 3x} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_{ - \pi }^\pi {\sqrt {1 + 36{{\sin }^2}3x} dx} \cr} $$
