Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {2\cos 3x} dx \cr
& u = 3x,{\text{ then }}du = 3dx,{\text{ }}\frac{1}{3}du = dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \pi /6,{\text{ }}u = 3\left( {\frac{\pi }{6}} \right) = \frac{\pi }{2} \cr
& {\text{if }}x = 0,{\text{ }}u = 3\left( 0 \right) = 0 \cr
& {\text{so}} \cr
& \int_0^{\pi /6} {2\cos 3x} dx = \int_0^{\pi /2} {2\cos u} \left( {\frac{1}{3}du} \right) \cr
& = \frac{2}{3}\int_0^{\pi /2} {cosudu} \cr
& {\text{find the antiderivative}} \cr
& = \frac{2}{3}\left( {\sin u} \right)_0^{\pi /2} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{2}{3}\left( {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( 0 \right)} \right) \cr
& = \frac{2}{3}\left( {1 - 0} \right) \cr
& = \frac{2}{3} \cr} $$
