Answer
$$\frac{1}{3}\left( {\sqrt 3 - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_{\pi /12}^{\pi /9} {{{\sec }^2}3\theta } d\theta \cr
& u = 3\theta ,{\text{ then }}du = 3d\theta dx,{\text{ }} \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \pi /9,{\text{ }}u = 3\left( {\pi /9} \right) = \pi /3 \cr
& {\text{if }}x = \pi /12,{\text{ }}u = 3\left( {\pi /12} \right) = \pi /4 \cr
& {\text{so}} \cr
& \int_{\pi /12}^{\pi /9} {{{\sec }^2}3\theta } d\theta = \int_{\pi /4}^{\pi /3} {se{c^2}u} \left( {\frac{1}{3}du} \right) \cr
& = \frac{1}{3}\int_{\pi /4}^{\pi /3} {se{c^2}u} du \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{3}\left( {\tan u} \right)_{\pi /4}^{\pi /3} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{3}\left( {\tan \left( {\frac{\pi }{3}} \right) - \tan \left( {\frac{\pi }{4}} \right)} \right) \cr
& = \frac{1}{3}\left( {\sqrt 3 - 1} \right) \cr} $$
