# Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9 - Page 341: 31

$$\frac{1}{3}\left( {\sqrt 3 - 1} \right)$$

#### 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}

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