Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 46

Answer

$$\frac{7}{6}$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /6} {{{\sec }^3}2\theta } \tan 2\theta d\theta \cr & {\text{Write the integrand as}} \cr & = \int_0^{\pi /6} {{{\sec }^{2 + 1}}2\theta } \tan 2\theta d\theta \cr & = \int_0^{\pi /6} {{{\sec }^2}2\theta } \sec 2\theta \tan 2\theta d\theta \cr & = \frac{1}{2}\int_0^{\pi /6} {{{\sec }^2}2\theta } \left[ {\sec 2\theta \tan 2\theta \left( 2 \right)} \right]d\theta \cr & {\text{Integrate}} \cr & = \frac{1}{2}\left( {\frac{{{{\sec }^3}2\theta }}{3}} \right)_0^{\pi /6} \cr & = \frac{1}{2}\left( {\frac{{{{\sec }^3}2\left( {\pi /6} \right)}}{3} - \frac{{{{\sec }^3}2\left( 0 \right)}}{3}} \right) \cr & = \frac{1}{2}\left( {\frac{{{{\sec }^3}\left( {\pi /3} \right)}}{3} - \frac{{{{\sec }^3}\left( 0 \right)}}{3}} \right) \cr & = \frac{1}{2}\left( {\frac{8}{3} - \frac{1}{3}} \right) \cr & = \frac{1}{2}\left( {\frac{7}{3}} \right) \cr & = \frac{7}{6} \cr} $$
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