Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 51

Answer

$$\frac{1}{3} + \frac{1}{2}\ln \left( {\sqrt 3 } \right)$$

Work Step by Step

$$\eqalign{ & \int_0^{1/\sqrt 3 } {\sqrt {{x^2} + 1} dx} \cr & x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\tan ^{ - 1}}\left( x \right) \cr & x = 0,{\text{ }}\theta = 0 \cr & x = 1/\sqrt 3 ,{\text{ }}\theta = \pi /6 \cr & {\text{we have}} \cr & \int_0^{1/\sqrt 3 } {\sqrt {{x^2} + 1} dx} = \int_0^{\pi /6} {\sqrt {1 + {{\tan }^2}\theta } {{\sec }^2}\theta d\theta } \cr & {\text{pythagorean identity}} \cr & = \int_0^{\pi /6} {\sqrt {{{\sec }^2}\theta } {{\sec }^2}\theta d\theta } = \int_0^{\pi /6} {{{\sec }^3}\theta } d\theta \cr & {\text{integrating}} \cr & = \left[ {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right]_0^{\pi /6} \cr & {\text{evaluate limits}} \cr & = \left[ {\frac{1}{2}\sec \left( {\frac{\pi }{6}} \right)\tan \left( {\frac{\pi }{6}} \right) + \frac{1}{2}\ln \left| {\sec \left( {\frac{\pi }{6}} \right) + \tan \left( {\frac{\pi }{6}} \right)} \right|} \right] - \left( 0 \right) \cr & {\text{simplify}} \cr & = \left[ {\frac{1}{2}\left( {\frac{{2\sqrt 3 }}{3}} \right)\left( {\frac{{\sqrt 3 }}{3}} \right) + \frac{1}{2}\ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right|} \right] - \left( 0 \right) \cr & = \frac{1}{3} + \frac{1}{2}\ln \left( {\sqrt 3 } \right) \cr} $$
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