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 - 7.4 Trigonometric Substitutions - Exercises Set 7.4 - Page 513: 20

Answer

$${\sin ^{ - 1}}\left( {\frac{{\sin \theta }}{2}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\cos \theta }}{{\sqrt {2 - {{\sin }^2}\theta } }}} d\theta \cr & \cr & {\text{substitute }}t = \sin \theta ,{\text{ }}dt = \sin \theta d\theta \cr & \int {\frac{{\cos \theta }}{{\sqrt {2 - {{\sin }^2}\theta } }}} d\theta = \int {\frac{{dt}}{{\sqrt {2 - {t^2}} }}} \cr & \cr & {\text{Let sin}}\alpha = \frac{t}{{\sqrt 2 }},{\text{ }}dt = \sqrt 2 \cos \alpha d\alpha \cr & \int {\frac{{dt}}{{\sqrt {2 - {t^2}} }}} = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt {2 - 2{{\sin }^2}\alpha } }}} \cr & = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt 2 \sqrt {1 - {{\sin }^2}\alpha } }}} \cr & \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt 2 \sqrt {{{\cos }^2}\alpha } }}} \cr & = \int {d\alpha } \cr & = \alpha + C \cr & \cr & {\text{Where }}\alpha = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) \cr & = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) + C \cr & {\text{ replace }}t = \sin \theta \cr & = {\sin ^{ - 1}}\left( {\frac{{\sin \theta }}{2}} \right) + C \cr} $$
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