Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 83

Answer

$$\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^3}}}{3} + C $$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx}}{{\sqrt {1 - {x^2}} }}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = {\sin ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & {\text{then}} \cr & \int {\frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx}}{{\sqrt {1 - {x^2}} }}} = \int {{u^2}} du \cr & {\text{integrate by the power rule}} \cr & = \frac{{{u^3}}}{3} + C \cr & {\text{write in terms of }}x;{\text{ replace }}{\sin ^{ - 1}}x{\text{ for }}u \cr & = \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^3}}}{3} + C \cr} $$
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