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: 44

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {1 - 4{x^2}} }}} \cr & {\text{write }}1 - 4{x^2}{\text{ as }}1 - {\left( {2x} \right)^2} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = 2x,{\text{ so that }}du = 2dx,\,\,\,\,\,\,dx = \frac{{du}}{2} \cr & {\text{then}} \cr & \int {\frac{{dx}}{{\sqrt {1 - 4{x^2}} }}} = \int {\frac{{du/2}}{{\sqrt {1 - {u^2}} }}} \cr & = \frac{1}{2}\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr & {\text{intgrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 1 \cr & = \frac{1}{2}{\sin ^{ - 1}}\left( u \right) + C \cr & {\text{write in terms of }}x;{\text{ replace }}2x{\text{ for }}u \cr & = \frac{1}{2}{\sin ^{ - 1}}\left( {2x} \right) + C \cr} $$
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