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: 43

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} \cr & {\text{write }}9 - {x^2}{\text{ as }}{\left( 3 \right)^2} - {x^2} \cr & = \int {\frac{{dx}}{{\sqrt {{{\left( 3 \right)}^2} - {x^2}} }}} \cr & {\text{intgrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 3{\text{ and }}u = x,\,\,\,\,du = dx \cr & = \int {\frac{{dx}}{{\sqrt {{{\left( 3 \right)}^2} - {x^2}} }}} = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr} $$
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