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: 67

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt { - {x^2} + 4x - 3} }}} \cr & {\text{complete the square for }} - {x^2} + 4x - 3 \cr & - {x^2} + 4x - 3 = - 3 - \left( {{x^2} - 4x} \right) \cr & = - 3 - \left( {{x^2} - 4x + 4} \right) + 4 \cr & = 1 - \left( {{x^2} - 4x + 4} \right) \cr & = 1 - {\left( {x - 2} \right)^2} \cr & {\text{then}} \cr & = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x - 2} \right)}^2}} }}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = x - 2,{\text{ so that }}du = dx \cr & \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x - 2} \right)}^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {u^2}} }}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 1 \cr & = {\sin ^{ - 1}}\left( {\frac{u}{1}} \right) + C \cr & {\text{write in terms of }}y;{\text{ replace }}x - 2{\text{ for }}u \cr & = {\sin ^{ - 1}}\left( {x - 2} \right) + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.