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: 69

Answer

$$\pi $$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^0 {\frac{{6dt}}{{\sqrt {3 - 2t - {t^2}} }}} \cr & {\text{complete the square for }}3 - 2t - {t^2} \cr & 3 - 2t - {t^2} = 3 - \left( {{t^2} + 2t + 1} \right) + 1 \cr & = 4 - {\left( {t + 1} \right)^2} \cr & {\text{then}} \cr & \int_{ - 1}^0 {\frac{{6dt}}{{\sqrt {3 - 2t - {t^2}} }}} = \int_{ - 1}^0 {\frac{{6dt}}{{\sqrt {4 - {{\left( {t + 1} \right)}^2}} }}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = t + 1,{\text{ so that }}du = dt \cr & {\text{the new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}t = 0,{\text{ then }}u = 1 + 1 = 1 \cr & \,\,\,\,\,\,{\text{If }}t = - 1,{\text{ then }}u = - 1 + 1 = 0 \cr & {\text{then}} \cr & \int_{ - 1}^0 {\frac{{6dt}}{{\sqrt {4 - {{\left( {t + 1} \right)}^2}} }}} = \int_0^1 {\frac{{6du}}{{\sqrt {4 - {u^2}} }}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 2 \cr & = 6\left( {{{\sin }^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_0^1 \cr & = 6{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) - 6{\sin ^{ - 1}}\left( {\frac{0}{2}} \right) \cr & = \pi \cr} $$
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.