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: 70

Answer

$$\frac{\pi }{2}$$

Work Step by Step

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