Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 499: 46

Answer

$\frac{1}{2}\sin \left( {{t^2}} \right) - \frac{1}{3}{\sin ^3}\left( {{t^2}} \right) + \frac{1}{{10}}{\sin ^5}\left( {{t^2}} \right) + C$

Work Step by Step

$$\eqalign{ & \int {t{{\cos }^5}\left( {{t^2}} \right)} dt \cr & {\text{Split }}{\cos ^5}\left( {{t^2}} \right){\text{ as }}{\cos ^4}\left( {{t^2}} \right)\cos \left( {{t^2}} \right) \cr & \int {t{{\cos }^4}\left( {{t^2}} \right)\cos \left( {{t^2}} \right)} dt \cr & {\text{Use the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr & = \int {t{{\left[ {{{\cos }^2}\left( {{t^2}} \right)} \right]}^2}\cos \left( {{t^2}} \right)} dt \cr & {\text{Rewrite }}{\sin ^2}x{\text{ using }}{\sin ^2}x + {\cos ^2}x = 1 \cr & = \int {t{{\left[ {1 - {{\sin }^2}\left( {{t^2}} \right)} \right]}^2}\cos \left( {{t^2}} \right)} dt \cr & {\text{Expand the binomial}} \cr & = \int {t\left( {1 - 2{{\sin }^2}\left( {{t^2}} \right) + {{\sin }^4}\left( {{t^2}} \right)} \right)\cos \left( {{t^2}} \right)} dt \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \sin \left( {{t^2}} \right),{\text{ }}du = 2t\cos \left( {{t^2}} \right)dt \cr & = \int {\left( {1 - 2{u^2} + {u^4}} \right)\left( {\frac{1}{2}} \right)} du \cr & = \frac{1}{2}\int {\left( {1 - 2{u^2} + {u^4}} \right)} du \cr & {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr & = \frac{1}{2}\left( {u - \frac{2}{3}{u^3} + \frac{1}{5}{u^5}} \right) + C \cr & = \frac{1}{2}u - \frac{1}{3}{u^3} + \frac{1}{{10}}{u^5} + C \cr & {\text{Write in terms of }}t,{\text{ substitute }}\sin \left( {{t^2}} \right){\text{ for }}u \cr & = \frac{1}{2}\sin \left( {{t^2}} \right) - \frac{1}{3}{\sin ^3}\left( {{t^2}} \right) + \frac{1}{{10}}{\sin ^5}\left( {{t^2}} \right) + C \cr} $$
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