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 498: 16

Answer

$ - \frac{1}{4}{\csc ^4}\theta + \frac{1}{2}{\csc ^2}\theta + C$

Work Step by Step

$$\eqalign{ & \text{Let }I=\int {{{\csc }^5}\theta } {\cos ^3}\theta d\theta \cr & {\text{Use the reciprocal identity }}\csc x = \frac{1}{{\sin x}} \cr & I = \int {\left( {\frac{1}{{{{\sin }^5}\theta }}} \right)} {\cos ^3}\theta d\theta \cr & {\text{Split }}{\cos ^3}\theta {\text{ as }}{\cos ^2}\theta \cos \theta \cr & I = \int {\left( {\frac{1}{{{{\sin }^5}\theta }}} \right)} {\cos ^2}\theta \cos \theta d\theta \cr & {\text{Rewrite }}{\cos ^2}\theta {\text{ using }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & I = \int {\left( {\frac{1}{{{{\sin }^5}\theta }}} \right)} \left( {1 - {{\sin }^2}\theta } \right)\cos \theta d\theta \cr & = \int {\left( {\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^5}\theta }}} \right)} \cos \theta d\theta \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \sin \theta ,{\text{ }}du = \cos \theta d\theta \cr & {\text{Substituting}} \cr & I = \int {\left( {\frac{{1 - {u^2}}}{{{u^5}}}} \right)} du \cr & = \int {\left( {\frac{1}{{{u^5}}} - \frac{{{u^2}}}{{{u^5}}}} \right)} du \cr & = \int {\left( {{u^{ - 5}} - {u^{ - 3}}} \right)} du \cr & {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr & I = \frac{{{u^{ - 4}}}}{{ - 4}} - \frac{{{u^{ - 2}}}}{{ - 2}} + C \cr & = - \frac{1}{{4{u^4}}} + \frac{1}{{2{u^2}}} + C \cr & {\text{Write in terms of }}\theta ,{\text{ substitute }}\sin \theta {\text{ for }}u \cr & I = - \frac{1}{{4{{\sin }^4}\theta }} + \frac{1}{{2{{\sin }^2}\theta }} + C \cr & = - \frac{1}{4}{\csc ^4}\theta + \frac{1}{2}{\csc ^2}\theta + C \cr} $$
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