Calculus: Early Transcendentals 9th Edition

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

Chapter 7 - Section 7.3 - Trigonometric Substitution - 7.3 Exercises - Page 505: 4

Answer

$$\eqalign{ & \left( {\text{a}} \right)x = \frac{3}{2}\sin \theta \cr & \left( {\text{b}} \right) - \frac{9}{{16}}\int {\left( {{u^{ - 2}} - 1} \right)} du \cr} $$

Work Step by Step

$$\eqalign{ & \int {\frac{{{x^3}}}{{{{\left( {9 - 4{x^2}} \right)}^{3/2}}}}} dx \cr & {\text{We have the expression }}9 - 4{x^2} = {a^2} - {u^2},{\text{ }} \cr & {\text{set }}x = \frac{3}{2}\sin \theta \to dx = \frac{3}{2}\cos \theta d\theta \cr & {\text{Substituting}} \cr & \int {\frac{{{x^3}}}{{{{\left( {9 - 4{x^2}} \right)}^{3/2}}}}} dx = \int {\frac{{{{\left( {\frac{3}{2}\sin \theta } \right)}^3}}}{{{{\left( {9 - 4{{\left( {\frac{3}{2}\sin \theta } \right)}^2}} \right)}^{3/2}}}}} \left( {\frac{3}{2}\cos \theta } \right)d\theta \cr & {\text{Simplify the integrand}} \cr & = \frac{{81}}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\left( {9 - \frac{9}{4}{{\sin }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr & = \frac{{81}}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\left( {9 - 9{{\sin }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr & = \frac{9}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr & {\text{Use the identity si}}{{\text{n}}^2}\theta + {\cos ^2}\theta = 1 \cr & = \frac{9}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr & = \frac{9}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}} \left( {\cos \theta } \right)d\theta \cr & = \frac{9}{{16}}\int {\frac{{{{\sin }^3}\theta }}{{{{\cos }^2}\theta }}} d\theta \cr & {\text{Split }}{\sin ^3}\theta {\text{ as }}{\sin ^2}\theta \sin \theta \cr & = \frac{9}{{16}}\int {\frac{{{{\sin }^2}\theta \sin \theta }}{{{{\cos }^2}\theta }}} d\theta \cr & {\text{Use the identity si}}{{\text{n}}^2}\theta + {\cos ^2}\theta = 1 \cr & = \frac{9}{{16}}\int {\frac{{\left( {1 - {{\cos }^2}\theta } \right)\sin \theta }}{{{{\cos }^2}\theta }}} d\theta \cr & = \frac{9}{{16}}\int {\left( {\frac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \left( {\sin \theta } \right)d\theta \cr & = \frac{9}{{16}}\int {\left( {\frac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \left( {\sin \theta } \right)d\theta \cr & {\text{Let }}u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr & = - \frac{9}{{16}}\int {\left( {\frac{{1 - {u^2}}}{{{u^2}}}} \right)} du \cr & = - \frac{9}{{16}}\int {\left( {{u^{ - 2}} - 1} \right)} du \cr & {\text{This integral can be solved using the power rule}}{\text{, then}} \cr & {\text{an appropiate substitution is }}x = \frac{3}{2}\sin \theta \cr & \cr & \left( {\text{a}} \right)x = \frac{3}{2}\sin \theta \cr & \left( {\text{b}} \right) - \frac{9}{{16}}\int {\left( {{u^{ - 2}} - 1} \right)} du \cr} $$
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