Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 56

Answer

$$\frac{{{{\csc }^{10}}x}}{{10}} - \frac{{{{\csc }^{12}}x}}{{12}} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\csc }^{10}}x{{\cot }^3}xdx} \cr & {\text{split the integrand}} \cr & = \int {{{\csc }^9}x{{\cot }^2}x\csc x\cot xdx} \cr & {\text{pythagorean identity }}{\cot ^2}x = {\csc ^2}x - 1 \cr & = \int {{{\csc }^9}x\left( {{{\csc }^2}x - 1} \right)\csc x\cot xdx} \cr & = \int {\left( {{{\csc }^{11}}x - {{\csc }^9}x} \right)\csc x\cot xdx} \cr & u = \csc x,{\text{ }}du = - \csc x\cot xdx \cr & = - \int {\left( {{u^{11}} - {u^9}} \right)du} \cr & {\text{by the power rule}} \cr & = \frac{{{u^{10}}}}{{10}} - \frac{{{u^{12}}}}{{12}} + C \cr & {\text{replace }}u = \csc x \cr & = \frac{{{{\csc }^{10}}x}}{{10}} - \frac{{{{\csc }^{12}}x}}{{12}} + C \cr} $$
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