University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 41


$$\int\frac{\cos ydy}{\sin^2y+\sin y-6}=\frac{1}{5}\ln|\sin y-2|-\frac{1}{5}\ln|\sin y+3|+C$$

Work Step by Step

$$I=\int\frac{\cos ydy}{\sin^2y+\sin y-6}$$ Set $u=\sin y$, which means $$du=\cos ydy$$ We then have, $$I=\int\frac{du}{u^2+u-6}$$ $$I=\int\frac{du}{(u-2)(u+3)}$$ $$I=\frac{1}{5}\int\frac{5}{(u-2)(u+3)}du$$ $$I=\frac{1}{5}\int\frac{(u+3)-(u-2)}{(u-2)(u+3)}du$$ $$I=\frac{1}{5}\int\frac{1}{u-2}du-\frac{1}{5}\int\frac{1}{u+3}du$$ $$I=\frac{1}{5}\ln|u-2|-\frac{1}{5}\ln|u+3|+C$$ $$I=\frac{1}{5}\ln|\sin y-2|-\frac{1}{5}\ln|\sin y+3|+C$$
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