Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 49

Answer

$$\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t=\ln \frac{ |\tan t+1|}{ |\tan t+2|}+C$$

Work Step by Step

Given $$\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t$$ Let \begin{array}{l}{ u=\tan t, \ \ \ d u=\sec ^{2} t \ d t . \text { Then }\\ \begin{aligned}I&=\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t\\ &=\int \frac{1}{u^{2}+3 u+2} d u\\ &=\int \frac{1}{(u+1)(u+2)} d u\end{aligned}} \\ {\text { Now: }\\ \frac{1}{(u+1)(u+2)}=\frac{A}{u+1}+\frac{B}{u+2} \Rightarrow 1=A(u+2)+B(u+1)} \\ {\text { Setting } u=-2\Rightarrow 1=-B, \Rightarrow B=-1}\\{ \text { Setting } u=-1\Rightarrow A=1} \\ {\text { Thus, } \\ \begin{aligned} I&=\int \frac{1}{(u+1)(u+2)} d u\\ &=\int\left(\frac{1}{u+1}-\frac{1}{u+2}\right) d u\\ &=\ln |u+1|-\ln |u+2|+C\\ &=\ln |\tan t+1|-\ln |\tan t+2|+C \\ &=\ln \frac{ |\tan t+1|}{ |\tan t+2|}+C\end{aligned}}\end{array}
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