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 502: 61

Answer

$ \int\frac{1}{3sinx -4cosx}dx=\frac{1}{5}\ln|2(tan\frac{x}{2})-1|-\frac{1}{5}\ln|tan\frac{x}{2}+2|+C$

Work Step by Step

$Ɪ=\int\frac{1}{3sinx -4cosx}dx$ Using the substitution $t=tan\frac{x}{2}$ then: $sinx =\frac{2t}{1+t^{2}}$ and $cosx=\frac{1-t^{2}}{1+t^{2}}$ and $dx=\frac{2}{1+t^{2}}dt$ then the integral becomes Ɪ=$\int\frac{1}{3(\frac{2t}{1+t^{2}}) -4(\frac{1-t^{2}}{1+t^{2}})}\frac{2}{1+t^{2}}dt$ $ Ɪ=\int\frac{1}{2t^{2}+3t-2}dt=\int\frac{1}{(2t-1)(t+2)}dt$ the integrand is a partial fraction can be written as: $\frac{1}{2t^{2}+3t-2}=\frac{A}{2t-1}+\frac{B}{t+2}$ Multiplying by $(2t-1)(t+2)$ $A(t+2)+B(2t-1)=1$ Or $2A-B=1$ and $A+2B=0$ solving the equation we found that $A=\frac{2}{5}$ and $B=\frac{-1}{5}$ then the integral is $ Ɪ=\int\frac{1}{(2t-1)(t+2)}dt=\frac{2}{5}\int\frac{dt}{2t-1}-\frac{1}{5}\int\frac{dt}{t+2}$ substituting $u=2t-1$ in the first integral we find $\int\frac{dt}{2t-1}=\frac{1}{2}\int\frac{1}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|2t-1|+C1$ the second integral is $\int\frac{dt}{t+2}=\ln|t+2|+C2$ then $ Ɪ=\frac{2}{5}\int\frac{dt}{2t-1}-\frac{1}{5}\int\frac{dt}{t+2}=\frac{1}{5}\ln|2t-1|-\frac{1}{5}\ln|t+2|+C$ since $t=tan\frac{x}{2}$ then $ Ɪ=\frac{1}{5}\ln|2(tan\frac{x}{2})-1|-\frac{1}{5}\ln|tan\frac{x}{2}+2|+C$
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