Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 83

Answer

(a) By using the method of differentiating right side of the equation , we conclude that $\int cotxdx=ln|sinx|+C$. (b) By using the method of substitution, we prove that $\int cotxdx=ln|sinx|+C$.

Work Step by Step

Show that $\int cotxdx=ln|sinx|+C$ (a) By using the method of differentiating right side of the equation as follows: Since, $\frac{d}{dx}[A(x)+c]=B(x)$ Then $\int B(x) =A(x)+C$ This implies $[\frac{d}{dx}ln|sinx|+C]=\frac{d}{dx}ln|sinx|+\frac{d}{dx}C$ $=\frac{1}{sinx}.(cosx)+0$ $=cotx$ Therefore, by using the method of differentiating right side of the equation , we conclude that $\int cotxdx=ln|sinx|+C$. (b) By using the method of substitution. For this, first we write $cotx$ in terms of sine and cosine such as: $\int cotx dx=\int\frac{cosx}{sinx}dx$ This suggests that we should substitute $u=sinx$, since then $du=cosxdx$ Thus, $\int cotx dx=\int\frac{cosx}{sinx}dx=\int\frac{1}{u}du$ This implies $ln|u|+C=ln|sinx|c$ This result can also be written as $\int cotxdx=ln|sinx|+C$ Hence, the result is proved.
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