Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 73

Answer

$\displaystyle \frac{12}{\pi}\ln(2+\sqrt{3})$

Work Step by Step

$A=\displaystyle \int_{a}^{b}f(x)dx$ $A=2\displaystyle \int_{0}^{2}\sec(\frac{\pi x}{6})dx$ Find the indefinite integral. Use the table on page 333. $\displaystyle \int\sec(\frac{\pi x}{6})dx=\left[\begin{array}{ll} u=\frac{\pi}{6}x & \\ du=\frac{\pi}{6}dx & dx=\frac{6}{\pi}du \end{array}\right]=\displaystyle \frac{6}{\pi}\int\sec udu$ $=\displaystyle \frac{6}{\pi}\ln|\sec u+\tan u|+C$ $=\displaystyle \frac{6}{\pi}\ln|\sec(\frac{\pi}{6}x)+\tan(\frac{\pi}{6}x)|+C$ $A=2\left[\frac{6}{\pi}\ln|\sec(\frac{\pi}{6}x)+\tan(\frac{\pi}{6}x)|\right]_{0}^{2}$ $=\displaystyle \frac{12}{\pi}[ \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi}{3})| - \ln|\sec(0)+\tan(0)|]$ $=\displaystyle \frac{12}{\pi}[ \ln|2+\sqrt{3}| - \ln|1+0|]$ $= \displaystyle \frac{12}{\pi}\ln(2+\sqrt{3})$ Verified with online calculator (desmos.com):
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