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.
$\displaystyle \int\sec(\frac{\pi x}{6})dx=\left[\begin{array}{ll}
u=\frac{\pi}{6}x & \\
du=\frac{\pi}{6}dx & dx=\frac{6}{\pi}du
\end{array}\right]=\displaystyle \frac{6}{\pi}\int\sec udu$
$=\displaystyle \frac{6}{\pi}\ln|\sec u+\tan u|+C$
$=\displaystyle \frac{6}{\pi}\ln|\sec(\frac{\pi}{6}x)+\tan(\frac{\pi}{6}x)|+C$
$A=2\left[\frac{6}{\pi}\ln|\sec(\frac{\pi}{6}x)+\tan(\frac{\pi}{6}x)|\right]_{0}^{2}$
$=\displaystyle \frac{12}{\pi}[ \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi}{3})| - \ln|\sec(0)+\tan(0)|]$
$=\displaystyle \frac{12}{\pi}[ \ln|2+\sqrt{3}| - \ln|1+0|]$
$= \displaystyle \frac{12}{\pi}\ln(2+\sqrt{3})$
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