Answer
$\displaystyle \frac{1}{\pi}+\frac{6}{\pi}\ln\frac{\sqrt{3}}{2}$
Work Step by Step
Formula 69.
$\displaystyle \int\tan^{3}udu=\frac{1}{2}\tan^{2}u+\ln|\cos u|+C$
---
$\displaystyle \int_{0}^{1}\tan^{3}(\frac{\pi}{6}x)dx=\quad \left[\begin{array}{ll}
u =(\pi/6)x, & du =(\pi/6)dx\\
x=0\rightarrow u=0, & x=1\rightarrow u=\pi/6
\end{array}\right]$
$=\displaystyle \frac{6}{\pi}\int_{0}^{\pi/6}\tan^{3}udu \quad ... $apply formula 69, .
$=\displaystyle \frac{6}{\pi}[\frac{1}{2}\tan^{2}u+\ln|\cos u|]_{0}^{\pi/6}$
$=\displaystyle \frac{6}{\pi}[(\frac{1}{2}(\frac{1}{\sqrt{3}})^{2}+\ln\frac{\sqrt{3}}{2})-(0+\ln 1)]$
$=\displaystyle \frac{1}{\pi}+\frac{6}{\pi}\ln\frac{\sqrt{3}}{2}$
