$$\\int_{0}^{\\frac{\\pi}{6}}\\tan^{3}\\phi\\sec^{2}\\phi d\\phi$$

Explanation:

Step1: Use substitution

Let (u = \tan\phi), then (du=\sec^{2}\phi d\phi). When (\phi = 0), (u=\tan(0) = 0); when (\phi=\frac{\pi}{6}), (u=\tan(\frac{\pi}{6})=\frac{\sqrt{3}}{3}). The integral (\int_{0}^{\frac{\pi}{6}}\tan^{3}\phi\sec^{2}\phi d\phi) becomes (\int_{0}^{\frac{\sqrt{3}}{3}}u^{3}du).

Step2: Integrate (u^{3})

Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), for (n = 3), (\int u^{3}du=\frac{u^{4}}{4}+C). Evaluating (\left[\frac{u^{4}}{4}\right]_{0}^{\frac{\sqrt{3}}{3}}).

Step3: Substitute the limits

(\frac{(\frac{\sqrt{3}}{3})^{4}}{4}-\frac{0^{4}}{4}). Since ((\frac{\sqrt{3}}{3})^{4}=\frac{9}{81}=\frac{1}{9}), then (\frac{1}{9\times4}-0).

Answer:

(\frac{1}{36})