With trig sub it whittlles down nicely.
Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$
Make the subs and you get:
$\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$
$\displaystyle u^2 = x^2 - 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow dx = \frac{u}{x} \, du$
$\displaystyle x = 1 \Rightarrow u^2 = 0 \Rightarrow u = 0$. $\displaystyle x = 2 \Rightarrow u^2 = 3 \Rightarrow u = \sqrt{3}$ where the positive root is used.
Then:
$\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u}{x} \frac{u}{x} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{x^2} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du$
$\displaystyle = \int_{0}^{\sqrt{3}} \frac{(u^2 + 1) - 1}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$,
which integrates easily to give $\displaystyle I = \sqrt{3} - \frac{\pi}{3}$.