Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 531: 58

$$\int_{0}^{\pi / 3} \tan ^{2} x d x =\sqrt{3}-\pi /3$$

$$ \int_{0}^{\pi / 3} \tan ^{2} x d x $$ Since $$1+\tan^2x=\sec^2x$$ Then \begin{align*} \int_{0}^{\pi / 3} \tan ^{2} x d x&=\int_{0}^{\pi / 3}(\sec^2x-1) d x\\ &=(\tan x -x)\bigg|_{0}^{\pi / 3}\\ &=(\tan (\pi/3) -(\pi/3))- (\tan (0) -(0))\\ &=\sqrt{3}-\pi /3 \end{align*}