Area bounded by the curves y = tan x, and y = tan²x in between $x∈(-\frac{π}{3},\frac{π}{3})$ is equal to:

$(\frac{-π}{6}+ln\,2+2\sqrt{3}-2)$sq.units

$\int\limits_{-π/3}^{π/3}|tan^2x-tanx|dx=\int\limits_{-π/3}^{0}(tan^2x-tanx)dx+\int\limits_{0}^{π/4}(tanx-tan^2x)dx+\int\limits_{π/4}^{π/3}(tan^2x-tanx)dx$

As $\int(tan^2x-tanx)dx=(tanx-x-log|sec x|)$

Required area = $-\frac{\pi}{6}+\log(2)+2(\sqrt{3}-1)$