$\int\limits_0^{\pi /4} {2{{\tan }^3}xdx = 1 - \log 2}$

: $\int\limits_0^{\pi /4} {2{{\tan }^3}xdx} = \int\limits_0^{\pi /4} {2{\mathop{\rm tanx}\nolimits} \cdot {{\tan }^2}} dx$

$= \int\limits_0^{\pi /4} {2\tan x\left( {{{\sec }^2}x - 1} \right)dx} = 2\int\limits_0^{\pi /4} {\left( {\tan x} \right){{\sec }^2}xdx} - 2\int\limits_0^{\pi /4} {\tan x\,dx}$

$= 2\left[ {\cfrac{{{{\tan }^2}x}}{2}} \right]_0^{\pi /4} - 2\left[ { - \log \left| {\cos \,x} \right|} \right]_0^{\pi /4}$

$= \left( {{{\tan }^2}\cfrac{\pi }{4} - {{\tan }^2}0} \right) + 2\left( {\log \cos \cfrac{\pi }{4} - \log \cos 0} \right)$

$= \left( {1 - 0} \right) + 2\left( {\log \cfrac{1}{{\sqrt 2 }} - \log 1} \right)$

$= 1 + 2\left( {\log 1 - \log \sqrt 2 - \log 1} \right) = 1 + 2 \times \left( { - \cfrac{1}{2}\log 2} \right) = 1 - \log 2$