Chọn A

Ta có:

$\int_{0}^{\frac{\pi }{4}}{{{\tan }^{3}}xdx}=\int_{0}^{\frac{\pi }{4}}{\left[ \tan x\left( 1+{{\tan }^{2}}x \right)-\tan x \right]dx}=\int_{0}^{\frac{\pi }{4}}{\tan xd\tan x}-\int_{0}^{\frac{\pi }{4}}{\tan xdx}$

$=\left. \frac{{{\left( \tan x \right)}^{2}}}{2} \right|_{0}^{\frac{\pi }{4}}+\int_{0}^{\frac{\pi }{4}}{\frac{d\cos x}{\cos x}}=\left. \frac{1}{2}+\ln \left| \cos x \right| \right|_{0}^{\frac{\pi }{4}}=\frac{1}{2}+\ln \frac{1}{\sqrt{2}}=\frac{1}{2}\left( 1-\ln 2 \right)$

Vậy $\int_{0}^{\frac{\pi }{4}}{{{\tan }^{3}}xdx}=\frac{1}{2}\left( 1-\ln 2 \right)$