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$\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$


Let $I=\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x$

$I=2 \int_{0}^{\frac{\pi}{4}} \tan ^{2} x \tan x d x=2 \int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) \tan x d x$

$=2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x \tan x d x-2 \int_{0}^{\frac{\pi}{4}} \tan x d x$

$=2\left[\frac{\tan ^{2} x}{2}\right]_{0}^{\frac{\pi}{4}}+2[\log \cos x]_{0}^{\frac{\pi}{4}}$

$=1+2\left[\log \cos \frac{\pi}{4}-\log \cos 0\right]$

$=1+2\left[\log \frac{1}{\sqrt{2}}-\log 1\right]$

$=1-\log 2-\log 1=1-\log 2$

Hence, the given result is proved.

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