Prove that $\int_{0}^{\frac{\pi}{4}} 2 \tan^{3} x \, dx = 1 - \log 2$.

(N/A) Let $I = \int_{0}^{\frac{\pi}{4}} 2 \tan^{3} x \, dx$.
We can write $\tan^{3} x$ as $\tan x (\sec^{2} x - 1)$.
So,$I = 2 \int_{0}^{\frac{\pi}{4}} \tan x (\sec^{2} x - 1) \, dx$.
$I = 2 \int_{0}^{\frac{\pi}{4}} \tan x \sec^{2} x \, dx - 2 \int_{0}^{\frac{\pi}{4}} \tan x \, dx$.
For the first integral,let $u = \tan x$,then $du = \sec^{2} x \, dx$. When $x = 0, u = 0$ and when $x = \frac{\pi}{4}, u = 1$.
$2 \int_{0}^{1} u \, du = 2 \left[ \frac{u^{2}}{2} \right]_{0}^{1} = 1$.
For the second integral,$\int \tan x \, dx = \log |\sec x| = -\log |\cos x|$.
So,$-2 \int_{0}^{\frac{\pi}{4}} \tan x \, dx = 2 [\log |\cos x|]_{0}^{\frac{\pi}{4}} = 2 (\log \frac{1}{\sqrt{2}} - \log 1) = 2 (\log 2^{-1/2} - 0) = 2 \times (-\frac{1}{2}) \log 2 = -\log 2$.
Combining these,$I = 1 - \log 2$.
Hence,the result is proved.