Answer
$$\int_{0}^{\pi / 4} 6 \tan ^{3} x d x = 3-3\ln \left(2\right) $$
Work Step by Step
$$
\int_{0}^{\pi / 4} 6 \tan ^{3} x d x
$$
Since $$ 1+\tan^2x=\sec^2 x$$
Then
\begin{align*}
\int_{0}^{\pi / 4} 6 \tan ^{3} x d x&=\int_{0}^{\pi / 4} 6 \tan ^{2} x \tan x d x\\
&=\int_{0}^{\pi / 4} 6 (\sec ^{2} x-1) \tan x d x\\
&=\int_{0}^{\pi / 4} 6 (\sec ^{2} x\tan x-\tan x) d x\\
&=3\tan^2x-6\ln |\sec x|\bigg|_{0}^{\pi / 4}\\
&= 3-3\ln \left(2\right)
\end{align*}
