Answer
$$3 \tan ^{4} x-2 \tan ^{3} x+C$$
Work Step by Step
Given $$\int \sec ^{2}(x) \cdot\left(12 \tan ^{3} x-6 \tan ^{2} x\right) d x $$
Let
$$ u=\tan x \ \ \ \ \Rightarrow \ \ \ du = \sec^2 x dx$$
Then
\begin{aligned} \int \sec ^{2}(x) \cdot\left(12 \tan ^{3} x-6 \tan ^{2} x\right) d x &=\int\left(12 u^{3}-6 u^{2}\right) d u \\ &=3 u^{4}-2 u^{3}+C \\ &=3 \tan ^{4} x-2 \tan ^{3} x+C \end{aligned}
