Prove

$\tan ^{4} x$

$=\tan ^{2} x \cdot \tan ^{2} x$

$=\left(\sec ^{2} x-1\right) \tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-\tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-\left(\sec ^{2} x-1\right)$

$=\sec ^{2} x \tan ^{2} x-\sec ^{2} x+1$

$\begin{aligned} \therefore \int \tan ^{4} x d x &=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 \cdot d x \\ &=\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+\mathrm{C} \end{aligned}$    ...(1)

Consider $\int \sec ^{2} x \tan ^{2} x d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

From equation (1), we obtain

$\int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+\mathrm{C}$