Write a value

let $\tan x=t$

Differentiating on both sides we get,

$\sec ^{2} x d x=d t$

Substituting above equation in $\int \tan ^{3} x \sec ^{2} x d x$ we get,

$=\int \mathrm{t}^{3} \mathrm{dt}$

$=\frac{t^{4}}{4}+c$

$=\frac{\tan ^{4} x}{4}+c$