Write a value

given $\int \tan x \sec ^{3} x d x$

$=\int(\tan x \sec x) \sec ^{2} x d x$

Let $\sec x=t$

Differentiating on both sides we get,

$\tan x \sec x d x=d t$

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

$=\int t^{2} d t$

$=\frac{t^{3}}{3}+c$

$=\frac{\sec ^{3} x}{3}+c$