Solution:

Assume $I=\int \sec ^{6} x \tan x d x$
We can write the above equation as
$\Rightarrow I=\int \sec ^{5} x(\sec x \tan x) d x$
Substituting, $\sec x=t \Rightarrow \sec x \tan x d x=d t$, we get
$\Rightarrow \mathrm{I}=\int \mathrm{t}^{5} \mathrm{dt}$
On integrating we obtain
$\Rightarrow \mathrm{I}=\frac{\mathrm{t}^{6}}{6}+\mathrm{c}$
Substituting the values of $t$ we obtain
$\Rightarrow \mathrm{I}=\frac{\sec ^{6} \mathrm{x}}{6}+\mathrm{c}$
Hence, $\int \sec ^{5} x(\sec x \tan x) d x=\frac{\sec ^{6} x}{6}+c$