Find $\int {\sec }^{3}x \mathrm{dx}$

Let $\mathrm{I}=\int {\sec }^3\mathrm{xdx}$
$\begin{array}{l}=\int \mathrm{secx} .{\sec }^2 \mathrm{x}\cdot \mathrm{dx}\\ =\sec x\int {\sec }^2 \mathrm{x}\cdot \mathrm{dx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}(\sec x)\int {\sec }^2 \mathrm{x}\cdot \mathrm{dx}\right]\cdot \mathrm{dx}\\ =\sec \mathrm{xtan} x-\int (\sec \mathrm{xtan} x)(\tan x)\cdot \mathrm{dx}\\ =\sec \mathrm{xtan} x-\int \sec {\mathrm{xtan}}^2 \mathrm{x}\cdot \mathrm{dx}\\ =\sec \mathrm{xtan} x-\int \sec x\left({\sec }^2 \mathrm{x}-1\right)\cdot \mathrm{dx}\\ =\sec \mathrm{xtan} x-\int {\sec }^3 \mathrm{x}\cdot \mathrm{dx}+\int \sec x\cdot \mathrm{dx}\end{array}$
$\begin{array}{l}\therefore \mathrm{I}=\sec \mathrm{xtan} x-I+\log |\sec x+\tan x|\\ \therefore 2\mathrm{I}=\sec \mathrm{xtan} x+\log |\sec x+\tan x|\\ \therefore \mathrm{I}=\frac{1}{2}[\sec \mathrm{xtan} x+\log |\sec x+\tan x\left|\right]+\mathrm{C}\end{array}$
Therefore, $\int {\sec }^3 \mathrm{xdx}=\frac{1}{2}[\sec \mathrm{xtan} \mathrm{x}+\log |\sec \mathrm{x}+\tan \mathrm{x}\left|\right]+\mathrm{C}$