Answer
$$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
Work Step by Step
Given $$\int \sec ^{m} x d x =\int \sec ^{m-2} x\sec^2 x d x $$
Let
\begin{align*}
u&= \sec ^{m-2}x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sec^2 x d x\\
du&= (m-2)\sec^{m-2}x\tan x \ \ \ \ \ \ \ v=\tan x
\end{align*}
Then
\begin{align*}
\int \sec ^{m} x d x&= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m-2}x\tan^2 xdx\\
&= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m-2 }x(\sec^2 x-1)dx\\
&= \tan x \sec ^{m-2}x-(m-2)\int \sec^{m }x dx +(m-2)\int \sec^{m-2 }xdx\\
\end{align*}
Hence
$$\int \sec ^{m} x d x=\frac{\tan x \sec ^{m-2} x}{m-1}+\frac{m-2}{m-1} \int \sec ^{m-2} x d x$$
