Integration by reduction I have learnt how to integrate by reduction formula but this one seems to give me hell someone to lift me by telling me what to do or simply to solve it.
\begin{equation}
I_n=\int\sec^n x\,dx
\end{equation}

Write the integral as
\begin{align}
\int\sec^n x\,dx=\int\sec^{n-2}x\sec^2  x\,dx
\end{align}
Then use integration by parts. Let
\begin{align}
u&=\sec^{n-2}x\\
du&=(n-2)\sec^{n-3}x\sec x\tan x\,dx=(n-2)\sec^{n-2}x\tan x\,dx\\
\end{align}
and
\begin{align}
dv&=\sec^{2}x\\
v&=\int\sec^{2}x\,dx=\int\,d(\tan x)=\tan x\\
\end{align}
Hence
\begin{align}
\int\sec^{n-2}x\sec^2  x\,dx&=\sec^{n-2}x\tan x-(n-2)\int\tan x\sec^{n-2}x\tan x\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x\tan^2x\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x(\sec^2 x-1)\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n}x\,dx+(n-2)\int\sec^{n-2}x\,dx\\
I_n&=\sec^{n-2}x\tan x-(n-2)I_n+(n-2)I_{n-2}\\
I_n+(n-2)I_n&=\sec^{n-2}x\tan x+(n-2)I_{n-2}\\
(n-1)I_n&=\sec^{n-2}x\tan x+(n-2)I_{n-2}\\
I_n&=\frac{\sec^{n-2}x\tan x+(n-2)I_{n-2}}{n-1}
\end{align}
