It can be tedious to integrate the powers of the secant, especially for powers that are large. For \(k>2\). The reduction formula is \(\int \sec^kxdx=\frac{1}{k-1}\sec^{k-2}x\tan x+\frac{k-2}{k-1}\int \sec^{k-2}xdx\).

(a) Use the given reduction formula to find \(\int \sec^3xdx\) and \(\int \sec^7xdx\).

(b) Use integration by parts to prove the reduction formula. (Hint: Choose \(dv = sec^2x dx\).

Consider the integral \(\int \sec^7xdx\) and \(\int \sec^3xdx\).

Substitute \(k=7\) into reduction formula and yields,

\(\begin{align*}\int \sec^7xdx&=\frac{1}{7-1}\sec^{7-2}x\tan x+\frac{7-2}{7-1}\int\sec^{7-2}xdx\\&=\frac{1}{6}\sec^5x\tan x+\frac{5}{6}\left ( \frac{1}{4}\sec^3x\tan x+\frac{3}{4}\int\sec^3xdx \right )\\&=\frac{1}{6}\sec^5x\tan x+\frac{5}{24}\sec^3x\tan x+\frac{5}{8}\left ( \frac{1}{2}\sec x\tan x+\frac{1}{2}\int \sec xdx \right )\\&=\frac{1}{6}\sec^5x\tan x+\frac{5}{24}\sec^3x\tan x+\frac{5}{16}\sec x\tan x+\frac{5}{16}\log|\sec x+\tan x|+C\end{align*}\)

Substitute \(k=3\) into the reduction formula.

\(\begin{align*}\int \sec^3xdx&=\frac{1}{2}\sec x\tan x+\frac{1}{2}\int \sec xdx\\&=\frac{1}{2}\sec x\tan x+\frac{1}{2}\log|\sec x+\tan x|+C\end{align*}\).

Let \(I_k=\int \sec^kxdx\).

Rewrite the integral as \(I_k=\int \sec^{k-2}x\sec^2xdx\).

Using by part: \(\int uv'=uv-\int u'v\)

Let \(u=\sec^{k-2}x,v'=\sec^2x\)

\(u'=(k-2)\sec^{k-3}\sec x\tan x, v=\tan x\).

\(\begin{align*}&I_k=\int \sec^{k-2}x\sec^2xdx\\&I_k=\sec^{k-2}x\tan x-\int (k-2)\sec^{k-3}x\sec x\tan x\tan xdx\\&I_k=\sec^{k-2}x\tan x-(k-2)\int \sec^{k-2}\tan^2xdx\\&I_k=\sec^{k-2}x\tan x-(k-2)\int\sec^{k-2}(\sec^2x-1)dx\\&I_k=\sec^{k-2}x\tan x-(k-2)\int \sec^kdx+(k-2)\int\sec^{k-2}xdx\\&I_k=\sec^{k-2}x\tan x-(k-2)I_k+(k-2)\int\sec^{k-2}xdx\\I_k+(k-2)I_k&=\sec^{k-2}x\tan x+(k-2)\int \sec^{k-2}xdx\\(k-1)I_k&=\sec^{k-2}x\tan x+(k-2)\int \sec^{k-2}xdx\\&I_k=\frac{1}{k-1}\sec^{k-2}x\tan x+\frac{(k-2)}{(k-1)}\int \sec^{k-2}xdx \end{align*}\)