Answer
$\displaystyle \frac{1}{4}\tan x\sec^{3}x+\frac{3}{8}\tan x\sec x+\frac{3}{8}\ln|\sec x+\tan x|+C$
Work Step by Step
$\displaystyle \int\sec^{5}xdx=$
Table of integrals:
$\color{blue}{ 77. \quad \displaystyle \int\sec^{n}udu=\frac{1}{n-1}\tan u\sec^{n-2}u+\frac{n-2}{n-1}\int\sec^{n-2}udu }$
$=\displaystyle \frac{1}{4}\tan x\sec^{3}x+\frac{3}{4}\int\sec^{3}xdx=$
... apply formula 77 again
$=\displaystyle \frac{1}{4}\tan x\sec^{3}x+\frac{3}{4}(\frac{1}{2}\tan x\sec x+\frac{1}{2}\int\sec xdx)$
... Table formula 14 for the integral....
$=\displaystyle \frac{1}{4}\tan x\sec^{3}x+\frac{3}{8}\tan x\sec x+\frac{3}{8}\ln|\sec x+\tan x|+C$
