Question

Solve
$\int {\sec }^{3}xdx$.

Answer 1

$I={\int }^{}{\sec }^{3}xdx$

$={\int }^{}\sec x.{\sec }^{2}xdx$

$=\sec x.\tan x-{\int }^{}\sec x\tan x.\tan xdx$

$=\sec x.\tan x-{\int }^{}\sec x.{\tan }^{2}xdx$

$=\sec x.\tan x-{\int }^{}\left({\sec }^{3}x-\sec x\right)dx$

$=\sec x.\tan x-I+{\int }^{}\sec xdx$

$=\sec x.\tan x-I+\log \left|\sec x+\tan x\right|$

$2I=\sec x.\tan x+\log \left|\sec x+\tan x\right|+C$

We get,

$I=\frac{1}{2}\sec x.\tan x+\frac{1}{2}\log \left|\sec x+\tan x\right|+C$