Question

Integrate $\int {\sec }^{3}xdx$

Answer 1

Consider the given integral.

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

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

We know that,

$\int u.v=u\int vdx-\int (\frac{d}{dx}\left(u\right).\int vdx)dx$

Therefore,

$I=\sec x\tan x-\int \sec x\tan x\tan xdx$

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

$I=\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx$

$I=\sec x\tan x-\int ({\sec }^{3}x-\sec x)dx$

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

$I=\sec x\tan x-I+\int \sec xdx$

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

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

$I=\frac{1}{2}[\sec x\tan x+\log |\sec x+\tan x|]+C$

Hence, the value of integral is $.\frac{1}{2}[\sec x\tan x+\log |\sec x+\tan x|]+C$.