Question

Evaluate the integral `int (sec 2x - 1)/(sec 2x + 1) dx`?

Answer 1

`int \ (sec 2x - 1)/(sec 2x + 1) \ dx = tan x -x + c`

Explanation:

Let:

`I = int \ (sec 2x - 1)/(sec 2x + 1) \ dx`

We can simplify the denominator by using the definition of `secx`

`I = int \ (sec 2x - 1)/(sec 2x + 1) \ dx`
`\ \ = int \ (1/(cos 2x) - 1)/(1/(cos2x) + 1) \ dx`
`\ \ = int \ (1/(cos 2x) - 1)/(1/(cos2x) + 1) * (cos2x)/(cos2x) \ dx`
`\ \ = int \ (1-cos 2x)/(1+cos2x) \ dx`

Using the identity `cos2A -= cos^2 A - sin^2A` we then have:

`I = int \ (1+sin^2x-cos^2x)/(1+cos^2 x - sin^2x) \ dx`

Using the identity `sin^2A+cos^2A-=1` we then have:

`I = int \ (sin^2x+sin^2x)/(cos^2 x + cos^2x) \ dx`
`\ \ = int \ (2sin^2x)/(2cos^2x) \ dx`
`\ \ = int \ tan^2x \ dx`

We can now easily integrate by using the identity `tan^2A+1=sec^2A` to get:

`I = int \ sec^2x -1 \ dx`
`\ \ = tan x -x + c`