prove that the integral of arcsec(x) is equal to xarcsec(x) - ln|x+ sqrt((x^2)-1)

Answer 1

Let x = secθ
dx = secθtanθ dθ
θ = arcsec x

Int(arcsec(x) dx)
= Int(θ secθtanθ dθ)

now integrate by parts

u = θ
du = dθ

dv = secθtanθ dθ
v = secθ

Int(θ secθtanθ) = θsecθ - Int(secθ dθ)

To integrate secθ you have to resort to a trick:

secθ (secθ + tanθ)/(secθ + tanθ) dθ

now the top is sec^2θ + secθtanθ
let u = secθ + tanθ and we have

1/u du

so, Int(secθ dθ) is ln|secθ + tanθ|
and we end up with

Int(θ secθtanθ) = θsecθ - ln|secInt(θ secθtanθ) = θsecθ - Int(secθ dθ) + tanInt(θ secθtanθ)
= θsecθ - Int(secθ dθ)|

Now, what does all that equal in x's?

θ = arcsec(x)
secθ = x
tanθ = √(x^2-1)

and you have your answer.

Answer 2

copy-paste error. That last complicated line should read:

Int(θ secθtanθ) = θsecθ - Int(secθ dθ)
= θsecθ - ln|secθ + tanθ|