Calculus
posted by Shayna .
prove that the integral of arcsec(x) is equal to xarcsec(x)  lnx+ sqrt((x^2)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 lnsecθ + tanθ
and we end up with
Int(θ secθtanθ) = θsecθ  lnsecInt(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^21)
and you have your answer. 
copypaste error. That last complicated line should read:
Int(θ secθtanθ) = θsecθ  Int(secθ dθ)
= θsecθ  lnsecθ + tanθ