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October 24, 2014

October 24, 2014

Posted by **Shayna** on Wednesday, January 18, 2012 at 8:45pm.

- Calculus -
**Steve**, Thursday, January 19, 2012 at 4:48amLet 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.

- Calculus - PS -
**Steve**, Thursday, January 19, 2012 at 4:52amcopy-paste error. That last complicated line should read:

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

= θsecθ - ln|secθ + tanθ|

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