I have tried two expressions.(version 10)

[1]

 Integrate[1/Sqrt[-1 + 2^2*Sech[x]^2], {x, 0, ArcCosh[2]}] 

$\left(\frac{1}{2}-i\right) \pi$

[2]

$Assumptions = {a > 1}; Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], {x, 0, ArcCosh[a]}] 

$\frac{\pi }{2}$

1. What difference does it make it?

2. The first computing makes the imaginary term. - pi i I don't know why it did such result?

(version 9)

 Integrate[1/Sqrt[-1 + 2^2*Sech[x]^2], {x, 0, ArcCosh[2]}] 

$\frac{3 \pi }{2}$

3. If possible, I want to know mathematica's detail process.