Edit

The define integral is solved using substitution method as Dr. Wolfgang Hintze says like this.

$u$ =$\frac{\cosh ^2(x)-1}{a^2-1}$

$dx$ = $\frac{\left(a^2-1\right)}{2 \sinh (x) \cosh (x)}du$

$\int_0^1 \frac{a^2-1}{2 \sinh (x) \cosh (x) \sqrt{a^2 \text{sech}^2(x)-1}} \, du$

$\frac{1}{2} \int_0^1 \frac{a^2-1}{\sqrt{a^2 \sinh ^2(x)-\sinh ^2(x) \cosh ^2(x)}} \, du$

$\frac{1}{2} \int_0^1 \frac{1}{\sqrt{\frac{\left(\cosh ^2(x)-1\right) \left(a^2-\cosh ^2(x)+1-1\right)}{\left(a^2-1\right) \left(a^2-1\right)}}} \, du$

$\frac{1}{2} \int_0^1 \frac{1}{\sqrt{u (1-u)}} \, du=\frac{\pi }{2}$

It is solved in the real number region. And ArcCos[2]is also real number. But I don't konw why mathematica make $\int_0^{\cosh ^{-1}(2)} \frac{1}{\sqrt{2^2 \text{sech}^2(x)-1}}\, dx$ appear a imaginary term like $\left(\frac{1}{2}-i\right) \pi$ .

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 [1] makes the imaginary term. - I π. 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.