Update

If it is a bug, I think the bug is not a simple case but is living proof of critical something problems of Mathematica's integral mechanism as version 9, version 10..., what's your opinion about this?

Edit

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

Origin

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.