integration - How to show that this integral equals $fracpi2$?

While solving a physical problem from Landau, Lifshitz "Mechanics" book, I came across an integral:

$$\int_0^\delta \frac{du}{\sqrt{\left(\frac{\cosh\delta}{\cosh u}\right)^2-1}}.$$

In the book only the final answer for the problem is given, from which I deduce that this integral must be $\frac\pi2$.

I've tried feeding it to Wolfram Mathematica, but it wasn't able to evaluate it, returning unevaluated result. Evaluating it numerically confirms that this is a likely answer, but I haven't been able to prove this.

I've tried making a substitution $v=\frac{\cosh\delta}{\cosh u}$ and got this integral instead:

$$\gamma \int_1^\gamma \frac{dv}{v\sqrt{(\gamma^2-v^2)(v^2-1)}},$$

where $\gamma=\cosh\delta$, but still this doesn't give me a clue how to proceed. Also, I can't seem to eliminate the parameter ($\delta$ or $\gamma$), which shouldn't affect the result at all.

So, the question is: how can one evaluate this integral or at least prove that it's equal $\frac\pi2$?

Answer

It's actually a lot simpler than this. Rewrite the integral as

$$\int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\cosh^2{\delta}-\cosh^2{u}}} = \int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\sinh^2{\delta}-\sinh^2{u}}}$$

Sub $y=\sinh{u}$ and the integral becomes

$$\int_0^{\sinh{\delta}} \frac{dy}{\sqrt{\sinh^2{\delta}-y^2}}$$

Now sub $y=\sinh{\delta}\, \sin{t}$ and the integral is

$$\int_0^{\pi/2} dt = \frac{\pi}{2}$$