Evaluating $\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(x)}{ \cos(a \cos(x) \cos(y))} dx dy $ Can we avoid the use of the geometric interpretation combined with polar coordinates change of variable for proving that 
$$\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(x)}{ \cos(a \cos(x)  \cos(y))} d x dy =\frac{\pi}{ 2a}\log\left( \frac{\displaystyle 1+ \tan \left(\frac{a}{2}\right)}{ \displaystyle 1-\tan\left(\frac{a}{2}\right)}\right)$$  ?
EDIT
What if we go further and we also consider the case  
$$\int_0^{\pi/2}\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(x)}{ \cos(a \cos(x)  \cos(y) \cos(z))} d x dy \ dz$$
? What can we say about the closed form of this one? 
 A: The simplest way I can see doing these integrals is to expand the secant term into a Taylor series, say
$$\sec{q} = \sum_{k=0}^{\infty} s_{2 k} q^{2 k}$$
where $s_{2 k}$ are known but not really important here.  By using this expansion, however, we see that the double integral is
$$\sum_{k=0}^{\infty} s_{2 k} a^{2 k} \int_0^{\pi/2} dx \, \cos^{2 k+1}{x} \, \int_0^{\pi/2} dy \, \cos^{2 k}{y} $$
These integrals are easily done and may be found here.  (Scroll down to the Addendum.)  The binomial coefficient terms cancel and we get, for the integral,
$$\frac{\pi}{2} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k}}{2 k+1}$$
which we may rewrite as
$$\frac{\pi}{2 a} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k+1}}{2 k+1}$$
If we differentiate the sum, we get the series back for $\sec{a}$!  So really, the integral is simply $\pi/(2 a)$ times the antiderivative of $\sec{a}$ (zero when $a=0$), so we get that the integral is
$$\frac{\pi}{2 a} \log{(\sec{a}+\tan{a})}$$
which is equivalent to the stated result.
The 3d case, is conceptually simple at this point; again, it is a sum:
$$\frac{\pi^2}{4} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k}}{2 k+1} \frac1{2^{2 k}} \binom{2 k}{k}$$
You would now need to know that $s_{2 k}=|E_{2 k}|/(2 k)!$, where $E_{2 k}$ are the Euler numbers.  I am not sure there is an easy way to do this sum, but perhaps there is. 
ADDENDUM
It should be clear that this problem is a special case of the following result: let $f(x)$ be an even-valued function over the reals.  Then
$$\int_0^{\pi/2} dx  \, \int_0^{\pi/2} dy \, \cos{y} \: f(a \cos{x} \cos{y}) = \frac{\pi}{2 a} F(a)$$
where $F'(x) = f(x)$ and $F(0)=0$.
