Check if $f(x,y)=\frac{1}{\cos(x)\cos(y)}$ is integrable on $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2$ I started like this:
$$f(x,y)=\frac{1}{\cos(x)\cos(y)}$$
Notice $f(-x,-y)=f(x,y)$ so and $f(x,y)$ is continuous on the square $A=\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2$ so if $f$ is integrable then $\int_A df =4\int_0^{\pi/2}\int_0^{\pi/2} f(x,y) \,dx\,dy$.
I wanted to bound $f$ from below because I have a feeling it is divergent so $\cos(x)\leq \frac{4x}{\pi}+2$ so $$f(x,y)\geq  \frac{1}{\left(-\frac{4x}{\pi}+2\right)\left(-\frac{4y}{\pi}+2\right)}=\frac{1}{\frac{16}{\pi^2}xy-\frac{8}{\pi}(x+y)+4}\geq\frac{1}{xy+4}$$ and here I am stuck because this bound it didnt give me too much. So I need different one that goes to zero as $x$ approaches $\pi/2$ and it cannot be linear in $x$ because cos is convex.
 A: We have
\begin{align}
\int_0^{\pi/2} \frac{dx}{\cos x} &= \int_0^{\pi/2} \frac{\cos x}{\cos^2 x}\,dx \\
&=  \int_0^{\pi/2} \frac{\cos x}{1-\sin^2 x}\,dx \\
&= \begin{bmatrix} t = \sin x \\ dt = \cos x\,dx\end{bmatrix}\\
&= \int_0^{1} \frac{dt}{1-t^2}\\
&= \frac12\int_0^1 \frac{dt}{1-t} + \frac12\int_0^1 \frac{dt}{1+t}\\
&= \left[-\frac12 \ln(1-t) + \frac12 \ln(1+t)\right]_0^1\\
&= \frac12 \ln \left(\frac{1+t}{1-t}\right)\Bigg|_0^1\\
&= +\infty
\end{align}
so $$\int_{\left(\frac{-\pi}{2},\frac{\pi}{2}\right)^2} \frac{dx}{\cos x\cos y} = 4\int_0^{\pi/2} \left(\int_0^{\pi/2} \frac{dx}{\cos x}\right)\frac{dy}{\cos y} = +\infty$$
A: The trouble is at $\pi/2,$ where $\cos =0.$ The MVT shows
$$\cos y = \cos y-\cos (\pi/2) = -\sin (c_y)\cdot(y-\pi/2) = \sin (c_y)\cdot(\pi/2-y).$$
Here $y\in (0,\pi/2)$ and $c_y\in (y,\pi/2).$ Because $0<\sin c_y <1,$ we have
$$0<\cos y < \pi/2-y \implies \frac{1}{\cos y}>\frac{1}{\pi/2-y}.$$
Since $\int_0^{\pi/2} \dfrac{1}{\pi/2-y}\,dy = \infty,$ $\int_0^{\pi/2}\dfrac{1}{\cos y}\, dy =\infty,$ and this implies your double integral diverges.
