Step 1: Use symmetry to simplify the integral.
We have $k = \int_0^\pi \frac{dx}{1 + 2\sin^2 x}$. Since $\sin(\pi - x) = \sin x$, the integrand is symmetric about $x = \frac{\pi}{2}$: \[k = 2\int_0^{\pi/2} \frac{dx}{1 + 2\sin^2 x}\]
Step 2: Apply the standard formula for this type of integral.
The standard result (derivable via the Weierstrass substitution $t = \tan x$) is: \[\int_0^{\pi/2} \frac{dx}{a + b\sin^2 x} = \frac{\pi}{2\sqrt{a(a+b)}}\] Here $a = 1, b = 2$, so: \[\int_0^{\pi/2} \frac{dx}{1 + 2\sin^2 x} = \frac{\pi}{2\sqrt{1 \cdot 3}} = \frac{\pi}{2\sqrt{3}}\]
Step 3: Compute k.
\[k = 2 \cdot \frac{\pi}{2\sqrt{3}} = \frac{\pi}{\sqrt{3}}\]
Step 4: Find the decimal approximation of k.
Using $\pi \approx 3.1416$ and $\sqrt{3} \approx 1.7321$: \[k \approx \frac{3.1416}{1.7321} \approx 1.814\]
Step 5: Apply the floor function.
Since $1 < 1.814 < 2$, the greatest integer not exceeding $k$ is $1$. The floor function $\lfloor k \rfloor = 1$.
Step 6: State the final answer.
The greatest integer less than or equal to $k$ is \[\boxed{1}\]