This may be hard to believe, but I promise this question is doable via only 4U methods.
(Won't provide hints unless someone actually attempts it.)
It's been a while since I did maths on AN, but I spent too long on this to not cash in
Hopefully there aren't any mistakes!
Let \(k = 1009\). Then the integral becomes
\[
\int_{-\pi/(4k)}^{\pi/(4k)} \frac{1}{(2k)^{2kx}+1} \frac{\cos^{2k}(2kx)}{\sin^{2k}(2kx) + \cos^{2k}(2kx)} \, dx.
\]
Make the change of variables \(x = r/(2k)\). Then \(dx/dr = 1/(2k)\) and the integral becomes
\[
\int_{-\pi/2}^{\pi/2} \frac{1}{(2k)^{r}+1} \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r} \frac{1}{2k} \, dr.
\]
Define the functions \(f\colon \mathbb{R} \to \mathbb{R}\) and \(g\colon \mathbb{R} \to \mathbb{R}\) by
\[
f(r) = \frac{1}{(2k)^r + 1}
\quad \text{and} \quad
g(r) = \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r}
\quad \forall r \in \mathbb{R}.
\]
Then the integral we wish to compute is
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx.
\]
Hint 1 (for my particular solution)
Try to use \(f(x) g(x) = [1 - f(x)] [1 - g(x)] + f(x) + g(x) - 1\).
Hint 2 (for my particular solution)
As is usual with these types of questions, we wish to make use of identities/symmetries satisfied by \(f\) and \(g\). For this particular problem, we will use the result that for every \(x \in \mathbb{R}\) we have
I leave it to the reader to verify that these hold. It is perhaps a bit arbitrary to begin with these, and in fact these are not identities that come out of thin air; rather they are motivated by the observation in Hint 1.
Solution
We integrate the equation in Hint 1 in \(x\) over the interval \([-\pi/2, \pi/2]\) to obtain
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \underbrace{\int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx}_{I_1} + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Let \(I_1\) be the integral as indicated above. We use a change of variables \(x = -r\). Then \(dx/dr = -1\) and \(I_1\) becomes
\[
I_1
= \int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx
= \int_{\pi/2}^{-\pi/2} \underbrace{[1 - f(-r)]}_{{} = f(r)} [1 - \underbrace{g(-r)}_{{} = g(r)}] (-1) \, dr
= \int_{-\pi/2}^{\pi/2} f(r) [1 - g(r)] \, dr.
\]
Note that we have used some of the identities in Hint 2. We now have
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \int_{-\pi/2}^{\pi/2} f(x) [1 - g(x)] \, dx + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Adding \(\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx\) to both sides, we obtain
\[
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= 2\int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi
= 2\int_{-\pi/2}^{\pi/2} \underbrace{\bigg[f(x) - \frac{1}{2}\bigg]}_{\text{odd in \(x\)}} \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx
= \int_{-\pi/2}^{\pi/2} g(x) \, dx.
\]
Note that the fact that \(f(x) - 1/2\) is odd in \(x\) follows from an identity in Hint 2. We now wish to compute \(\int_{-\pi/2}^{\pi/2} g(x) \, dx\). By evenness in \(x\) of \(g(x)\), this is the same as \(2\int_{0}^{\pi/2} g(x) \, dx\), which is also the same as \(2\int_{-\pi/2}^{0} g(x) \, dx\). Since \(g(x) = 1 - g(x + \pi/2)\) for all real \(x\), making the substitution \(x = r - \pi/2\), we have
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= 2\int_{-\pi/2}^{0} g(x) \, dx
= 2\int_{-\pi/2}^{0} [1 - g(x + \pi/2)] \, dx
= 2\int_{0}^{\pi/2} [1 - g(r)] \, dr.
\]
We now write
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= \frac{1}{2} \Bigg[\int_{-\pi/2}^{\pi/2} g(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx\Bigg]
= \frac{1}{2} \Bigg[2\int_{0}^{\pi/2} g(x) \, dx + 2\int_{0}^{\pi/2} [1 - g(x)] \, dx\Bigg]
= \frac{\pi}{2}
\quad \implies \quad
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{2}.
\]
Dividing both sides by \(4k\), we obtain
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{8k}.
\]
Recalling that \(k = 1009\), we have
\[
\int_{-\pi/4036}^{\pi/4036} \frac{1}{2018^{2018x}+1} \frac{\cos^{2018}(2018x)}{\sin^{2018}(2018x) + \cos^{2018}(2018x)} \, dx
= \frac{\pi}{8072}.
\]
Comments
This holds for all positive integers \(k\). Integrality of \(k\) is used when the identities regarding the function \(g\) in Hint 2 are derived.
I changed the integration variable from \(r\) to \(x\) multiple times without comment. I'm not sure if it's possible for marks to be lost for this under a HSC marking scheme, so be careful.