Show that $\int_{0}^{\pi\over 2}\arctan(\tan^8{(\pi^2{x}}))\mathrm dx={5\over 4}$ Can anyone help to provide a proof for $(1)$? Pleases! Thank you.
$$\int_{0}^{\pi\over 2}\arctan(\tan^8{(\pi^2{x}}))\mathrm dx={5\over 4}\tag1$$
Enforcing $u=\tan^8{(\pi^2{x})}$ then $du={8\over \pi^2}\tan^7{(\pi^2{x})}\sec^2{(\pi^2{x})}dx$
Recall: $1+\tan^2{x}=\sec^2{x}$
$du={8\over \pi^2}\tan^7{(\pi^2{x})}+{8\over \pi^2}\tan^8{(\pi^2{x})}dx$
$${\pi^2\over 8}\int_{0}^{k}\arctan{u}\cdot{\mathrm du\over u+u^{7/8}}$$
$k=\arctan{\left(\tan^8{\left(\pi^3\over 2\right)}\right)}$
This is where I got so far. Can't go any further.
Extra note
I think this is the correct version 
$$\lim_{n\to \infty}\int_{0}^{\pi\over 2}\arctan(\tan^{2n}{(\pi^2{x}}))\mathrm dx={5\over 4}\tag2$$

We can solve this using our friend, the dominated convergence theorem. $\tan^{-1}(x) \leq \frac{\pi}{2}$, so the integrand satisfies the conditions for the theorem. Let $u = \pi^2x$ and apply d.c. by moving the limit in.
$$\lim_{n\to\infty}\int_0^{\frac{\pi}{2}} \tan^{-1}(\tan^{2n}(\pi^2 x)) dx = \lim_{n\to\infty}\frac{1}{\pi^2}\int_0^{\frac{\pi^3}{2}} \tan^{-1}(\tan^{2n}(u)) du$$
$$= \frac{1}{\pi^2}\int_0^{\frac{\pi^3}{2}} \tan^{-1}(\lim_{n\to\infty}(\tan^{2}(u))^n) du$$
Depending on the value of u, the inside of the arctangent will either be $0$, $1$, or $\infty$. In other words, the function we are integrating only takes on the constant values $0$, $\frac{\pi}{4}$, or $\frac{\pi}{2}$ (we actually don't care about the middle value since it only happens on a set of measure zero). Taking the limit, we get a rectangle function centered at every odd multiple of $\frac{\pi}{2}$, with height and width $\frac{\pi}{2}$. Such a function is $\pi$-periodic. $$\frac{9\pi}{2} < \frac{\pi^3}{2} < 5\pi$$ so we obtain the full area of 5 of the squares (the only piece cut off has height zero). Thus
$$= \frac{1}{\pi^2}5*(\frac{\pi}{2})^2 = \frac{5}{4}$$
