MIT Integration Bee 2017 problem:$\int_0^{\pi/2}\frac 1 {1+\tan^{2017} x} \, dx$ : Need hints This is a problem from MIT integration bee 2017.
$$\int_0^{\pi/2} \frac 1 {1+\tan^{2017} x} \, dx$$
I have tried substitution method, multiplying numerator and denominator with $\sec^2x$, breaking the numerator in terms of linear combination of the denominator and the derivative of it. None of these methods work.
Some hints please?

Setting the change of variable: $u=\frac\pi2-x $ and since, $\tan x =\cot(\frac\pi2 -x)$ we have, 
\begin{align}
& \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx = \int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} (\frac\pi2-u) } \, du \\[10pt]
= {} & \int_0^{\frac\pi2}\frac{1}{1+\cot^{2017}u} \, du = \int_0^{\frac\pi2}\frac{\tan^{2017} u}{1+\tan^{2017} u} \, du \color{red}{= \frac{\pi}{2} -\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} u} \, du}
\end{align}
That is $$\int_0^{\frac\pi2}\frac{1}{1+\tan^{2017} x} \, dx =\frac\pi4$$
