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?

Try using $$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$
and the fact that $\tan(\frac{\pi}{2}-x) = \cot(x)$ to convert to get the following:
$$I = \int_0^{\pi / 2} \frac 1 {1+ \tan^{2017}(x)} \, dx = 
 \int_0^{\pi / 2} \frac 1 {1+ \tan^{2017}(\pi/2-x)} \, dx \\
 =  \int_0^{\pi / 2} \frac 1 {1+ \cot^{2017}(x)} \, dx =  \int_0^{\pi / 2} \frac{\tan^{2017}(x)}{1+ \tan^{2017}(x)} \, dx$$
Hence
$$2I = \int_{0}^{\pi / 2}  dx = \frac{\pi}{2}$$
