Help on how to proceed on this trig integral i would appreciate if you could help me with this problem.
$$I=\int_{0}^{\pi}\frac{x \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}dx$$
I am completely overwhelmed on how to proceed with this and i am stuck. so i would appreciate some tips to start
after reading some tips i have done this
$$\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}dx$$
after that i distributed it
$$\int_{0}^{\pi}\frac{\pi\sin^{2018}(\pi-x)-x\sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}=\int_{0}^{\pi}\frac{\pi\sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}-I$$
But after that i'm lost again, i know i should be look for a way to make some terms cancel but that's as far as i got
please don't give me the direct solution, i would like to solve it by myself
Thanks in advance!

HINT
Note that for even $n$ (as in your case) we have
$$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$

As asked for, this is a hint :) If you need any more help please don't hesitate to ask.

This is how I'd continue. We know that
$$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$
Adding, we find that
$$2\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx+\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx=\pi$$
So $$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\frac{\pi}{2}$$
and in particular $$\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx=\frac{\pi}{2}$$
We now know that
$$I=\pi\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx-I=\pi\times\frac{\pi}{2}-I$$
Hence $2I=\pi\times\frac{\pi}{2}=\frac{\pi^2}{2}$ and $I=\frac{\pi^2}{4}$.
