Tips on how to start calculating a definite integral i would appreciate if you could help me with this problem.
I don't want the answer only how should i proceed with solving an integral like this
$$I=\int_{0}^{\pi}\frac{x \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}dx$$
I am completely overwhelmed on how to proceed with this and i am stuck. so i would appreciate some tips to start.
after reading the 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!

Start with $$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$

Edit: Since you are still stuck after performing the above step, you get $$I=\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}\, \,\mathrm dx$$
and add this to $I$ to get $$2I=\pi \int_{0}^{\pi}\frac{ \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}\,\mathrm dx $$
If you are still stuck, use the property

$\displaystyle \int_0^{2a}f(x)\,\mathrm dx=2\int_0^a f(x)\,\mathrm dx$, when $f(2a-x)=f(x)$. If you are unaware of this property, you may see it's proof in this answer.

I hope, now you can take it from here.
