The value of
$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ is equal to :

I = $\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ .....(1) <br><br>I = $$\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ ......(2) <br><br>Adding (1) and (2) <br><br>2I = $$\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ <br><br>$\Rightarrow$ I = $$2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ <br><br>$\Rightarrow$ I = $$2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$$ <br><br>= ${2\pi \int\limits_0^{\pi /2} 1 dx}$ <br><br>= $2\pi .{\pi \over 2}$ = ${\pi ^2}$