Evaluating $ \int_{-\pi /2014}^{\pi /2014}\frac{1}{2014^{x}+1}\left( \frac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx $

let $$f(x)=\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}\Longrightarrow f(x)=f(-x)$$ $$I=\int_{-a}^{a}\dfrac{f(x)}{1+2014^x}dx=\int_{-a}^{a}\dfrac{f(-x)}{1+2014^{-x}}dx=\int_{-a}^{a}\dfrac{f(x)}{1+2014^{-x}}$$ so $$2I=\int_{-a}^{a}f(x)\left(\dfrac{1}{1+2014^x}+\dfrac{1}{1+2014^{-x}}\right)dx=\int_{-a}^{a}f(x)dx$$ where $a=\dfrac{\pi}{2014}$ so we only find $$I'=\int_{-\frac{\pi}{2014}}^{\frac{\pi}{2014}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx$$ if $\dfrac{\pi}{2014}$ replace $\dfrac{\pi}{2}$,we have simple result because $$I''=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=2\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=2I'''$$ and let $x\to \frac{\pi}{2}-x$,then we have $$I'''=\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx$$ so $$2I'''=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}+\cos^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}$$

Let $\displaystyle I = \int \limits_{-\pi /2014}^{\pi /2014}\dfrac{1}{2014^{x}+1}\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx .$

Applying $\displaystyle \int_a^bf(x)\;{dx} = \int_a^b f(a+b-x)\;{dx}$, so we have

$\displaystyle I = \int \limits_{-\pi /2014}^{\pi /2014}\dfrac{1}{2014^{-x}+1}\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx $

Adding and noting that $\frac{1}{2014^{x}+1}+\frac{1}{2014^{-x}+1} = 1$ we have

$\begin{aligned} \displaystyle 2I & = \int \limits_{-\pi /2014}^{\pi /2014} \left(\dfrac{1}{2014^{-x}+1}+\frac{1}{2014^{-x}+1}\right)\left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx \\& = \int \limits_{-\pi /2014}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx = 2\int \limits_{0}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right)\;{dx}\end{aligned} $

because an integral of an even function over $[a, -a]$ is twice over $[0, a]$; thus

$\begin{aligned} \displaystyle I & = \int \limits_{0}^{\pi /2014} \left( \dfrac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right)\;{dx}\end{aligned} $

Whoever created the question was trying to create something slightly different, I bet.