Evaluating $ \int_{-\pi /2014}^{\pi /2014}\frac{1}{2014^{x}+1}\left( \frac{\sin ^{2014}x}{\sin ^{2014}x+\cos ^{2014}x}\right) dx $ The following integration problem appears in our calculus assignment: 

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

But the problem is I have no idea how to begin this problem. Could anyone give me some help ?
Any hints/ideas are much appreciated.
 A: 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. 
A: 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
$$2I'''=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{2014}{x}+\cos^{2014}{x}}{\sin^{2014}{x}+\cos^{2014}{x}}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}$$
