<s>[INST] Solve trigonometric integral $\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx $ Please help me to solve the following integral:
$$\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx$$
I have tried a lot, but no results. I only transformed this integral to the following also not easy integral:
$$\int \frac{1}{(1+t^2)(1+t^{2014})} dt.$$
 [/INST] Consider the integral
\begin{align}
I = \int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx.
\end{align}
This may also be seen as
\begin{align}
I &= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx + \int_{-\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\
&= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx - \int_{\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\
&= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx
\end{align}
where $x \to -x$ was made in the second integral. Now make the substitution $x = t-\pi/2$ to obtain
\begin{align}
I = 2 \, \int_{-\pi/2}^{0} \frac{\cos^{2a}t}{\sin^{2a}t + \cos^{2a}t} \, dt.
\end{align}
Now let $t \to -x$ to obtain
\begin{align}
I = 2 \, \int_{0}^{\pi/2} \frac{\cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx.
\end{align}
Adding the two integral expressions leads to
\begin{align}
2 I &= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x + \cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = 2 \, \int_{0}^{\pi/2} dx = \pi.
\end{align}
It can now be stated that
\begin{align}
\int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = \frac{\pi}{2}.
\end{align}
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