How to find this integral $I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$ Question:

Find this integral
  $$I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$$

let $x\to -x$,so
$$I=\int_{-\pi}^{\pi}\dfrac{x\sin(x) \cot^{-1}{(2014^{-x})}}{1+\cos^4(x)}dx$$
and note
$$\cot\cot^{-1}{2014^x}+\cot\cot^{-1}{2014^{-x}}=\dfrac{\pi}{2}$$
so
$$2I=\dfrac{\pi}{2}\int_{-\pi}^{\pi}\dfrac{x\sin(x)}{1+(\cos{x})^4}dx=\pi\int_{0}^{\pi}\dfrac{x\sin{x}}{1+(\cos{x})^4}dx=\pi^2\int_{0}^{\pi/2}\dfrac{\sin{x}}{1+(\cos{x})^4}dx$$
so
$$I=\dfrac{\pi^2}{2}\int_{0}^{1}\dfrac{1}{1+x^4}dx$$
I feel it's very ugly. Could you please help me using easy methods? It is said that we can use the Gamma function?
 A: Substitution by $x+x^{-1}=u$ and $x-x^{-1}=v$ clean things up a bit.
\begin{align}
2\int_0^1\frac{1}{1+x^4}\,\mathrm{d}x&=\int_0^1\frac{1+x^{-2}}{x^{-2}+x^2}\,\mathrm{d}x-\int_0^1\frac{1-x^{-2}}{x^{-2}+x^2}\,\mathrm{d}x\\
&=\int_{-\infty}^0\frac{1}{v^2+2}\,\mathrm{d}v+\int_{2}^{\infty}\frac{1}{u^2-2}\,\mathrm{d}u\\
&=\frac{\pi}{2\sqrt{2}}+\frac{\log(\sqrt{2}+1)}{\sqrt{2}}
\end{align}
A: We have
\begin{equation}
I=\frac{\pi^2}{2}\int_0^1\frac{1}{1+x^4}dx\tag{1}
\end{equation}
Substituting $t=x^4$ yields
\begin{equation}
I=\frac{\pi^2}{8}\int_0^1\frac{t^{-\frac{3}{4}}}{1+t}dt
\end{equation}
According to Part 11. Scientia 18, 2009, 61-75 by Khristo N. Boyadzhiev, Luis A. Medina, and Victor H. Moll, the above integral is defined as the incomplete beta function
\begin{equation}
\beta(a)=\int_0^1\frac{t^{a-1}}{1+t}dt=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}\right) \right]
\end{equation}
It can easily be proved by multiplying the integrand by $\dfrac{1-t}{1-t}$ and expanding each of the integrands as a geometric series. The details proof can be seen in the cited paper. So
\begin{align}
I&=\frac{\pi^2}{8}\beta\left(\frac{1}{4}\right)=\frac{\pi^2}{16}\left[\psi\left(\frac{5}{8}\right)-\psi\left(\frac{1}{8}\right) \right]\tag{2}
\end{align}
Both results in $(1)$ and $(2)$ are numerically agreed to 50 digits precision
\begin{equation}
I\approx4.2783402057377873840521430671212757030095999944703
\end{equation}
