What is $\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x$? Could you prove that: 
\begin{align} \displaystyle 2\left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{7} + & 53 \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{5} \\ + \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{4} + & \quad \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{3} \\ + & 19\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2014 \end{align}
Just for the fun of it. 
Observe that $2\times53\times1\times1\times19=2014.$ 
EDIT: A different proof of
$$
\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2
$$
may be found here. Thanks.
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$\ds{\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x = 2:
     \ {\large ?}}$

\begin{align}&\color{#c00000}{%
\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x}
=-\,\Im\int_{0}^{\pi}{\dd x \over x\ic + \ln\pars{2\sin\pars{x}}}
\\[3mm]&=-\,\Im
\int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}}
{\dd z/\pars{\ic z}\over
\ln\pars{z} + \ln\pars{2\bracks{z^{2} - 1}/\bracks{2\ic z}}}
\\[3mm]&=\Re
\int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}}
{1 \over \ln\pars{\bracks{1 - z^{2}}\ic}}\,{\dd z \over z}
\end{align}

We close the contour with the 'line segment'
$\ds{\braces{\pars{x,0}\ \mid\ x \in \pars{-1,1}}}$ with 'indented points' at $\ds{z = -1,0,1}$. I t turns out that the whole contribution to the integral arises from the 'indented point' at $\ds{z = 0}$: See my detailed calculation in one of  my previous answers which is very close to the present one. Then,

\begin{align}&\color{#66f}{\large%
\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x}
=\left. -\lim_{\epsilon\ \to\ 0^{+}}\Re\int_{\pi}^{0}
{1 \over \ln\pars{\bracks{1 - z^{2}}\ic}}\,{\dd z \over z}
\right\vert_{\,z\ \equiv\ \epsilon\expo{\ic\theta}}
\\[3mm]&=-\,\Re\int_{\pi}^{0}{1 \over \ln\pars{\ic}}\,\ic\,\dd\theta
=-\,\Re\int_{\pi}^{0}{1 \over \pi\,\ic/2}\,\ic\,\dd\theta
=-\,{2 \over \pi}\int_{\pi}^{0}\dd\theta
=\color{#66f}{\Large 2}
\end{align}

A: Here is a proof, using complex analysis, that the integral is equal to $2$. Put
$$
f(z) = {1\over\log{(i(1-e^{i2z}))}},
$$
the logarithm being the principal branch. As can be deduced from the comments, the original integral is equal to
$$
\int_0^\pi \operatorname{Re}{f(x)}\,dx.
$$
Next consider, for $R>\epsilon > 0$, the following contour:

It is straightforward to check that for each fixed $\epsilon$ and $R$, the function $f$ is analytic on an open set containing this contour (just consider where $i(1-e^{i2z})\leq0$; this can only happen when $\operatorname{Im}{z}<0$ or when $\operatorname{Im}{z} = 0$ and $\operatorname{Re}{z}$ is an integer multiple of $\pi$). It then follows from Cauchy's theorem that $f$ integrates to zero over it. First let's see that the integrals over the quarter-circular portions of the contour vanish in the limit $\epsilon \to 0$. I'll look at the quarter circle near $\pi$ (near the bottom right corner of the contour) but the one near zero is similar, if not easier. Writing $i(1-e^{i2z}) = i(1-e^{i2(z-\pi)})$, it is clear that $i(1-e^{i2z}) = O(z-\pi)$ as $z\to\pi$. It follows from this that $f(z) = O(1/\log{|z-\pi|})$ as $z\to0$, and therefore that the integral of $f$ over the bottom right quarter circle is $O(\epsilon/\log{\epsilon})$ as $\epsilon \to0$, hence it vanishes in the limit as claimed.
Thus for fixed $R$, we can let $\epsilon \to 0$ to see that $f$ integrates to zero over the rectangle with corners $0,\pi, \pi +iR,$ and $iR$. Now the vertical sides of this rectangle give the contribution
\begin{align*}
-\int_0^R f(iy)\,idy + \int_0^R f(iy+\pi)\,idy = 0,
\end{align*}
since $f(iy) = f(iy+\pi)$. It follows at once that for each $R$, we can set the contribution from the horizontal sides equal to zero, giving
\begin{align*}
\int_0^{\pi} f(x)\,dx - \int_0^\pi f(x+iR)\,dx=0, \qquad R>0. \tag{1}
\end{align*}
(Note that the above equation implies that the integral in $x$ of $f(x+iR)$ over the interval $[0,\pi]$ is constant as a function of $R$; another way to evaluate the integral is to prove this directly, which I'll add below in a moment.) Now $f(x+iR) \to 1/\log{i} = 2/\pi$ as $R\to\infty$, uniformly for $x\in[0,\pi]$. Thus
\begin{align*}
\lim_{R\to\infty} \int_0^\pi f(x+iR)\,dx = \pi\cdot {2\over \pi} = 2,
\end{align*}
and it follows from $(1)$ that the original integral is equal to $2$ as well.
(By the way, the above idea is basically a replication of the technique I used here and here, and which I originally got from Ahlfors' book on complex analysis.)
