计算
\[ I=\int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx} \]

(tian_275461)
解
\[ I=\int_{0}^{\pi}{xd\arctan{(\sin{x})}}=-\int_{0}^{\pi}{\arctan{(\sin{x})}dx}=-2\int_{0}^{\frac{\pi}{2}}{\arctan{(\sin{x})}dx}\]

注意到
\[ \arctan{\left(\frac{\sin{x}}{+\infty}\right)}-\arctan{\left(\frac{\sin{x}}{1}\right)}=-\int_{1}^{+\infty}{\frac{\sin{x}}{y^2+\sin^2{x}}dy}\]
故
\begin{align*}
I&=-2\int_{0}^{\frac{\pi}{2}}{\int_{1}^{+\infty}{\frac{\sin{x}}{y^2+\sin^2{x}}dy}dx}\\
&=-2\int_{1}^{+\infty}{\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{y^2+\sin^2{x}}dx}dy}\\
&=-2\int_{1}^{+\infty}{\int_{0}^{1}{\frac{1}{y^2+1-t^2}dt}dy}\qquad (t=\cos{x}) \\
&=-\int_{1}^{+\infty}{\frac{1}{\sqrt{y^2+1}}\ln{\left(\frac{\sqrt{y^2+1}+1}{\sqrt{y^2+1}-1}  \right)}dy}\\
&=-\int_{\text{arcsinh}1}^{+\infty}{\ln\left(\frac{\cosh{z}+1}{\cosh{z}-1}\right)dz}\qquad (y=\sinh{z})\\
&=2\int_{\text{arcsinh}1}^{+\infty}{\ln\left(\frac{1-e^{-z}}{1+e^{-z}}\right)dz}\\
&=2\int_{0}^{\sqrt{2}-1}{\frac{\ln{(1-t)}-\ln(1+t)}{t}dt}\qquad (t=e^{-z})\\
&=2\text{Li}_{2}(1-\sqrt{2})-2\text{Li}_{2}(\sqrt{2}-1)
\end{align*}
套用多重对数函数的性质(2)(3)(4)
(2)
\[ \text{Li}_{2}(1-x)+\text{Li}_{2}\left(1-\frac{1}{x}\right)=-\frac{1}{2}\ln^2{x} \]
(3)
\[ \text{Li}_{2}(x)+\text{Li}_{2}(1-x)=\frac{1}{6}\pi^2-\ln{x}\cdot\ln(1-x) \]
(4)
\[ \text{Li}_{2}(-x)-\text{Li}_{2}(1-x)+\frac{1}{2}\text{Li}_{2}(1-x^2)=-\frac{1}{12}\pi^2-\ln{x}\cdot\ln{(x+1)}\]
得到
\begin{align*}
   \text{Li}_{2}(2-\sqrt{2})+ \text{Li}_{2}(-\sqrt{2})&=-\frac{1}{2}\ln^2(\sqrt{2}-1) \qquad (\text{用$x=\sqrt{2}-1$套第2个})\\
    \text{Li}_{2}(2-\sqrt{2})+ \text{Li}_{2}(\sqrt{2}-1)&=\frac{\pi^2}{6}-\ln{(\sqrt{2}-1)}\cdot\ln{(2-\sqrt{2})}\qquad (\text{用$x=\sqrt{2}-1$套第3个})\\
     \text{Li}_{2}(-\sqrt{2})- \text{Li}_{2}(1-\sqrt{2})+\frac{1}{2} \text{Li}_{2}(1-(-\sqrt{2})^2)&=-\frac{\pi^2}{12}-\ln{\sqrt{2}}\cdot\ln{(1+\sqrt{2})}\qquad (\text{用$x=-\sqrt{2}$套第4 个})
\end{align*}
用第2个式子减去第1个式子加上第3个式子，得到
\[  \text{Li}_{2}(\sqrt{2}-1)- \text{Li}_{2}(1-\sqrt{2})=\frac{\pi^2}{8}-\frac{1}{2}\ln^2(\sqrt{2}+1)\]
故
\[ I=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4} \]

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以上是tian_275461的解法，在stackexchange上看到sos440表示

\[\int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = 4 \chi_{2}(1-\sqrt{2})\]

where

\[\chi_{2}(z) = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)^2}\]

is the Legendre chi function of order 2. By exploiting some identities involving dilogarithm, we find that

\[\chi_{2}(1-\sqrt{2}) = \frac{1}{4} \log^2 (1+\sqrt{2}) - \frac{3}{8}\zeta(2).\]

This gives the answer

\[\int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = \log^2(1+\sqrt{2}) - \frac{\pi^2}{4}.\]

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reference：

http://math.stackexchange.com/questions/323603/integrate-displaystyle-int-0-pi-fracx-cosx1-sin2xdx

http://sos440.tistory.com/83