Trig integral equals alternate inverse reciprocal binomial series

In this entry we will solve the following integral



\begin{align*} \int_0^{\pi}\frac{x\cos(x)}{1+\sin^2(x)}\,dx&= \ln^2(1+\sqrt{2})-\frac{\pi^2}{4} \end{align*}


to this end we should recall a previous established result. Click here for the proof of the integral.


                     \begin{align*} \sum_{n=0}^\infty\frac{ 2^{2n}}{\binom{2n}{n}}\frac{(-1)^n}{\left(2n+1\right)^2}= \frac{\pi^2}{8}-\frac{\ln^2\left(1+\sqrt{2} \right)}{2} \end{align*}


Proof of the binomial series here .








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