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Show that $$ \int_0^\pi \frac{x\,dx}{1+\cos^2(x)} = \frac{\pi^2}{2\sqrt{2}} $$ I tried using change of variable $x = \pi-y$ and then ended up with integral $\int_0^\pi \frac{1}{1+\cos^2(y)}dy$ which I think doesn't make thing easier. I am wondering if there is any other clever change of variable or some trick to compute this original integral.

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I think your start is good. I add that part for completeness. $$ \begin{align} I & =\int_0^{\pi}\frac{x}{1+\cos^2x}\,dx=[y=\pi-x]\\ &=\int_0^\pi\frac{\pi-y}{1+\cos^2y}\,dy\\ &=\pi\int_0^\pi\frac{1}{1+\cos^2y}\,dy-I, \end{align} $$ and so $$ I=\frac{\pi}{2}\int_0^\pi \frac{1}{1+\cos^2y}\,dy. $$ This way we got rid of the $x$ in the numerator.

Then, let us use this trick: $$ \frac{1}{1+\cos^2y}=\frac{1}{\sin^2y+2\cos^2y}=\frac{1}{\sin^2y}\frac{1}{1+2\cot^2y}. $$ Let $u=\cot y$. This leads to $$ \int_{-\infty}^{+\infty}\frac{1}{1+2u^2}\,du, $$ which I'm sure you can calculate.

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  • $\begingroup$ Thank you mickep. And yes I there is a factor of $\pi/2$ in front of the second integral. $\endgroup$ – Paul555 Jan 20 '15 at 21:18
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\pi}\frac{x\,\dd x}{1 + \cos^{2}\pars{x}} =\frac{\pi^{2}}{2\root{2}}:\ {\large ?}}$. \begin{align}&\color{#66f}{\large% \int_{0}^{\pi}\frac{x\,\dd x}{1 + \cos^{2}\pars{x}}} =\int_{-\pi/2}^{\pi/2}\frac{x + \pi/2}{1 + \sin^{2}\pars{x}}\,\dd x =\pi\int_{0}^{\pi/2}\frac{\dd x}{1 + \sin^{2}\pars{x}} =\pi\int_{0}^{\pi/2}\frac{\dd x}{2 - \cos^{2}\pars{x}} \\[5mm]&=\pi\int_{0}^{\pi/2}\frac{\sec^{2}\pars{x}\,\dd x}{2\sec^{2}\pars{x} - 1} ={\pi \over \root{2}}\int_{0}^{\pi/2} \frac{\root{2}\sec^{2}\pars{x}\,\dd x}{\bracks{\root{2}\tan\pars{x}}^{2} + 1} ={\pi \over \root{2}}\int_{0}^{\infty}\frac{\dd x}{x^{2} + 1} \\[5mm]&={\pi \over \root{2}}\,{\pi \over 2} =\color{#66f}{\large\frac{\pi^{2}}{2\root{2}}} \end{align}

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I think your change of variables is wrong, but anyway. Split into integral of $x$ and integral of $$x(\tan^2(x) +1)$$ then use the identity $$\frac{1}{\cos^2(x)}= \tan^2(x) + 1$$ First term you can compute and for the second term just use integration by parts noting that $\tan^2(x) + 1$ is the derivative of $\tan(x)$.

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