The value of the integral

$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} $$ is :

$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x$ <br/><br/>Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ <br/><br/>$$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x $$ <br/><br/>$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$ <br/><br/>Now, <br/><br/>$$ \begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned} $$