${\displaystyle \underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{\left(\pi +4x^3\right)dx}{2-\cos \left(\left|x\right|+\frac{\pi }{3}\right)}=}$

$I_1=\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{\pi dx}{2-\cos (\left|x\right|+\frac{\pi }{3})},$ dropping the odd term

         $=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{dx}{2-\cos \left(x+\frac{\pi }{3}\right)},x+\frac{\pi }{3}=t}$

         $=2\pi {\displaystyle \underset{\frac{\pi }{3}}{\overset{\frac{2\pi }{3}}{\int }}\frac{dt}{2-\cos t},u=\tan \frac{t}{2}}$

         $=4\pi {\displaystyle \underset{\frac{1}{\sqrt{3}}}{\overset{\sqrt{3}}{\int }}\frac{du}{2\left(1+u^2\right)-\left(1-u^2\right)}}$

          $=4{\pi {\displaystyle \underset{\frac{1}{\sqrt{3}}}{\overset{\sqrt{3}}{\int }}\frac{du}{1+3u^2}=\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\sqrt{3}u}|}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$

            $=\frac{4\pi }{\sqrt{3}}\left({\tan }^{-1}3-{\tan }^{-1}1\right)=\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{2}\right)$