Question

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

Answer 1

$\begin{array}{c}I={\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{\left(\pi +4x^3\right)dx}{2-\cos \left(\left|x\right|+\frac{\pi }{3}\right)}\\ ={\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{\pi }{2-\cos \left(\left|x\right|+\frac{\pi }{3}\right)}dx+{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{4x^{3}dx}{2-\cos \left(\left|x\right|+\frac{\pi }{3}\right)}\\ =2{\int }_0^{\frac{\pi }{3}}\frac{\pi dx}{2-\cos \left(x+\frac{\pi }{3}\right)}+0\\ =2\pi {\int }_0^{\frac{\pi }{3}}\frac{dx}{2-\cos \left(x+\frac{\pi }{3}\right)}\\ putx+\frac{\pi }{3}=t\\ dx=dt\\ \therefore I=2\pi {\int }_{\frac{\pi }{3}}^{\frac{2\pi }{3}}\frac{dt}{2-\cos t}\\ =2\pi \times \frac{1}{\frac{1}{\sqrt{3}}}{\left\{{\tan }^{-1}\left(\frac{\tan \frac{x}{2}}{\frac{1}{\sqrt{3}}}\right)\right\}}_{\frac{\pi }{3}}^{\frac{2\pi }{3}}\\ =\frac{4\pi }{\sqrt{3}}\left({\tan }^{-1}3-\frac{\pi }{4}\right)\end{array}$