Question

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

Answer 1

$\stackrel{a}{\underset{-a}{\int }}f\left(x\right)dx=\stackrel{a}{\underset{0}{\int }}\left(f\right(x)+f(-x\left)\right)dx$

$\stackrel{\frac{\pi }{3}}{\underset{-\frac{\pi }{3}}{\int }}\frac{\pi +4x^3}{2-\cos (|x|+\frac{\pi }{3})}=\stackrel{\frac{\pi }{3}}{\underset{0}{\int }}\frac{\pi +4x^3}{2-\cos \left(|x|\frac{\pi }{3}\right)}+\frac{\pi -4x^3}{2-\cos (|x|+\frac{\pi }{3})}$

$I=2\pi \stackrel{\frac{\pi }{3}}{\underset{0}{\int }}\frac{dx}{2-(x+\frac{\pi }{3})}$

$x+\frac{\pi }{3}=z$

$I=2\pi \stackrel{\frac{2\pi }{3}}{\underset{\frac{\pi }{3}}{\int }}\frac{dz}{2-\cos Z}$

$=2\pi \stackrel{\frac{2\pi }{3}}{\underset{\frac{\pi }{3}}{\int }}\frac{dz}{1+2{\sin }^2\frac{z}{2}}$

$2\pi \stackrel{2\frac{\pi }{2}}{\underset{\frac{\pi }{3}}{\int }}\frac{\sec \frac{z}{2}}{{\sec }^2\frac{z}{2}+2{\tan }^2\frac{z}{2}}=\stackrel{2\frac{\pi }{3}}{\underset{\frac{2}{3}}{\int }}\frac{{\sec }^2\frac{z}{2}dz}{1+3{\tan }^{\frac{z}{2}}}$

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

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

${\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(1\right)$

${\tan }^{-1}\theta ={\tan }^{-1}\left(\frac{3-1}{1+3}\right)={\tan }^{-1}\frac{1}{2}$

$\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\frac{1}{2}$.