How to prove $\int_0^{4 \pi }\left(\frac{2-\cos x}{\sqrt{3}}\right)^{-m}dx=\int_0^{4 \pi }\left(\frac{2+\sin x}{\sqrt{3}}\right)^{m-1}dx$

Question

How to prove that: $$\int_0^{4 \pi }\left(\frac{2-\cos x}{\sqrt{3}}\right)^{-m}dx=\int_0^{4 \pi }\left(\frac{2+\sin x}{\sqrt{3}}\right)^{m-1}dx$$ Here $x,m \in \mathbb{R}$.

Answer 1

Put \begin{equation*} I_1=\int_{0}^{4\pi}\left(\dfrac{2-\cos x}{\sqrt{3}}\right)^{-m}\,\mathrm{d}x \end{equation*} and \begin{equation*} I_2=\int_{0}^{4\pi}\left(\dfrac{2+\sin x}{\sqrt{3}}\right)^{m-1}\,\mathrm{d}x . \end{equation*} Both integrands are periodic functions with period $2\pi$. Consequently we can integrate over any interval of length $2\pi$. We make some substitutions. \begin{gather*} I_1=2\int_{-\pi}^{\pi}\left(\dfrac{2-\cos x}{\sqrt{3}}\right)^{-m}\,\mathrm{d}x =\left[t=\tan\frac{x}{2}\right] = 2\int_{-\infty}^{\infty}\left(\dfrac{2-\frac{1-t^2}{1+t^2}}{\sqrt{3}}\right)^{-m}\dfrac{2}{1+t^2}\,\mathrm{d}t =\\[2ex] 4\sqrt{3}^{m}\int_{-\infty}^{\infty}\dfrac{(1+t^2)^{m-1}}{(1+3t^2)^{m}}\,\mathrm{d}t . \end{gather*} Furthermore \begin{gather*} I_2=\int_{0}^{4\pi}\left(\dfrac{2+\sin x}{\sqrt{3}}\right)^{m-1}\,\mathrm{d}x =\left[x=\frac{\pi}{2}-y\right] = \int_{0}^{4\pi}\left(\dfrac{2+\cos y}{\sqrt{3}}\right)^{m-1}\,\mathrm{d}y =\\[2ex] \left[s=\tan\frac{y}{2}\right] = 2\int_{-\infty}^{\infty}\left(\dfrac{2+\frac{1-s^2}{1+s^2}}{\sqrt{3}}\right)^{m-1}\dfrac{2}{1+s^2}\,\mathrm{d}s =4\sqrt{3}^{1-m}\int_{-\infty}^{\infty}\dfrac{(3+s^2)^{m-1}}{(1+s^2)^{m}}\,\mathrm{d}s =\\[2ex] \left[s=\sqrt{3}u\right]=4\sqrt{3}^{m}\int_{-\infty}^{\infty}\dfrac{(1+u^2)^{m-1}}{(1+3u^2)^{m}}\,\mathrm{d}u = I_1. \end{gather*}