Trigonometric Integral

[jacks]

Evaluation of $\displaystyle \int_{0}^{\pi}\frac{1}{(2+\cos x)^3}dx$

[Tolaso J Kos]

Evaluation of $\displaystyle \int_{0}^{\pi}\frac{1}{(2+\cos x)^3}dx$

Hello Jacks,

here is a solution.

\begin{align*}
\int_{0}^{\pi} \frac{{\rm d}x}{\left (2+\cos x \right )^3} &=\frac{1}{2}\int_{0}^{2\pi} \frac{{\rm d}x}{\left ( 2+\cos x \right )^3} \\
&\overset{\begin{subarray}{c} z=e^{ix} \\
{\rm d}z = iz \, {\rm d}x \end{subarray}}{=\! =\! =\! =\! =\! =\!} \frac{1}{2}\oint \limits_{\left | z \right |=1} \frac{{\rm d}z}{iz \left ( 2+ \frac{1}{2}\left ( z+z^{-1} \right ) \right )^3} \\
&= 4\oint \limits_{\left | z \right |=1} \frac{z^2}{i \left ( 4z+z^2+1 \right )^3}\, {\rm d}z\\
&= \frac{4}{i} 2\pi i \mathfrak{Res}\left ( \frac{z^2}{\left ( 4z+z^2+1 \right )^3}; \sqrt{3}-2 \right ) \\
&= 8 \pi \cdot \frac{1}{16 \sqrt{3}} \\
&=\frac{\pi}{2\sqrt{3}}
\end{align*}

.... and happy holidays Jacks ...