I need assistance with calculating the following integral using the Cauchy formula. I have been encountering incorrect results and would greatly appreciate your help in identifying the mistake in my calculations.
$$\int_{0}^{2\pi}\frac{\cos t}{5-3\cos t}dt$$
My calculation:
$$\int_{0}^{2\pi}\frac{\cos t}{5-3\cos t}dt = \int_{0}^{2\pi}\frac{\frac{e^{it} + e^{-it}}{2}}{5 - 3\frac{e^{it} + e^{-it}}{2}}dt = \int_{0}^{2\pi}\frac{(e^{it})^2 + 1}{-3(e^{it})^{2} + 10e^{it} -3}dt = \int_{0}^{2\pi}\frac{(e^{it})^2 + 1}{-3(e^{it})^{2} + 10e^{it} -3}\frac{d(e^{it})}{ie^{it}} = \int_{0}^{2\pi}\frac{z^{2} + 1}{-3z^{2} + 10z - 3}\frac{d(z)}{iz} = -i\int_{0}^{2\pi}\frac{z^{2} + 1}{z(-3z^{2} + 10z - 3)}dz$$
Where $\gamma (t)=e^{it}$ is Line (In terms of line integral)
I have calculated the roots and got: $$-i\oint_{\gamma}^{}\frac{z^{2} + 1}{z(z-3)(z-\frac{1}{3})}dz = -i\oint_{\gamma}^{}\frac{\frac{z^{2} + 1}{z - 3}}{z(z-\frac{1}{3})}$$
After this I have calculated partial fraction decomposition and used Cauchy formula:
$$-i\oint{}^{}3\frac{\frac{z^{2} + 1}{z - 3}}{z} - 3\frac{\frac{z^{2} + 1}{z - 3}}{z - \frac{1}{3}} = -3i\cdot2\pi i\cdot \frac{0 + 1}{0 - 3} + 3i\cdot2\pi i \cdot \frac{\frac{1}{9} + 1}{\frac{1}{3} - 3} =\frac{\pi}{2}$$
But the correct result is: $\frac{\pi}{6}$
The underlying mistake is in the factorization of the denominator and subsequent partial fraction expansion:
$$\require{cancel} \begin{align*} & \int_0^{2\pi} \frac{\cos t}{5-3\cos t} \, dt \\ &= \oint\limits_{\gamma:\lvert z\rvert=1} \frac{z+\frac1z}{10-3\left(z+\frac1z\right)} \, \frac{dz}{iz} \\ &= -\frac i{\color{red}{3}} \oint_\gamma \frac{z^2+1}{z \left(z-\frac13\right) (z-3)} \, dz \\ &= -\frac i{\cancel3} \oint_\gamma \left(\frac{\cancel3}{z-\frac13} - \frac{\cancel3}z\right)\underbrace{\frac{z^2+1}{z-3}}_{=:f(z)} \, dz \\ \\ &= -i \cdot 2\pi i \left(f\left(\frac13\right) - f(0)\right) = \boxed{\frac\pi6} \end{align*}$$