1. **Problem statement:** Evaluate the integral $$\int_0^{2\pi} \frac{\cos 3\theta}{5 + 4 \cos \theta} d\theta$$ using contour integration. 2. **Formula and approach:** We use the substitution $z = e^{i\theta}$, so that $\cos \theta = \frac{z + z^{-1}}{2}$ and $d\theta = \frac{dz}{iz}$. The integral over $\theta$ from $0$ to $2\pi$ corresponds to a contour integral over the unit circle $|z|=1$ in the complex plane. 3. **Rewrite the integral:** $$\cos 3\theta = \frac{e^{i3\theta} + e^{-i3\theta}}{2} = \frac{z^3 + z^{-3}}{2}$$ The denominator: $$5 + 4 \cos \theta = 5 + 2(z + z^{-1})$$ So the integral becomes $$\int_0^{2\pi} \frac{\cos 3\theta}{5 + 4 \cos \theta} d\theta = \oint_{|z|=1} \frac{\frac{z^3 + z^{-3}}{2}}{5 + 2(z + z^{-1})} \cdot \frac{dz}{iz}$$ 4. **Simplify the integrand:** Multiply numerator and denominator by $z^3$ to clear negative powers: $$\frac{z^3 + z^{-3}}{5 + 2(z + z^{-1})} \cdot \frac{1}{iz} = \frac{z^3 + z^{-3}}{5 + 2(z + z^{-1})} \cdot \frac{1}{iz} = \frac{z^3 + z^{-3}}{5 + 2(z + z^{-1})} \cdot \frac{1}{iz}$$ Rewrite denominator: $$5 + 2(z + z^{-1}) = \frac{5z + 2(z^2 + 1)}{z} = \frac{2z^2 + 5z + 2}{z}$$ So the integrand is: $$\frac{z^3 + z^{-3}}{5 + 2(z + z^{-1})} \cdot \frac{1}{iz} = \frac{z^3 + z^{-3}}{\frac{2z^2 + 5z + 2}{z}} \cdot \frac{1}{iz} = \frac{z^3 + z^{-3}}{1} \cdot \frac{z}{2z^2 + 5z + 2} \cdot \frac{1}{iz} = \frac{z^3 + z^{-3}}{iz} \cdot \frac{z}{2z^2 + 5z + 2} = \frac{z^3 + z^{-3}}{i(2z^2 + 5z + 2)}$$ Multiply numerator by $z^3$ to clear negative powers: $$z^3 + z^{-3} = z^3 + \frac{1}{z^3}$$ So the integrand is: $$\frac{z^3 + z^{-3}}{i(2z^2 + 5z + 2)} = \frac{z^6 + 1}{i z^3 (2z^2 + 5z + 2)}$$ 5. **Integral becomes:** $$\oint_{|z|=1} \frac{z^6 + 1}{i z^3 (2z^2 + 5z + 2)} dz$$ 6. **Factor denominator quadratic:** $$2z^2 + 5z + 2 = (2z + 1)(z + 2)$$ 7. **Poles inside unit circle:** - $z=0$ (pole order 3) - $z = -\frac{1}{2}$ (inside unit circle since $| -\frac{1}{2} | = 0.5 < 1$) - $z = -2$ (outside unit circle) 8. **Calculate residues at poles inside $|z|=1$:** - At $z=0$ (pole order 3): Residue calculation involves the coefficient of $\frac{1}{z}$ term in Laurent expansion. - At $z = -\frac{1}{2}$ (simple pole): Residue = $$\lim_{z \to -\frac{1}{2}} (z + \frac{1}{2}) \frac{z^6 + 1}{i z^3 (2z + 1)(z + 2)} = \frac{(-\frac{1}{2})^6 + 1}{i (-\frac{1}{2})^3 (2(-\frac{1}{2}) + 1)((-\frac{1}{2}) + 2)}$$ Calculate numerator: $$(-\frac{1}{2})^6 = \frac{1}{64}$$ So numerator = $$\frac{1}{64} + 1 = \frac{65}{64}$$ Denominator: $$i (-\frac{1}{2})^3 (2(-\frac{1}{2}) + 1)((-\frac{1}{2}) + 2) = i (-\frac{1}{8}) ( -1 + 1)(\frac{3}{2}) = i (-\frac{1}{8}) (0)(\frac{3}{2}) = 0$$ Since denominator zero, re-check factorization: The factor $(2z + 1)$ zero at $z = -\frac{1}{2}$, so the pole is simple. Residue at simple pole $z_0$ is: $$\lim_{z \to z_0} (z - z_0) f(z)$$ Calculate: $$\text{Residue} = \lim_{z \to -\frac{1}{2}} (z + \frac{1}{2}) \frac{z^6 + 1}{i z^3 (2z + 1)(z + 2)} = \lim_{z \to -\frac{1}{2}} \frac{z^6 + 1}{i z^3 (z + 2)}$$ Plug in $z = -\frac{1}{2}$: $$\frac{\left(-\frac{1}{2}\right)^6 + 1}{i \left(-\frac{1}{2}\right)^3 \left(-\frac{1}{2} + 2\right)} = \frac{\frac{1}{64} + 1}{i \left(-\frac{1}{8}\right) \left(\frac{3}{2}\right)} = \frac{\frac{65}{64}}{i \left(-\frac{3}{16}\right)} = \frac{\frac{65}{64}}{-\frac{3i}{16}} = \frac{65}{64} \cdot \frac{16}{-3i} = \frac{65 \cdot 16}{64 \cdot (-3i)} = \frac{65 \cdot 16}{64 \cdot (-3i)}$$ Simplify numerator and denominator: $$\frac{65 \cdot 16}{64} = 65 \cdot \frac{16}{64} = 65 \cdot \frac{1}{4} = \frac{65}{4}$$ So residue = $$\frac{65/4}{-3i} = -\frac{65}{4 \cdot 3 i} = -\frac{65}{12 i} = \frac{65 i}{12}$$ 9. **Residue at $z=0$ (pole order 3):** Write integrand as: $$f(z) = \frac{z^6 + 1}{i z^3 (2z + 1)(z + 2)} = \frac{z^6 + 1}{i z^3 (2z + 1)(z + 2)}$$ Expand denominator near $z=0$: $$(2z + 1)(z + 2) = 1 + 2z + 2z + 4z^2 = 1 + 4z + 4z^2$$ So near $z=0$: $$f(z) = \frac{z^6 + 1}{i z^3 (1 + 4z + 4z^2)} = \frac{1 + z^6}{i z^3 (1 + 4z + 4z^2)}$$ Rewrite as: $$f(z) = \frac{1 + z^6}{i z^3} \cdot \frac{1}{1 + 4z + 4z^2}$$ Use series expansion for denominator: $$\frac{1}{1 + 4z + 4z^2} = 1 - 4z + 12 z^2 + \cdots$$ Multiply numerator and denominator expansions: $$f(z) = \frac{1 + z^6}{i z^3} (1 - 4z + 12 z^2 + \cdots) = \frac{1}{i z^3} (1 - 4z + 12 z^2 + \cdots) + \frac{z^6}{i z^3} (1 - 4z + 12 z^2 + \cdots)$$ Simplify: $$= \frac{1}{i z^3} (1 - 4z + 12 z^2) + \frac{z^3}{i} (1 - 4z + 12 z^2) + \cdots$$ Focus on terms with $z^{-1}$ (coefficient of $\frac{1}{z}$) for residue: From first term: $$\frac{1}{i z^3} (1 - 4z + 12 z^2) = \frac{1}{i} (z^{-3} - 4 z^{-2} + 12 z^{-1})$$ The coefficient of $z^{-1}$ is $$\frac{12}{i} = -12 i$$ From second term and higher powers, no $z^{-1}$ terms appear. So residue at $z=0$ is $$-12 i$$ 10. **Sum of residues inside unit circle:** $$\text{Residue sum} = -12 i + \frac{65 i}{12} = i \left(-12 + \frac{65}{12}\right) = i \left(-\frac{144}{12} + \frac{65}{12}\right) = i \left(-\frac{79}{12}\right) = -\frac{79 i}{12}$$ 11. **Apply residue theorem:** $$\oint_{|z|=1} f(z) dz = 2 \pi i \times \text{(sum of residues inside)} = 2 \pi i \times \left(-\frac{79 i}{12}\right) = 2 \pi i \times -\frac{79 i}{12}$$ Since $i \times i = -1$: $$= 2 \pi \times \frac{79}{12} = \frac{79 \pi}{6}$$ 12. **Final answer:** $$\boxed{\int_0^{2\pi} \frac{\cos 3\theta}{5 + 4 \cos \theta} d\theta = \frac{79 \pi}{6}}$$