Here is the first payoff of the residue theorem: real integrals of trigonometric expressions over a full turn, which are hard by real methods, become contour integrals around the unit circle.
Idea clave
For $\displaystyle\int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta$, put $z = e^{i\theta}$. As $\theta$ runs from $0$ to $2\pi$, $z$ goes once around $|z| = 1$, and
$$\cos\theta = \frac{z + z^{-1}}{2}, \qquad \sin\theta = \frac{z - z^{-1}}{2i}, \qquad d\theta = \frac{dz}{iz}.$$
The integral becomes $\oint_{|z| = 1}(\text{a rational function of } z)\,dz = 2\pi i\,\times$ (sum of the residues inside the unit circle). A standard result: for $a > |b|$,
$$\int_0^{2\pi}\frac{d\theta}{a + b\cos\theta} = \frac{2\pi}{\sqrt{a^2 - b^2}}.$$
Por qué funciona
The substitutions come from Euler's formula: $z + \frac{1}{z} = e^{i\theta} + e^{-i\theta} = 2\cos\theta$, and similarly for sine. From $z = e^{i\theta}$, $dz = ie^{i\theta}d\theta = iz\,d\theta$, so $d\theta = \frac{dz}{iz}$. For the standard result, $a + b\frac{z + z^{-1}}{2} = \frac{bz^2 + 2az + b}{2z}$, so the integrand becomes $\frac{2}{i(bz^2 + 2az + b)}$. The two roots of $bz^2 + 2az + b$ multiply to $1$; the one inside the unit circle is $z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}$, with residue $\frac{2}{ib(z_1 - z_2)} = \frac{1}{i\sqrt{a^2 - b^2}}$. Times $2\pi i$ gives the result.
Worked examples
Ejemplo resuelto: The standard integral
Evaluate $\displaystyle\int_0^{2\pi}\frac{d\theta}{5 + 3\cos\theta}$.
- With $z = e^{i\theta}$: $\frac{d\theta}{5 + 3\cos\theta} = \frac{dz}{iz}\cdot\frac{2z}{3z^2 + 10z + 3} = \frac{2\,dz}{i(3z^2 + 10z + 3)}$.
- $3z^2 + 10z + 3 = (3z + 1)(z + 3)$: roots $-\frac{1}{3}$ (inside) and $-3$ (outside).
- Residue at $-\frac{1}{3}$: $\frac{2}{i \cdot 3\left(-\frac{1}{3} + 3\right)} = \frac{2}{8i} = \frac{1}{4i}$.
- $2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2}$, which matches $\frac{2\pi}{\sqrt{25 - 9}}$.
Respuesta: $\frac{\pi}{2}$
Ejemplo resuelto: With sine
Evaluate $\displaystyle\int_0^{2\pi}\frac{d\theta}{13 + 5\sin\theta}$.
- The same method (or the same formula with $\sin$, since shifting $\theta$ by $\frac{\pi}{2}$ turns $\cos$ into $\sin$ over a full period) gives $\frac{2\pi}{\sqrt{169 - 25}}$.
- $\sqrt{144} = 12$, so the integral is $\frac{2\pi}{12} = \frac{\pi}{6}$.
Respuesta: $\frac{\pi}{6}$
Ejemplo resuelto: A double pole inside
Evaluate $\displaystyle\int_0^{2\pi}\frac{d\theta}{(5 + 3\cos\theta)^2}$.
- The integrand becomes $\frac{dz}{iz}\cdot\frac{4z^2}{(3z^2 + 10z + 3)^2} = \frac{4z\,dz}{9i(z + \frac{1}{3})^2(z + 3)^2}$, with a double pole at $-\frac{1}{3}$ inside.
- Residue: $\frac{4}{9i}\frac{d}{dz}\frac{z}{(z + 3)^2}\Big|_{-1/3} = \frac{4}{9i}\cdot\frac{3 - z}{(z + 3)^3}\Big|_{-1/3} = \frac{4}{9i}\cdot\frac{10/3}{(8/3)^3} = \frac{5}{64i}$.
- $2\pi i\cdot\frac{5}{64i} = \frac{5\pi}{32}$ (in general $\frac{2\pi a}{(a^2 - b^2)^{3/2}}$).
Respuesta: $\frac{5\pi}{32}$
- Replace $\cos\theta$, $\sin\theta$ and $d\theta$ using $z = e^{i\theta}$, and simplify to one fraction in $z$.
- Find the poles and keep those with $|z| < 1$.
- Add their residues and multiply by $2\pi i$. The result must be real; if it is not, recheck the factor $\frac{1}{iz}$.
The method needs a full period $0$ to $2\pi$. For $\int_0^{\pi}$, use symmetry first (for example, an even function of $\cos\theta$ over $[0, \pi]$ is half of the integral over $[0, 2\pi]$).
Error común
Forgetting $d\theta = \frac{dz}{iz}$, or counting poles outside the circle. Leaving out the factor $\frac{1}{iz}$ changes the integrand and its poles, and the answer comes out wrong (often not even real). And of the two roots of $bz^2 + 2az + b$, only the one with modulus less than $1$ is inside; including both gives $0$. Check: the original integral is a real number, and positive when the integrand is positive.