Trigonometric Integrals

Integrals of the form \displaystyle\int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta — a rational function of sine and cosine over a full period — turn into contour integrals around the unit circle. The bridge is Euler's formula: set z = e^{i\theta} and as \theta runs from 0 to 2\pi, the point z traces |z| = 1 exactly once.

With z = e^{i\theta} the trig functions become rational in z: \cos\theta = \frac{z + z^{-1}}{2}, \qquad \sin\theta = \frac{z - z^{-1}}{2i}.
The measure converts too: dz = i e^{i\theta}\,d\theta = i z\,d\theta, so d\theta = \dfrac{dz}{i z}.
The real integral becomes a contour integral over |z| = 1, evaluated by the residue theorem on the poles inside the unit circle.

Worked: the standard family

Evaluate \displaystyle\int_0^{2\pi} \frac{d\theta}{a + \cos\theta} for a > 1.

Step 1 — substitute. Replace \cos\theta = (z + z^{-1})/2 and d\theta = dz/(iz):

\int_0^{2\pi} \frac{d\theta}{a + \cos\theta} = \oint_{|z|=1} \frac{1}{a + \frac{z + z^{-1}}{2}}\,\frac{dz}{iz}.

Step 2 — clear the fractions. Multiply numerator and denominator inside by 2z so the integrand is rational in z:

= \oint_{|z|=1} \frac{2}{i\,(z^2 + 2az + 1)}\,dz.

Step 3 — locate the poles. Solve z^2 + 2az + 1 = 0:

z_\pm = -a \pm \sqrt{a^2 - 1}.

Since z_+ z_- = 1, exactly one root lies inside the unit circle: z_+ = -a + \sqrt{a^2 - 1} (with |z_+| < 1 for a > 1), while z_- sits outside.

Step 4 — take the residue at the interior pole. Write the denominator as (z - z_+)(z - z_-); the residue of the integrand \dfrac{2}{i\,(z - z_+)(z - z_-)} at z_+ is

\operatorname{Res} = \frac{2}{i\,(z_+ - z_-)} = \frac{2}{i \cdot 2\sqrt{a^2 - 1}} = \frac{1}{i\sqrt{a^2 - 1}}.

Step 5 — apply the residue theorem. Multiply by 2\pi i:

\int_0^{2\pi} \frac{d\theta}{a + \cos\theta} = 2\pi i \cdot \frac{1}{i\sqrt{a^2 - 1}} = \frac{2\pi}{\sqrt{a^2 - 1}}.

For example, with a = 2 the integral is \dfrac{2\pi}{\sqrt{3}}.

A definite integral over a full period [0, 2\pi] is secretly a loop, because e^{i\theta} returns to where it began. That single observation converts an awkward trigonometric average into a rational contour integral, and the residue theorem finishes it with one pole. The condition a > 1 is exactly what keeps the integrand finite on the real axis and pushes one root inside the circle and the other out — so the geometry of the poles encodes the analysis of the integral.

Slide the parameter and move the pole

The contour is the unit circle |z| = 1. The transformed integrand has poles at z_\pm = -a \pm \sqrt{a^2 - 1}; raise the parameter a with the slider and watch the interior pole z_+ drift while z_- stays outside. The readout reports the integral's value 2\pi/\sqrt{a^2 - 1}, which shrinks as a grows.