Trigonometric Integrals | Sahithyan's Notes

Real integrals of the form

\int_0^{2\pi} F(\cos\theta, \sin\theta)\,\text{d}\theta

can be evaluated using residues. Here $F$ is a rational function of $\cos\theta$ and $\sin\theta$ with no poles on the unit circle after the substitution below.

Substitution

Let

z = e^{i\theta} = \cos\theta + i\sin\theta, \quad 0 \le \theta \le 2\pi

Then $z$ moves once counterclockwise around the unit circle $|z| = 1$ , and

\cos\theta = \frac{1}{2}\left(z + \frac{1}{z}\right), \quad \sin\theta = \frac{1}{2i}\left(z - \frac{1}{z}\right), \quad \text{d}\theta = \frac{\text{d}z}{iz}

Thus

\int_0^{2\pi} F(\cos\theta, \sin\theta)\,\text{d}\theta = \oint_{|z|=1} F\left(\frac{1}{2}\left(z + \frac{1}{z}\right), \frac{1}{2i}\left(z - \frac{1}{z}\right)\right) \frac{\text{d}z}{iz}

The right side equals $2\pi i$ times the sum of the residues of the integrand at its poles inside the unit circle.

Examples:

$\int_0^{2\pi} \frac{\text{d}\theta}{1 + 3\cos^2\theta} = \pi$
$\int_0^{2\pi} \frac{\cos 2\theta}{5 - 4\cos\theta}\,\text{d}\theta = \frac{\pi}{6}$

Worked Examples

Even Integrand

Show that $\int_0^{2\pi} \frac{\text{d}\theta}{1 + 3\cos^2\theta} = \pi$ .

Substitute $\cos^2\theta = \frac{1}{4}\left(z + \frac{1}{z}\right)^2 = \frac{z^4 + 2z^2 + 1}{4z^2}$ .

1 + 3\cos^2\theta = \frac{3z^4 + 10z^2 + 3}{4z^2} = \frac{(3z^2+1)(z^2+3)}{4z^2}

\int_0^{2\pi} \frac{\text{d}\theta}{1 + 3\cos^2\theta} = \oint_{|z|=1} \frac{4z^2}{(3z^2+1)(z^2+3)} \cdot \frac{\text{d}z}{iz} = \oint_{|z|=1} \frac{4z}{i(3z^2+1)(z^2+3)}\,\text{d}z

The poles are $z = \pm \frac{i}{\sqrt{3}}$ and $z = \pm i\sqrt{3}$ . Only $\pm \frac{i}{\sqrt{3}}$ lie inside the unit circle.

For $z_0 = \pm \frac{i}{\sqrt{3}}$ , the factor $3z^2 + 1$ has derivative $6z$ , so

\text{Res} = \frac{4z_0}{i \cdot 6z_0 \cdot (z_0^2 + 3)} = \frac{4}{6i \cdot \frac{8}{3}} = \frac{1}{4i}

\int_0^{2\pi} \frac{\text{d}\theta}{1 + 3\cos^2\theta} = 2\pi i \left(\frac{1}{4i} + \frac{1}{4i}\right) = \pi

Higher Harmonic

Show that $\int_0^{2\pi} \frac{\cos 2\theta}{5 - 4\cos\theta}\,\text{d}\theta = \frac{\pi}{6}$ .

Use $\cos 2\theta = \text{Re}(e^{2i\theta}) = \frac{1}{2}\left(z^2 + z^{-2}\right) = \frac{z^4 + 1}{2z^2}$ .

5 - 4\cos\theta = 5 - 2\left(z + \frac{1}{z}\right) = -\frac{(2z-1)(z-2)}{z}

\int_0^{2\pi} \frac{\cos 2\theta}{5 - 4\cos\theta}\,\text{d}\theta = \oint_{|z|=1} g(z)\,\text{d}z, \quad g(z) = -\frac{z^4 + 1}{2i\, z^2 (2z-1)(z-2)}

The poles inside the unit circle are $z = 0$ (order 2) and $z = \frac{1}{2}$ (order 1). The pole $z = 2$ lies outside.

\text{Res}\left(g, \tfrac{1}{2}\right) = -\frac{\frac{1}{16} + 1}{2i \cdot \frac{1}{4} \cdot 2 \cdot \left(-\frac{3}{2}\right)} = \frac{17}{24i}

For $z = 0$ , differentiate $-\frac{z^4 + 1}{2i(2z^2 - 5z + 2)}$ at $0$ .

\text{Res}(g, 0) = -\frac{1}{2i} \cdot \frac{0 \cdot 2 - 1 \cdot (-5)}{2^2} = -\frac{5}{8i}

\int_0^{2\pi} \frac{\cos 2\theta}{5 - 4\cos\theta}\,\text{d}\theta = 2\pi i \left(\frac{17}{24i} - \frac{15}{24i}\right) = \frac{\pi}{6}

Linear Denominator

Evaluate $\int_0^{2\pi} \frac{\text{d}\theta}{2 + \cos\theta}$ .

2 + \cos\theta = \frac{z^2 + 4z + 1}{2z}

\int_0^{2\pi} \frac{\text{d}\theta}{2 + \cos\theta} = \oint_{|z|=1} \frac{2}{i(z^2 + 4z + 1)}\,\text{d}z

The roots of $z^2 + 4z + 1$ are $z_{1,2} = -2 \pm \sqrt{3}$ . Only $z_1 = -2 + \sqrt{3} \approx -0.27$ lies inside the unit circle.

\text{Res} = \frac{2}{i(z_1 - z_2)} = \frac{2}{2\sqrt{3}\, i} = \frac{1}{\sqrt{3}\, i}

\int_0^{2\pi} \frac{\text{d}\theta}{2 + \cos\theta} = 2\pi i \cdot \frac{1}{\sqrt{3}\, i} = \frac{2\pi}{\sqrt{3}}