Trigonometric polynomials.

Warning:

In this subsection you need some basic knowledge on complex numbers.

We call here a trigonometric polynomial a linear combination of terms of the form $\cos ^p x \cdot \sin ^q x$ , where $p$ and $q$ are non negative integers. To compute the integral of a trigonometric polynomial, we must first of all linearize it. For this purpose, we use the so-called Euler's formulae:

$$\begin{cases}\cos x = \frac 12 (e^{ix}+e^{-ix}) \sin x = \frac {1}{2i} (e^{ix}-e^{-ix}) \end{cases}$$

We will not develop the general case, but give some examples:

Example 8.2.12   Let $F(x)= \int \cos^2 x \; dx$ .

We have:

$\cos^2 x = \left( \frac 12 (e^{ix}+e^{-ix}) \right)^2 = \frac 14 ( e^{2ix} + 2 + e^{-2ix} ) = \frac (2 \cos 2x +2 ) = \frac 12 \cos 2x + \frac 12$ . Therefore: $F(x)= \int \cos^2 x \; dx=\frac 12 \int \cos 2x \; dx + \frac 12 \int dx =\frac 14 \sin 2x + \frac 12 x +C$ . Check this result either by differentiation, or by using standard trigonometric identities.

Example 8.2.13   Let $F(x)= \int \sin^3 x \; dx$ .

We have:

$$\sin^3 x = \left( \frac {1}{2i} (e^{ix}-e^{-ix}) \right)^3$$

$$\quad =-\frac {1}{8i} ( e^{3ix}-3e^{2ix}e^{-ix}+3e^{ix}e^{-2ix}-e^{-3ix})$$

$$\quad = -\frac {1}{8i} ( e^{3ix}-e^{-3ix} - 3e^{ix} + 3e^{-ix})$$

$$\quad = -\frac {1}{8i} ( 2i \sin 3x - 2i \sin x )$$

$$\quad =- \frac 14 \sin 3x + \frac 14 \sin x.$$

Therefore: $F(x)= \int\sin^3 x \; dx = - \frac 14 \int \sin 3x \; dx + \frac 14 \int \sin x \; dx =\frac {1}{12} \cos 3x - \frac 14 \cos x +C$ .