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

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:

$\displaystyle \sin^3 x$ $\displaystyle = \left( \frac {1}{2i} (e^{ix}-e^{-ix}) \right)^3$    
$\displaystyle \quad$ $\displaystyle =-\frac {1}{8i} ( e^{3ix}-3e^{2ix}e^{-ix}+3e^{ix}e^{-2ix}-e^{-3ix})$    
$\displaystyle \quad$ $\displaystyle = -\frac {1}{8i} ( e^{3ix}-e^{-3ix} - 3e^{ix} + 3e^{-ix})$    
$\displaystyle \quad$ $\displaystyle = -\frac {1}{8i} ( 2i \sin 3x - 2i \sin x )$    
$\displaystyle \quad$ $\displaystyle =- \frac 14 \sin 3x + \frac 14 \sin x.$    

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$ .

Noah Dana-Picard 2007-12-28