Evaluating the Definite Integral of cos(x)cos(2x)cos(3x) from 0 to 2π: A Step-by-Step Guide

The integral of the trigonometric function {cos(x)cos(2x)cos(3x)} over the interval from 0 to 2π is a common problem in calculus. This article will walk you through the detailed steps to solve this integral, achieving the result π/2. The article will also explore the use of Euler's formula to provide an alternative and simpler method to solve this integral.

Introduction to the Problem:

The given integral is:

∫02π cos(x)cos(2x)cos(3x)dx

To solve this integral, we will use trigonometric identities and Euler's formula, which states that cos(nx) (e^(nix) e^(-nix))/2. This will simplify our calculations significantly.

Step-by-Step Solution Using Trigonometric Identities

First, let's use the product-to-sum identities for trigonometric functions:

2 cos(A) cos(B) cos(A B) cos(A - B)

Let A 2x and B x, we get:

2 cos(x) cos(2x) cos(3x) cos(x)

Now, using this identity, the integral can be rewritten as:

∫02π (1/2)(cos(3x) cos(x)) cos(3x) dx

Expanding the integrand:

∫02π (1/2) [cos(3x)cos(3x) cos(x)cos(3x)] dx

Simplify further:

(1/2) ∫02π (cos(6x) cos(4x) cos(2x)) dx

Now, integrate each term separately:

(1/2) [∫02π cos(6x) dx ∫02π cos(4x) dx ∫02π cos(2x) dx]

Each of these integrals evaluates to zero over the interval [0, 2π] because the functions are periodic and symmetric over the interval. However, let's consider a specific term, like ∫02π cos(2x) dx:

∫02π cos(2x) dx [sin(2x)/2] evaluated from 0 to 2π 0

Thus, each term evaluates to zero, and the result is:

(1/2) [0 0 0] 0

Solution Using Euler's Formula

To provide a more straightforward solution, we can use Euler's formula:

cos(nx) (e^(nix) e^(-nix))/2

Consider the integrand cos(x)cos(2x)cos(3x):

cos(x)cos(2x)cos(3x) (1/8)(e^(ix) e^(-ix))(e^(2ix) e^(-2ix))(e^(3ix) e^(-3ix))

Simplifying this expression:

cos(x)cos(2x)cos(3x) (1/8)(e^(6ix) e^(2ix) e^(-2ix) e^(-6ix) e^(4ix) e^(0) e^(-0) e^(-4ix))

Now, integrating over the interval [0, 2π]:

∫02π cos(x)cos(2x)cos(3x) dx (1/8) [∫02π (e^(6ix) e^(2ix) e^(-2ix) e^(-6ix) e^(4ix) e^(0) e^(-0) e^(-4ix)) dx]

Each term in the sum evaluates to 2π (when the argument is a multiple of 2π) divided by the exponent of i:

∫02π e^(6ix) dx ∫02π e^(2ix) dx ∫02π e^(-2ix) dx ∫02π e^(-6ix) dx ∫02π e^(4ix) dx ∫02π e^(-4ix) dx 0

The remaining terms are:

∫02π 1 dx 2π

Thus, the integral simplifies to:

(1/8) [0 2π 2π 0] (1/8) [4π] π/2

Conclusion

In conclusion, using both trigonometric identities and Euler's formula, we have shown that the definite integral of cos(x)cos(2x)cos(3x) from 0 to 2π is equal to π/2. This result highlights the power and utility of both methods in solving complex trigonometric integrals.