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Integrating products of trigonometric functions

Integrating products of trigonometric functions

Products of trigonometric functions usually refers to the integral of a product of trigonometric functions over a specified range. It can be interpreted as finding area under the curve made by a trigonometric function. For example the function:

$$\int sin(x)cos(x)dx$$

The equation represents an area under the curve formed by the product of sine and cosine functions. The integral results to

$$\int -cos(x)sin(x)dx$$

in a geometrical terms, finding the integral of products of trigonometric functions can be described as finding an area under the curve defined by those functions over a certain interval. for example:

$$\int_{0}^{\pi} sin(x)cos(x)dx = -\frac{1}{4}cos(2x)+C$$

represents the area between the product of sin(x) and cos (x) functions and the x-axis from x =0 to x=π.

The result of the above integral is 0.25

The topic of integrals trigonometric products is important because these integrals are useful in areas like:

  • Fourier series analysis
  • modelling of wave interference patterns in physics
  • describing mechanical vibrations and oscillations
  • used to describe AC circuits and their oscillatory behaviors

Trigonometric functions of sine and cosines

products of sines and cosines are of the form:

$$\int sin^m(x)cos^n(x)dx$$

Integrating products trigonometric functions involves use of trigonometric identities like double angle identities depending on the form of the integral.

case 1

At least one of the indices m and n is an odd positive integer. If m is an odd positive integer , then we isolate the one one sine.

consider the identity:

$$sin^2x = 1-cos^2x$$

we express the remaining sinm x as sin(m-1) x and then express it in terms of cos x. That is:

$$\int sin^m(x)cos^n(x)dx = \int sin^{m-1}(x)sin(x)cos^n(x)dx$$

As an example, consider the following expression.

$$\int sin^3(x)cos^2(x)dx = \int sin^2(x)sin(x)cos^2(x)dx = \int(1-cos^2x)cos^2xsinxdx$$
$$=\int sinxcos^2xdx – \int sinxcos^4xdx = -frac{1}{3}cos^3x + \frac{1}{5}cos^5x+c$$
Integrating products of trigonometric functions

case 2

If both m and n are non-negative even integer, we use the half angle formula which states:

$$sin^2 \theta = \frac{1}{2}(1-cos2 \theta)$$
$$xcos^2 \theta = \frac{1}{2}(1+cos2 \theta)$$

As as an example, consider the integral:

$$sin(x)cos^2(x)dx = \int \frac{1}{2}(1-cos2x)\frac{1}{2}(1+cos2x)dx$$

please note:

(1−cos2x)(1+cos2x)=1−cos2x+cos2x–cos22x=1−cos22x

therefore:

$$\frac{1}{4}(1-cos^22x)dx = \frac{1}{4} \int (1-\frac{1}{2}(1+cos4x)dx$$
$$\frac{1}{4} \int (1-\frac{1}{2}(1+cos4x)dx=\frac{1}{4}dx – \frac{1}{8}dx-\frac{1}{8}cos4xdx$$

14∫(1−12(1+cos4x)dx=14∫dx–18∫dx–18∫cos4xdx=14x–18x–133sin4x+c=18(x−14sin4x)+c

$$\frac{1}{4}x – \frac{1}{8}x – \frac{1}{32}sin4x + c = \frac{1}{8}(x-\frac{1}{4}sin4x)+c $$

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