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application of products of trigonometric functions

Integrating products of trigonometric functions 1

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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:

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

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

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

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

Integrating products of trigonometric functions of sine and cosines

products of sines and cosines are of the form:

$$\int sin^mx cos^nxdx$$

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.

$$\int sin^3xcos^2x = \int sin^2x sinx cos^2x dx = \int (1-cos^x)cos^2x sin x dx $$
$$\int sinxcos^2x dx – \int sinx cos^4xdx = -\frac{1}{3}cos^3x+\frac{1}{5}cos^5x+c$$

case 2

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

$$sin^2θ = \frac{1}{2}(1-cos2θ)$$ $$cos^2θ = \frac{1}{2}(1+cos2θ)$$

As as an example, consider the integral:

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

please note:

$$(1-cos 2x)(1+cos2x) = 1- cos 2x + cos 2x – cos^2 2x = 1- cos^2 2x $$

therefore:

$$=\frac{1}{4} \int (1-cos^2 2x) dx = \frac{1}{4} \int (1- \frac{1}{2}(1+cos 4x)dx$$
$$ \frac{1}{4} \int (1- \frac{1}{2}(1+cos 4x)dx = \frac{1}{4} \int dx – \frac{1}{8} \int dx – \frac{1}{8} \int cos 4x dx$$ $$=\frac{1}{4}x – \frac{1}{8}x – \frac{1}{33} sin 4x + c = \frac{1}{8}(x- \frac{1}{4}sin 4x) +c$$

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