Problem 25 For any positive integers $m$ ... [FREE SOLUTION]

Chapter 2: Problem 25

For any positive integers $m$ and $n,$ show that $$\int_{0}^{2 \pi} \sin (m x) \cos (n x) d x=0 \text { , }$$ $\int_{0}^{2 \pi} \sin (m x) \sin (n x) d x=\left\\{\begin{array}{ll}0, & \text { if } m \neq n \\ \pi & \text { if } m=n\end{array}\right.$ and $\int_{0}^{2 \pi} \cos (m x) \cos (n x) d x=\left\\{\begin{array}{ll}0, & \text { if } m \neq n \\ \pi, & \text { if } m=n\end{array}\right.$

Short Answer

Expert verified

All integrals evaluate to zero unless $m = n$, except $\int \sin^2(mx)$ and $\int \cos^2(mx)$ which result in $\pi.$

Step by step solution

Understand the Integrals Involving Sine and Cosine

We need to compute the integrals of the following forms over the interval $[0, 2\pi]$: $\int_0^{2\pi} \sin(mx) \cos(nx) \, dx$, $\int_0^{2\pi} \sin(mx) \sin(nx) \, dx$, and $\int_0^{2\pi} \cos(mx) \cos(nx) \, dx$.

Evaluate the Mixed Sine and Cosine Integral

Consider $\int_0^{2\pi} \sin(mx) \cos(nx) \, dx$. This integral can be evaluated by using the product-to-sum identities. Specifically, use $\sin(mx) \cos(nx) = \frac{1}{2}[\sin((m+n)x) + \sin((m-n)x)]$. The integral becomes:\[\frac{1}{2}\int_0^{2\pi} [\sin((m+n)x) + \sin((m-n)x)] \, dx.\]Since the integral of any sine function over a complete period from 0 to $2\pi$ is zero, both terms yield zero:$\int_0^{2\pi} \sin(kx) \, dx = 0$ for any integer $k$. Thus, $\int_0^{2\pi} \sin(mx) \cos(nx) \, dx = 0$.

Evaluate the Sine-Sine Integral

Consider $\int_0^{2\pi} \sin(mx) \sin(nx) \, dx$. Use the identity $\sin(mx) \sin(nx) = \frac{1}{2}[\cos((m-n)x) - \cos((m+n)x)]$. The integral becomes:\[\frac{1}{2} \int_0^{2\pi} [\cos((m-n)x) - \cos((m+n)x)] \, dx.\]For $m eq n$, both $\cos((m-n)x)$ and $\cos((m+n)x)$ integrate to zero over $[0, 2\pi]$. Thus, the integral is zero.For $m = n$, the expression simplifies to $\frac{1}{2} \int_0^{2\pi} [1 - \cos(2mx)] \, dx$, which equals $\pi$.

Evaluate the Cosine-Cosine Integral

Consider $\int_0^{2\pi} \cos(mx) \cos(nx) \, dx$. Use the identity $\cos(mx) \cos(nx) = \frac{1}{2}[\cos((m-n)x) + \cos((m+n)x)]$. The integral becomes:\[\frac{1}{2} \int_0^{2\pi} [\cos((m-n)x) + \cos((m+n)x)] \, dx.\]For $m eq n$, both terms integrate to zero over $[0, 2\pi]$. Thus, the integral is zero.For $m = n$, the expression simplifies to $\frac{1}{2} \int_0^{2\pi} [1 + \cos(2mx)] \, dx$, which equals $\pi$.

Conclusion

In conclusion, for the given integrals, when evaluating over the interval $[0, 2\pi]$, the results are as follows:- $\int_{0}^{2\pi} \sin(mx) \cos(nx) \ dx = 0$.- $\int_{0}^{2\pi} \sin(mx) \sin(nx) \ dx = 0$ if $m eq n$ and $\pi$ if $m = n$.- $\int_{0}^{2\pi} \cos(mx) \cos(nx) \ dx = 0$ if $m eq n$ and $\pi$ if $m = n$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Integrals

When working with trigonometric functions, integrals involving products of sine and cosine over specific intervals come up frequently, particularly in problems involving orthogonality. The intervals are often chosen such that they represent the full period, like \[ \int_0^{2\pi} \sin(mx) \cos(nx) \, dx \, \\int_0^{2\pi} \sin(mx) \sin(nx) \, dx, \\text{and} \\int_0^{2\pi} \cos(mx) \cos(nx) \, dx, \]where $m$ and $n$ are positive integers. These integrals reveal important properties when $m eq n$ or $m = n$.

The sine-cosine integral evaluates to zero, because sine and cosine are orthogonal functions over a period from 0 to $2\pi$.
The sine-sine and cosine-cosine integrals evaluate to zero when $m eq n$, reinforcing the concept of orthogonality in pairs of sine or cosine functions of different frequency.
When $m = n$, the sine-sine and cosine-cosine integrals evaluate to $\pi$, reflecting a symmetry and uniform distribution over the interval.

Product-to-Sum Identities

The product-to-sum identities are invaluable in simplifying the integrals of trigonometric functions. These identities transform products of sines and cosines into sums or differences of sines and cosines. This transformation helps break down complex integrals into more manageable forms.
For example:

$ \sin(mx) \cos(nx) = \frac{1}{2}[\sin((m+n)x) + \sin((m-n)x)] $
$ \sin(mx) \sin(nx) = \frac{1}{2}[\cos((m-n)x) - \cos((m+n)x)] $
$ \cos(mx) \cos(nx) = \frac{1}{2}[\cos((m-n)x) + \cos((m+n)x)] $

These identities simplify evaluations because the integrals of $\sin(kx)$ and $\cos(kx)$ over $[0, 2\pi]$ are often zero, rendering the expressions easier to calculate.

Fourier Series

The study of trigonometric integrals is closely related to Fourier series, which decompose periodic functions into sums of sine and cosine terms. Fourier series rely on the orthogonality of trigonometric functions, which means the integral of the product of different sine and cosine functions over a full period is zero.
Fourier series help:

Represent more complex periodic functions simply as infinite sums of sines and cosines.
Break down functions or signals into components based on their frequency content, which is a principal idea in signal processing.
Utilize the coefficients of the sine and cosine terms that are computed using integrals like those explored in the exercise.

Understanding this concept reveals how integral evaluations contribute to practical applications like signal analysis and solving physical problems.

Trigonometric Identities

Trigonometric identities are essential tools in simplifying and solving problems involving trigonometric functions. These identities relate different trigonometric ratios to each other and include simplifications and transformations such as product-to-sum and sum-to-product identities.
Some commonly used identities include:

Pythagorean identities like $ \sin^2(x) + \cos^2(x) = 1 $.
Angle sum and difference identities, such as $ \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y $.
Double angle identities, such as $ \sin(2x) = 2\sin x \cos x $ and $ \cos(2x) = \cos^2 x - \sin^2 x $.

These identities make it easier to manipulate and integrate trigonometric functions, providing a foundation for deeper exploration into topics like the orthogonality of functions and Fourier series. By mastering these identities, you can efficiently solve a wide range of mathematical and real-world problems.

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Short Answer

Step by step solution

Understand the Integrals Involving Sine and Cosine

Evaluate the Mixed Sine and Cosine Integral

Evaluate the Sine-Sine Integral

Evaluate the Cosine-Cosine Integral

Conclusion

Key Concepts

Trigonometric Integrals

Product-to-Sum Identities

Fourier Series

Trigonometric Identities

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