Question

$$ \text { Find } \int_{-\pi}^{\pi} \cos m x \cos n x d x, m \neq n ; m, n \text { integers. } $$

Short answer

The integral evaluates to 0 when \(m \neq n\).

Step-by-step solution

Understand the integral

We need to evaluate the definite integral \( \int_{-\pi}^{\pi} \cos(mx)\cos(nx)\,dx \) where \(m eq n\) and both \(m\) and \(n\) are integers. This is a product of two cosine functions with different frequencies.

Use trigonometric identities

To simplify the integral, use the trigonometric identity: \[\cos A \cos B = \frac{1}{2}(\cos(A+B) + \cos(A-B))\]This means that\[\cos(mx)\cos(nx) = \frac{1}{2}(\cos((m+n)x) + \cos((m-n)x))\]

Split the integral

Substitute the identity into the integral:\[\int_{-\pi}^{\pi} \cos(mx)\cos(nx)\,dx = \frac{1}{2} \int_{-\pi}^{\pi} \cos((m+n)x)\,dx + \frac{1}{2} \int_{-\pi}^{\pi} \cos((m-n)x)\,dx\]

Evaluate each integral separately

We need to evaluate each term separately. First consider\[\int_{-\pi}^{\pi} \cos((m+n)x)\,dx\]Since \(\cos((m+n)x)\) is an odd function about the y-axis when \(m+n eq 0\), and the interval is symmetric about 0, this integral evaluates to 0.

Consider the second integral

Now consider\[\int_{-\pi}^{\pi} \cos((m-n)x)\,dx\]Similar reasoning applies here since \(m eq n\), and thus \(m-n eq 0\). Therefore, this integral also evaluates to 0.

Conclude the solution

Since both integrals evaluate to 0, the result of the entire integral is:\[\int_{-\pi}^{\pi} \cos(mx) \cos(nx) \, dx = 0\] when \(m eq n\).

Key concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions, like sine and cosine, and are true for all values of the variable. They are crucial in simplifying complex trigonometric expressions. In the context of integrals, these identities can transform a product of trigonometric functions into a simpler form that is often easier to integrate.

One essential identity used in this exercise is the product-to-sum formula. When dealing with the product of two cosine functions, the identity \[\cos A \cos B = \frac{1}{2}(\cos(A+B) + \cos(A-B))\] converts the product into a sum of cosines. This transformation is particularly helpful for integration purposes as it breaks down a complex product into simpler parts.

Understanding and applying these identities is a fundamental skill when working with trigonometric integrals. It allows for the dealing of integrals involving trigonometric functions more efficiently, especially in cases where directly integrating the product would be complicated or unwieldy.

Definite Integrals

A definite integral represents the area under a curve within a given interval, often providing useful quantities like total displacement, area, or even work, depending on the problem at hand. In this exercise, we focus on the definite integral from \(-\pi\) to \(\pi\).

The notation \(\int_{a}^{b} f(x)\,dx\) tells us to find the net area under the curve \(f(x)\) between the limits \(x = a\) and \(x = b\). This calculation accounts for the fact that area above the x-axis is considered positive, whereas area below is negative. This inherent property is essential for symmetry arguments, particularly in this exercise.

When evaluating the integral of trigonometric functions over symmetric intervals, such as \(-\pi\) to \(\pi\), properties of even and odd functions can significantly simplify the process. Here, since the cosine function forms part of the expression, understanding these properties helps quickly determine if the integral resolves to zero due to symmetry.

Orthogonality of Functions

Orthogonality in the context of functions, particularly trigonometric functions, plays a vital role in simplifying problems involving integrals. Two functions are considered orthogonal over a given interval if their product integrates to zero over that interval.

In this specific exercise, we investigate the orthogonal properties of cosine functions. When considering the integral \(\int_{-\pi}^{\pi} \cos(mx) \cos(nx) \, dx\) with \(m eq n\), we aim to show that these functions are orthogonal. The transformed integrals resulting from trigonometric identities yield zero due to this orthogonality property.

This principle of orthogonality is pivotal, especially in fields like signal processing and Fourier analysis, where distinct frequency components need to be separated clearly. By proving orthogonality through integration, we demonstrate that two different frequency cosine functions, when multiplied and integrated over a symmetric interval, cancel each other's contributions, producing a zero result.