Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} d x, m \neq n, m, n\) integers.

Step-by-Step Solution

Answer

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

Step 1: Understand the integral structure

The exercise requires finding the integral of \( \cos \left( \frac{m \pi x}{L} \right) \cos \left( \frac{n \pi x}{L} \right) \) over the interval \([-L, L]\). The function is a product of cosine functions with different frequencies.

Step 2: Apply trigonometric identity

Use the trigonometric identity for the product of cosines: \[ \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \]Substitute \( A = \frac{m \pi x}{L} \) and \( B = \frac{n \pi x}{L} \) into the identity. Then, the integral becomes:\[ \int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} \, dx = \frac{1}{2} \int_{-L}^{L} \left[ \cos \left( \frac{(m+n) \pi x}{L} \right) + \cos \left( \frac{(m-n) \pi x}{L} \right) \right] \, dx \]

Step 3: Evaluate the integral of each term

Evaluate the integrals separately:1. \( \int_{-L}^{L} \cos \left( \frac{(m+n) \pi x}{L} \right) \, dx \)2. \( \int_{-L}^{L} \cos \left( \frac{(m-n) \pi x}{L} \right) \, dx \)For \( \int \cos(ax) \, dx \), the result over symmetric limits is zero if \( a eq 0 \) because cosine is an odd periodic function centered around zero for these limits.

Step 4: Determine the result of the integrals

Evaluate each integral:1. \( \int_{-L}^{L} \cos \left( \frac{(m+n) \pi x}{L} \right) \, dx = 0 \) because cosine oscillates symmetrically over \([-L, L]\).2. \( \int_{-L}^{L} \cos \left( \frac{(m-n) \pi x}{L} \right) \, dx = 0 \) because cosine oscillates symmetrically over \([-L, L]\).Thus, the initial integral simplifies to:\[ \frac{1}{2} (0 + 0) = 0 \]

Step 5: Conclude the integral result

Since both component integrals separately evaluate to zero, the original integral is also zero. Therefore, \( \int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} \, dx = 0 \) for \( m eq n \).

Key Concepts

Trigonometric Identities

Trigonometric identities are powerful tools used to simplify expressions involving trigonometric functions. They allow us to transform complex trigonometric expressions into simpler forms. In this exercise, we use the identity for the product of cosines:

• \( \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \)

This identity helps break down the product of two cosine functions into a sum of two cosines. By substituting \(A\) and \(B\) with the respective angles \(\frac{m \pi x}{L}\) and \(\frac{n \pi x}{L}\), the integral of a product of cosines can be transformed into a more manageable form. Applying such identities is a key step in simplifying and solving integrals involving trigonometric functions.

Definite Integrals

Definite integrals allow us to calculate the area under a curve over a given interval. In this context, our integral \( \int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} \, dx \) represents the total signed area between the curve and the x-axis from \(-L\) to \(L\).Key points:

• The limits of integration \([-L, L]\) reveal symmetry because the function's behavior will be mirrored over the y-axis.
• Integration of periodic functions like cosine often involves examining symmetry properties, which can simplify calculations.

We're particularly interested in how the integral resolves when transformed by trigonometric identities, which often aids in interpreting or simplifying outcomes, such as determining when these integrals result in zero.

Product of Cosines

The product of cosines \( \cos \left( \frac{m \pi x}{L} \right) \cos \left( \frac{n \pi x}{L} \right) \) in this problem represents a challenge due to the different frequencies of cosine functions. The frequencies appear as multiples of \(\pi/L\) and determine how oscillations occur over the interval.Essential points include:

• Using trigonometric identities to rewrite the product as a sum of cosines: \[ \frac{1}{2} \int_{-L}^{L} \left[ \cos \left( \frac{(m+n) \pi x}{L} \right) + \cos \left( \frac{(m-n) \pi x}{L} \right) \right] \, dx \]
• The roles of \(m\) and \(n\) being unequal ensures that each integral components like \(m+n\) or \(m-n\) remain non-zero, maintaining a higher frequency and causing the integral pairs to evaluate to zero within symmetric limits.

Symmetric Limits

Symmetric limits refer to the interval \([-L, L]\), where a function is evaluated over a symmetric region about the origin. This symmetry often simplifies computations because certain properties of trigonometric functions can be effectively utilized.Key benefits of using symmetric limits include:

• When the expression of interest involves even functions, such as cosine, symmetry can lead directly to cancellation of positive and negative areas, often resulting in zero over the entire interval.
• In these limits, the integral of cosine terms oscillating over full periods averages to zero, if the period fits an integer number of oscillations. This is particularly seen when dealing with functions like \( \cos(ax) \) when evaluated on \([-T, T]\); they sum to zero if \(a eq 0\).

Therefore, leveraging symmetric limits is a valuable tactic in integral calculus to simplify outcomes, predict results, and offer clarity in solutions involving periodic functions.