Problem 29 - Calculus

Question

Use the table of integrals inside the back cover to show that the following statements are true for positive integers \(k\) and \(m\). $$\int_{-\pi}^{\pi} \sin (k x) \cos (m x) d x=0$$

Step-by-Step Solution

Answer

The integral equals zero due to the orthogonality of sine and cosine functions over the given interval.

Step 1: Define the Problem

We need to evaluate the integral \( \int_{-\pi}^{\pi} \sin (kx) \cos (mx) \, dx \) for positive integers \(k\) and \(m\), and show that it equals zero.

Step 2: Recall the Orthogonality Property of Sine and Cosine

The functions \(\sin(n x)\) and \(\cos(m x)\) are orthogonal over the interval \([-\pi, \pi]\). That means the integral of their product over this interval is zero unless \(n = m\). Since \(k\) and \(m\) are given as different positive integers, this property can be directly applied.

Step 3: Simplify the Integral Using Orthogonality

Apply the orthogonality property: \[\int_{-\pi}^{\pi} \sin(kx) \cos(mx) \, dx = \frac{1}{2} \int_{-\pi}^{\pi} \left( \sin((k+m)x) + \sin((k-m)x) \right) \, dx\]Both components \(\sin((k+m)x)\) and \(\sin((k-m)x)\) integrate to zero over the interval \([-\pi, \pi]\), confirming that the integral of their initial product is zero.

Step 4: Final Evaluation

Both parts of the integral \( \int_{-\pi}^{\pi} \sin(kx) \cos(mx) \, dx = 0 \) because each term in the expression \( \sin(Ax) \) integrates to zero over the period \([-\pi, \pi]\). Therefore, the entire integral evaluates to zero.

Key Concepts

Orthogonality Principle

The Orthogonality Principle is a fundamental concept in mathematics, especially in the study of trigonometric functions. When we say that functions are orthogonal, it means their inner product over a specific interval is zero. For - \(\sin(kx)\) and \(\cos(mx)\),this principle holds truefor an interval \[\[-\pi, \pi\]\]. If you compute the integral of the product of these functions within this interval, the result is- zero.Being orthogonal means no overlapping influence in the context of trigonometric integral evaluation. This property is incredibly useful because it simplifies many problems, allowing us to solve complex integrals by recognizing them as orthogonal pairs.

Trigonometric Integrals

Trigonometric Integrals involve the integral of trigonometric functions. Among the various properties that these integrals possess, one particularly handy property is their behavior under multiplication. When integrating a product of sine and cosine with differing frequencies:- \(\sin(kx)\) and \(\cos(mx)\),for positive integers \(k\) and \(m\), the values will mutually cancel each other over the interval. This cancellation arises due to their distinct cycles not being in phase with each other, leading to a net integral of zero. Understanding the behavior of these integrals reduces complicated expressions to simple solutions. In calculus, integrating such functions might seem tedious, but leveraging trigonometric identities can make our calculations far simpler.

Definite Integral Evaluation

Evaluating a Definite Integral means finding the signed area under a curve within specific bounds. For integrals such as- \(\int_{-\pi}^{\pi} \sin(kx) \\cos(mx) \dx\),definite bounds from \(-\pi\) to \(\pi\) mean calculating the area while considering both positive and negative contributions along the x-axis. Using their orthogonal property, these trig functions complete cycles within this interval, hence why integrals lead to zero: their effects across the interval cancel each other perfectly. Evaluating these integrals becomes a question of recognizing the periodic, repeatable patterns in sine and cosine functions, removing any labor from the computation except pattern recognition.