Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The integral is 0.

Step 1: Use Trigonometric Identity

To simplify the product of cosines, use the trigonometric identity:\[\cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B))\]In this problem, set \(A = mx\) and \(B = nx\), so\[\cos(mx)\cos(nx) = \frac{1}{2} (\cos((m+n)x) + \cos((m-n)x))\]

Step 2: Integrate Each Term

Split the integral using the identity from Step 1:\[\int_{-\pi}^{\pi} \cos mx \cos nx \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} \cos((m+n)x) \, dx + \int_{-\pi}^{\pi} \cos((m-n)x) \, dx \right)\]Now, integrate each term separately.

Step 3: Evaluate the First Integral

Consider the integral:\[\int_{-\pi}^{\pi} \cos((m+n)x) \, dx = \left[ \frac{\sin((m+n)x)}{m+n} \right]_{-\pi}^{\pi}\]Since \(\sin((m+n)\pi) = \sin(-(m+n)\pi) = 0\), the integral evaluates to 0.

Step 4: Evaluate the Second Integral

Consider the integral:\[\int_{-\pi}^{\pi} \cos((m-n)x) \, dx = \left[ \frac{\sin((m-n)x)}{m-n} \right]_{-\pi}^{\pi}\]Similarly, \(\sin((m-n)\pi) = \sin(-(m-n)\pi) = 0\), this integral also evaluates to 0.

Step 5: Final Evaluation

Using the results from Steps 3 and 4:\[\frac{1}{2} (0 + 0) = 0\]Thus, the integral evaluates to 0.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical formulas that relate the angles and ratios of right triangles. These identities are incredibly useful in simplifying expressions involving trigonometric functions like sine, cosine, and tangent. In this exercise, we utilized the identity

• \(\cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B))\)
to simplify a product of cosines into a sum of cosines. This transformation makes it easier to evaluate the integral since dealing with sums is often simpler than dealing with products. Breaking a complex expression into simpler components is a key strategy in calculus.

• Setting \(A = mx\) and \(B = nx\), we derived that \(\cos(mx)\cos(nx) = \frac{1}{2} (\cos((m+n)x) + \cos((m-n)x))\).

Understanding these identities is an essential skill for solving a vast array of problems in calculus and beyond, making them a fundamental part of any mathematician's toolkit.

Definite Integrals

Definite integrals are a core concept in calculus, representing the signed area under a curve over a specific interval. When computing a definite integral, two main steps are involved:

• First, find the antiderivative of the function.
• Then, evaluate it at the bounds of the interval.

In this exercise, the definite integral of \(\cos mx \cos nx\) is computed from \(-\pi\) to \(\pi\). The bounds make our calculations simpler since trigonometric functions like sine and cosine have known values at important points like \(\pi\) and \(-\pi\).
Given the properties of sine and cosine, these bounds often lead to nice cancellations, as they do here, resulting in an integral value of zero. Understanding the properties of definite integrals is crucial for interpreting and solving a wide variety of mathematical and real-world problems.

Orthogonality of Sine and Cosine

Orthogonality is a concept borrowed from linear algebra where two vectors are considered orthogonal if their dot product is zero. In the context of functions, particularly sine and cosine functions, orthogonality implies that the integral of their product over a symmetric interval centered at zero is zero, whenever the frequencies differ.This exercise showcases the orthogonality of cosine functions with different frequencies, \(m\) and \(n\), over the interval \([-\pi, \pi]\). Since \(m eq n\), the integral of their product evaluates to zero, demonstrating that these functions do not 'interfere' with each other over this interval.

• Sine and cosine functions with differing arguments are often orthogonal over such intervals, leading to significant simplifications in problems like signal processing and Fourier analysis.

Appreciating this concept allows mathematicians and engineers to efficiently decompose complex signals into simpler, non-interfering components.

Integration of Trigonometric Functions

Integration of trigonometric functions is a common task in calculus. While anti-derivatives of basic trigonometric functions are well-known, problems can become challenging when dealing with products or compositions of these functions.This exercise demonstrates how we transform a product of cosine functions into a sum, using known identities, to simplify integration. Once in this simplified form,

• Each cosine function’s integral is found using the antiderivative \(\frac{\sin kx}{k}\), where \(k\) represents the respective frequency \(m+n\) or \(m-n\).
• Boundary conditions, such as \(\sin(k\pi) = 0\) for integer \(k\), help evaluate these integrals to zero.

Mastering the integration of trigonometric functions involves recognizing when to apply identities and understanding periodic properties. These techniques are not only applicable in mathematics but in physics, engineering, and any field involving wave-like phenomena.