Problem 5
Question
Evaluate (be careful if \(n=m\) ) $$ \int_{0}^{L} \sin \frac{n \pi x}{L} \sin \frac{m \pi x}{L} d x \quad \text { for } n>0, m>0 . $$ Use the trigonometric identity $$ \sin a \sin b=\frac{1}{2}[\cos (a-b)-\cos (a+b)] . $$
Step-by-Step Solution
Verified Answer
The result is 0 when n is not equal to m and L/2 when n equals m. These results follow from the evaluation of the integral and considering the special case when n equals m.
1Step 1: Implement the trigonometric identity
First, replace \(sin(a)sin(b)\) in the integrand with the provided formula. Both \(a\) and \(b\) represent angles here. In this case, \(a = \frac{n\pi x}{L}\) and \(b = \frac{m\pi x}{L}\). Hence the integrand becomes \(\frac{1}{2}[cos(\frac{n\pi x}{L} - \frac{m\pi x}{L}) - cos(\frac{n\pi x}{L} + \frac{m\pi x}{L})]\).
2Step 2: Split the integral
The integral will be split into two parts corresponding to the two terms in the integrand: integral from 0 to L of \(\frac{1}{2}\cos(\frac{n\pi x}{L} - \frac{m\pi x}{L}) dx\) and integral from 0 to L of \(\frac{1}{2}\cos(\frac{n\pi x}{L} + \frac{m\pi x}{L}) dx\). The integration of each part is done separately.
3Step 3: Evaluate the integrals
The integral of cos(\(ax\)) from 0 to L is \(\frac{1}{a}\sin(ax)\) evaluated from 0 to L. Apply the rules of definite integral to each integral to calculate the result.
4Step 4: Consider the special case
If n=m, then the original integral reduces to integral from 0 to L of \(\sin^2(\frac{n\pi x}{L}) dx\). This requires a different derivation by using the identity \(\sin^2(a) = \frac{1-\cos(2a)}{2}\).
Key Concepts
Trigonometric IdentityDefinite IntegralsFourier Series
Trigonometric Identity
Understanding trigonometric identities is crucial when dealing with integrals involving trigonometric functions, as they often allow for simplification of the integrand. In the context of our problem, we are given the identity \( \sin a \sin b = \frac{1}{2}[\cos(a-b) - \cos(a+b)] \). This identity transforms the product of two sine functions into a sum of cosine functions, which are easier to integrate. To get a better grasp on this, let's take a closer look at how this identity is used.
When you have an integral like the one in the exercise involving \( \sin \frac{n \pi x}{L} \sin \frac{m \pi x}{L} \), it can appear challenging at first. However, by substituting \( a = \frac{n\pi x}{L} \) and \( b = \frac{m\pi x}{L} \), the identity simplifies the complexity. Now, instead of dealing with the product of two sines, you can handle two separate cosines after splitting them, which is a task one is often more familiar with.
When you have an integral like the one in the exercise involving \( \sin \frac{n \pi x}{L} \sin \frac{m \pi x}{L} \), it can appear challenging at first. However, by substituting \( a = \frac{n\pi x}{L} \) and \( b = \frac{m\pi x}{L} \), the identity simplifies the complexity. Now, instead of dealing with the product of two sines, you can handle two separate cosines after splitting them, which is a task one is often more familiar with.
Definite Integrals
Definite integrals are foundational in calculus, and understanding them is vital for solving many mathematical problems. These integrals allow you to calculate the accumulated quantity, such as area under a curve, when the limits of integration are specified. In our exercise, we are dealing with the definite integral from 0 to L, which implies we’re interested in the area under the curve between these two bounds.
The process generally involves finding the antiderivative of the function and then applying the Fundamental Theorem of Calculus, which states that you can evaluate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit. The technique becomes slightly more complex when dealing with trigonometric functions, but by recognizing patterns and applying known identities, as we've done with our trigonometric identity, this process can be made significantly more manageable. It’s like breaking down a larger, complex problem into smaller, more conquerable parts.
The process generally involves finding the antiderivative of the function and then applying the Fundamental Theorem of Calculus, which states that you can evaluate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit. The technique becomes slightly more complex when dealing with trigonometric functions, but by recognizing patterns and applying known identities, as we've done with our trigonometric identity, this process can be made significantly more manageable. It’s like breaking down a larger, complex problem into smaller, more conquerable parts.
Fourier Series
The Fourier series is an essential concept in mathematical analysis, particularly in the study of periodic functions. It expresses any periodic function as a sum of simple oscillating functions, namely sines and cosines. The integral we are evaluating in our exercise is closely related to the coefficients that appear in the Fourier series.
In essence, when we calculate the integrals of products of sines and cosines over a period, we are determining the strength of the corresponding frequency components that make up a complex signal. So, when you deal with exercises like this one, you're engaging with the building blocks of what is known as Fourier analysis. Fourier series enable us to decompose complex functions into fundamental frequencies, a principle that has numerous applications in fields like signal processing, physics, and engineering. Each integral you solve could represent a step towards deciphering how different waves can construct or approximate more complex patterns.
In essence, when we calculate the integrals of products of sines and cosines over a period, we are determining the strength of the corresponding frequency components that make up a complex signal. So, when you deal with exercises like this one, you're engaging with the building blocks of what is known as Fourier analysis. Fourier series enable us to decompose complex functions into fundamental frequencies, a principle that has numerous applications in fields like signal processing, physics, and engineering. Each integral you solve could represent a step towards deciphering how different waves can construct or approximate more complex patterns.
Other exercises in this chapter
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