Question

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

Short answer

The integral evaluates to 0.

Step-by-step solution

Express the Product of Cosines Using Trigonometric Identity

To solve the integral, use the product-to-sum identity for cosine. The identity is: \[ \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \]Substituting in the given function, we have:\[ \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} = \frac{1}{2} \left( \cos \frac{(m+n) \pi x}{L} + \cos \frac{(m-n) \pi x}{L} \right) \]

Write the Integral with the Product-to-Sum Identity

The original integral can now be expressed as:\[ \int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} \, dx = \frac{1}{2} \left( \int_{-L}^{L} \cos \frac{(m+n) \pi x}{L} \, dx + \int_{-L}^{L} \cos \frac{(m-n) \pi x}{L} \, dx \right) \]

Evaluate Each Integral Separately

Let's evaluate each integral separately. Consider:\[ \int_{-L}^{L} \cos \frac{(m+n) \pi x}{L} \, dx \]The integral of a cosine function over a symmetric interval about zero is zero if the argument's multiple of \(\pi\) is not zero.Since \(m eq n\), both \(\int_{-L}^{L} \cos \frac{(m+n) \pi x}{L} dx\) and \(\int_{-L}^{L} \cos \frac{(m-n) \pi x}{L} dx\) evaluate to zero because the integrands are full periods of a cosine function.

Conclude the Solution

As both integrals evaluate to zero:\[ \int_{-L}^{L} \cos \frac{m \pi x}{L} \cos \frac{n \pi x}{L} \, dx = \frac{1}{2} (0 + 0) = 0 \]Thus, the solution to the integral is zero.

Key concepts

Trigonometric Identities

Trigonometric identities are formulas that express relationships between the angles and side lengths of triangles. These identities are incredibly helpful when manipulating and simplifying trigonometric expressions in calculus and algebra. They allow for the transformation of complex expressions into more manageable forms.
A common identity used for cosine is the product-to-sum identity. This identity is particularly useful when dealing with the product of two cosines.
It expresses this product as a sum of two cosines:

• Product-to-Sum Identity: \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \)

Transforming product terms into sum terms simplifies integration. It reduces complex trigonometric integrals into simpler forms, often making calculus operations like definite integration more accessible.

Definite Integrals

Definite integrals are a fundamental concept in calculus used to calculate the accumulation of quantities, such as area under a curve.
Unlike indefinite integrals, which represent antiderivatives, definite integrals compute a specific area bounded by a function and the x-axis between two limits, typically denoted by \( a \) and \( b \).
The notation used is \( \int_{a}^{b} f(x) \, dx\).

• This operation effectively sums all infinitesimal values of \( f(x) \) over the interval \([a, b]\).
• The result is a real number, representing the net area, where areas above the x-axis contribute positively and those below negatively.

Definite integrals also have properties of symmetry. For instance, if \( f(x) \) is an even function, meaning \( f(x) = f(-x) \), and if the limits are symmetric about the origin (from \(-L\) to \(L\)), the integral simplifies significantly, often leading to zero when the integrand completes full wave cycles.

Product-to-Sum Formulas

Product-to-sum formulas are another key tool in integral calculus for simplifying the multiplication of trigonometric functions.
These formulas come in handy when attempting to integrate the product of sines and cosines, which often appear in physical applications and engineering problems.
Instead of struggling with complex products, you can convert these into simple sums or differences.

• For cosine and cosine, the transformation is \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \).
• This conversion simplifies integration by breaking down products into a linear combination of terms that are each easier to integrate.

Applying product-to-sum formulas during integration not only simplifies calculations but also frequently reveals symmetries.
These symmetries, especially in definite integrals over symmetric intervals, may indicate integrals evaluating to zero, as seen with the integral in this exercise.