Problem 56 - Calculus

Question

Integrals that occur frequently in applications are $$ \int_{0}^{2 \pi} \cos ^{2} x d x \text { and } \int_{0}^{2 \pi} \sin ^{2} x d x $$ (a) Using a trigonometric identity, show that $$ \int_{0}^{2 \pi}\left(\sin ^{2} x+\cos ^{2} x\right) d x=2 \pi $$ (b) Show from graphical considerations that $$ \int_{0}^{2 \pi} \cos ^{2} x d x=\int_{0}^{2 \pi} \sin ^{2} x d x $$ (c) Conclude that \(\int_{0}^{2 \pi} \cos ^{2} x d x=\int_{0}^{2 \pi} \sin ^{2} x d x=\pi\).

Step-by-Step Solution

Answer

Each integral equals \( \pi \).

Step 1: Use the Pythagorean Identity for Part (a)

The Pythagorean identity states that \( \sin^2 x + \cos^2 x = 1 \). Substituting this into the integral gives us \[ \int_{0}^{2 \pi} (\sin^2 x + \cos^2 x) \, dx = \int_{0}^{2 \pi} 1 \, dx \]. The integral of 1 over \([0, 2\pi]\) is simply the length of the interval, which is \(2\pi\). Therefore, \[ \int_{0}^{2 \pi} (\sin^2 x + \cos^2 x) \, dx = 2\pi \].

Step 2: Graphical Consideration for Part (b)

From the unit circle, \( \cos^2 x \) and \( \sin^2 x \) are both periodic functions with a period of \( \pi \). Over one complete cycle from \( 0 \) to \( 2\pi \), these functions are symmetric about \( \pi \), each occupying an equal area under the curve within this interval. This symmetry implies that \[ \int_{0}^{2 \pi} \cos^2 x \, dx = \int_{0}^{2 \pi} \sin^2 x \, dx \].

Step 3: Conclude the Integrals from Parts (a) and (b)

Since \( \int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx = 2\pi \) and \( \int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx \), let \( I = \int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx \). From these, \( 2I = 2\pi \). Solving for \( I \), we divide both sides by 2, resulting in \( I = \pi \). Therefore, \[ \int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx = \pi \].

Key Concepts

Pythagorean Identity

A fundamental concept in trigonometry is the Pythagorean Identity, which states that for any angle \( x \), the equation \( \sin^2 x + \cos^2 x = 1 \) holds true. This identity is called the Pythagorean Identity because it embodies the relationship similar to the Pythagorean Theorem in right triangles. Each point on the unit circle adheres to this principle because the radius of the unit circle is always 1.

In the context of our exercise, this identity simplifies the integration process. By replacing \( \sin^2 x + \cos^2 x \) with 1, we reframe the integral from an apparently complex form into a straightforward one:

• The integral of 1 over any interval simply equates to the length of that interval, making evaluation much simpler.
• For the given problem, integrating from 0 to \( 2\pi \) results in \( 2\pi \), reflecting the importance of knowing key trigonometric identities for simplifying complex integrals.

Graphical Symmetry

Graphical symmetry in trigonometric functions refers to their repeating and mirror-like visual patterns over specific intervals. Understanding these patterns helps us to compute integrals more intuitively, as seen in the problem statement.

Both \( \cos^2 x \) and \( \sin^2 x \) demonstrate periodicity and symmetry over the interval \( [0, 2\pi] \). This cycle repeats every \( \pi \), and within this framework, each function behaves predictably:

• The graphs of \( \cos^2 x \) and \( \sin^2 x \) are mirror images during each half of \( \pi \).
• This symmetry results in equal areas under the curve between successive peaks and troughs.
• Thus, when integrating over a full period \( [0, 2\pi] \), the areas under \( \cos^2 x \) and \( \sin^2 x \) are equal.

Graphical symmetry not only aids in solving algebraic equations but also in visual verification of solutions, leading to the realization that \( \int_{0}^{2 \pi} \cos^2 x \, dx = \int_{0}^{2 \pi} \sin^2 x \, dx \).

Definite Integration

Definite integration is the process of calculating the area under a curve between two points on the x-axis. In the realm of trigonometric functions, definite integrals can reveal insights about the behavior and properties of these functions over intervals.

From our example, we start by integrating \( \sin^2 x + \cos^2 x \), where the integration yields \( 2\pi \), due to the transformation simplified by the Pythagorean Identity. To incorporate the graphical insights:

• Since we derived that \( \int_{0}^{2\pi} \cos^2 x \) is equal to \( \int_{0}^{2\pi} \sin^2 x \), we assigned these results the common value \( I \).
• Hence, \( 2I = 2\pi \), confirming that each integral \( \int_{0}^{2\pi} \cos^2 x \) and \( \int_{0}^{2\pi} \sin^2 x \) represents an area of \( \pi \).

This calculation illustrates that the combined area under \( \cos^2 x \) and \( \sin^2 x \) within one full cycle is equal to the integral of unity (1) over that interval, demonstrating the power of symmetry and identity in simplifying integration. The results reflect a balanced distribution of area and confirm that integral calculus unravels the nuanced geometry encoded in trigonometric functions.