Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Integrals that occur frequently in applications are \(\int_{0}^{2 \pi} \cos ^{2} x d x\) 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: Apply the Pythagorean Identity

Begin by using the Pythagorean trigonometric identity: \(\sin^2 x + \cos^2 x = 1\). Substitute this identity into the integral \(\int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx\). The integral simplifies to \(\int_{0}^{2\pi} 1 \, dx\).

Step 2: Evaluate Constant Integral

Evaluate the integral \(\int_{0}^{2\pi} 1 \, dx\). The integral of 1 over the interval from 0 to \(2\pi\) is simply \(2\pi\). Therefore, \(\int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx = 2\pi\).

Step 3: Use Symmetry and Graphical Interpretation

Recognize that \(\cos^2 x\) and \(\sin^2 x\) are symmetric over the interval from 0 to \(2\pi\). Graphically, both functions cover equal areas under the curve since they are complementary, which implies that \(\int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx\).

Step 4: Conclude Equal Integrals

Since the total integral \(\int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx\) is \(2\pi\) and both \(\cos^2 x\) and \(\sin^2 x\) integrals are equal, each contributes half of the total integrals' value. Thus, \(\int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx = \pi\).

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry, expressing a fundamental relationship between sine and cosine. The identity states that for any angle \( x \), \( \sin^2 x + \cos^2 x = 1 \). This is incredibly useful in integral calculus as it allows us to simplify complex trigonometric expressions.

In the given exercise, we used the Pythagorean Identity to transform the integral \( \int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx \) into a much simpler expression: \( \int_{0}^{2\pi} 1 \, dx \). By recognizing that \( \sin^2 x + \cos^2 x \) simplifies to 1, we no longer have to deal with the separate sine and cosine terms.

To evaluate the integral of a constant over an interval, you multiply the constant by the length of the interval. Since the constant is 1 and the interval from 0 to \( 2\pi \) has a length of \( 2\pi \), we find that \( \int_{0}^{2\pi} 1 \, dx = 2\pi \). This application of the Pythagorean Identity quickly reduces the complexity of the integral.

Integral Calculus

Integral calculus is a branch of calculus focused on the accumulation of quantities and the areas under curves. In the context of this problem, we are dealing specifically with definite integrals, which compute the area under a curve between two points on the x-axis.

For example, we calculated \( \int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx \), which gives us the total area under the curve of the expression \( \sin^2 x + \cos^2 x \) over one full period. Our use of the Pythagorean Identity helped simplify this problem significantly.

Furthermore, understanding that \( \int_{0}^{2\pi} 1 \, dx = 2\pi \) reveals how constant functions translate into areas. This fundamental principle of integral calculus is pivotal when determining areas for more complex expressions. It's important to remember that such integrals not only find tangible areas when applied to physics and engineering but also help analyze periodic functions, as seen here with trigonometry.

Symmetry in Trigonometric Functions

Symmetry plays a key role in simplifying integrals of trigonometric functions. Both \( \cos^2 x \) and \( \sin^2 x \) exhibit symmetry over the interval from 0 to \( 2\pi \). When examining the graphs of these functions, it's evident that each one creates patterns that are mirror images when shifted appropriately along the x-axis.

This symmetry implies that the area under the curve for \( \cos^2 x \) from \( 0 \) to \( 2\pi \) is equal to that for \( \sin^2 x \) over the same interval. Because of this symmetry, our integrals for \( \cos^2 x \) and \( \sin^2 x \) yield the same value without needing detailed computation.

Graphically interpreting these functions shows how both cover equal areas despite their sine and cosine origins. Thus, the total integral \( \int_{0}^{2\pi} (\sin^2 x + \cos^2 x) \, dx = 2\pi \) is divided equally between them, leading to \( \int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx = \pi \). This conclusion leverages both graphical understanding and mathematical symmetry to simplify the analysis.