Problem 124 - Calculus Volume 1

Question

Suppose that \(A=\int_{0}^{2 \pi} \sin ^{2} t d t\) and \(B=\int_{0}^{2 \pi} \cos ^{2} t d t\). Show that \(A+B=2 \pi\) and \(A=B\).

Step-by-Step Solution

Answer

\(A + B = 2\pi\) and \(A = B = \pi\).

Step 1: Understand the Problem

We need to evaluate two integrals: \(A = \int_{0}^{2\pi} \sin^2 t \, dt\) and \(B = \int_{0}^{2\pi} \cos^2 t \, dt\). Then, we need to find \(A + B\) and show that \(A + B = 2\pi\) and \(A = B\).

Step 2: Use a Trigonometric Identity

Recall the trigonometric identity: \(\sin^2 t + \cos^2 t = 1\). This identity will help us simplify the expression for \(A + B\).

Step 3: Compute A + B

Using the trigonometric identity \(\sin^2 t + \cos^2 t = 1\), the integral \(A + B\) becomes \(\int_{0}^{2\pi} (\sin^2 t + \cos^2 t) \, dt = \int_{0}^{2\pi} 1 \, dt\).

Step 4: Evaluate the Integral of 1

The integral \(\int_{0}^{2\pi} 1 \, dt\) is simply \(2\pi\) because the integral of 1 over an interval \([0, 2\pi]\) results in the value of the interval's length. Therefore, \(A + B = 2\pi\).

Step 5: Use Symmetry Property

The functions \(\sin^2 t\) and \(\cos^2 t\) are symmetrical over the interval \([0, 2\pi]\). This symmetry implies that the areas under each curve over the interval are equal. Therefore, \(A = B\).

Step 6: Use Average Value of Trigonometric Functions

Since \(\sin^2 t\) and \(\cos^2 t\) have the same average value of \(\frac{1}{2}\) over the interval \([0, 2\pi]\), each integral evaluates to \(\pi\). Therefore, \(A = B = \pi\).

Key Concepts

Definite Integral

A definite integral represents the area under a curve between two points on the x-axis, providing a precise numerical value. We calculate definite integrals by integrating a function over a specific interval. In the context of trigonometric integration, a definite integral can help solve problems involving periodic functions, such as sine and cosine.
The problem at hand uses specific definite integrals:

• For the integral \(A = \int_{0}^{2\pi} \sin^2 t \, dt\), we are finding the area under the curve of \(\sin^2 t\) from 0 to \(2\pi\).
• Similarly, \(B = \int_{0}^{2\pi} \cos^2 t \, dt\) gives the area for \(\cos^2 t\).

To simplify these, we often utilize trigonometric identities. The challenge of definite integrals in trigonometry lies in their ability to transform periodic functions effectively, providing symmetry and uniformity across the interval.

Trigonometric Identities

Trigonometric identities are essential tools that simplify expressions involving trigonometric functions. In calculus, these identities are invaluable, especially for integrals involving sine and cosine. The core identity for this problem is \(\sin^2 t + \cos^2 t = 1\). This identity allows us to simplify the expression for \(A + B\).
Here's how it works:

• The expression \(\sin^2 t + \cos^2 t\) equals 1 for any \(t\), simplifying calculation of the integral.
• Using \(\sin^2 t + \cos^2 t = 1\), the problem becomes integrating \(1\) over \([0, 2\pi]\), which is much simpler.

These identities help streamline the integration process by reducing complex integrals to more manageable forms. By applying identities, we see that the integral becomes a matter of evaluating basic calculus operations, making it easier to determine that \(A + B = 2\pi\).

Symmetry in Calculus

Symmetry plays a crucial role in solving integrals, especially when periodic functions like sine and cosine are involved. Both \(\sin^2 t\) and \(\cos^2 t\) are symmetrical functions over the interval \([0, 2\pi]\). This means they exhibit repetitive patterns that simplify their integration.
By employing the concept of symmetry, we see that the areas under the curves for \(\sin^2 t\) and \(\cos^2 t\) are identical over one full period. This reveals:

• The symmetry ensures \(A = B\), indicating that both contribute equally to the total area.
• Moreover, the symmetry about \(\pi\) divides each function into two identical halves, further confirming the equality of \(A\) and \(B\).

Symmetrical behavior allows us to deduce properties about their integrals without detailed calculations, leading to quick solutions such as proving \(A = B = \pi\). This demonstrates symmetry's power as a problem-solving tool in calculus.