Problem 11 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-28\). $$\int_{\pi / 4}^{3 \pi / 4} \csc \theta \cot \theta d \theta$$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Identify the Integral Form

The given integral is \( \int_{\pi / 4}^{3 \pi / 4} \csc \theta \cot \theta \, d\theta \). The integrand \( \csc \theta \cot \theta \) is a standard derivative, which immediately suggests a specific antiderivative that we can use.

Step 2: Recall the Antiderivative Formula

Recall that the derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \). Thus, we know that \( \int \csc \theta \cot \theta \, d\theta = -\csc \theta + C \), where \( C \) is the constant of integration.

Step 3: Evaluate the Antiderivative Over the Given Limits

Using the antiderivative, we find \(-\csc \theta\), and evaluate it from \( \pi/4 \) to \( 3\pi/4 \). Thus, \[\left[ -\csc \theta \right]_{\pi/4}^{3\pi/4} = -\csc(3\pi/4) + \csc(\pi/4).\]

Step 4: Calculate \(\csc(\theta)\) at Specific Angles

We know that \( \csc(\pi/4) = \frac{1}{\sin(\pi/4)} = \frac{1}{\sqrt{2}/2} = \sqrt{2} \). Similarly, \( \csc(3\pi/4) = \frac{1}{\sin(3\pi/4)} = \frac{1}{\sqrt{2}/2} = \sqrt{2} \).

Step 5: Substitute and Simplify

Substitute the known values of \( \csc(\theta) \) back into the expression:\[-\csc(3\pi/4) + \csc(\pi/4) = -\sqrt{2} + \sqrt{2} = 0.\]Therefore, the value of the definite integral is 0.

Key Concepts

Antiderivatives

When working with integrals, one crucial concept is the antiderivative. An antiderivative, essentially, is a function whose derivative matches the given integrand. In the context of the exercise, we are dealing with the function \( \csc \theta \cot \theta \). By recognizing this function as the derivative of \( -\csc \theta \), we effectively find the antiderivative. This means:

• The integral \( \int \csc \theta \cot \theta \, d\theta \) equals \( -\csc \theta + C \).
• \( C \) is the constant of integration which is unnecessary when evaluating definite integrals owing to prescribed limits.

Antiderivatives are powerful because they provide a direct way to solve many integrals by shifting the problem into a simpler subtraction of function evaluations. For definite integrals, understanding their connection with antiderivatives enables us to calculate exact areas instead of needing to approximate them.

Trigonometric Integrals

Trigonometric integrals are integrals that involve trigonometric functions like sine, cosine, tangent, and their reciprocals. In this exercise, we work with \( \csc \theta \cot \theta \), which are reciprocals of sine and cosine respectively. Evaluating these integrals usually entails recognizing standard forms or relationships:

• For instance, \( \csc \theta \cot \theta \) is commonly known to be the derivative of \( -\csc \theta \).
• In this way, using trigonometric identities or derivatives helps quickly find antiderivatives for these integrals.

In this problem, applying trigonometric knowledge allows direct identification of the antiderivative, simplifying the process. Working with these functions can seem daunting initially due to their complexity and periodic nature, yet practice and familiarity with trigonometric properties streamline solving related integrals.

Integral Calculus

Integral calculus is the branch of mathematical analysis concerned with calculating integrals, primarily to determine the accumulation of quantities or areas under curves. The integral calculus divides into two main types: definite and indefinite integrals.

• Definite integrals provide numerical values and concern the area under a curve between two limits.
• Indefinite integrals give a general form of antiderivatives of a function, usually including the constant of integration, \( C \).

In this exercise, we focus on evaluating a definite integral. We use the antiderivative found via our trigonometric identities to compute the exact value between set limits \( \pi/4 \) and \( 3\pi/4 \). After evaluating \( -\csc(3\pi/4) + \csc(\pi/4) \), the result is zero, indicating that there is a perfect balance of area that cancels out over the specified interval. This integral calculus approach circumvents tedious geometric calculations, reflecting the elegance of calculus in bridging arithmetic operations with geometric interpretations.