Problem 41 - Thomas Calculus

Question

In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int \frac{\csc \theta \cot \theta}{2} d \theta$$

Step-by-Step Solution

Answer

The antiderivative is \(-\frac{1}{2} \csc \theta + C\).

Step 1: Identify the Basic Antiderivative

The expression given is \( \int \frac{\csc \theta \cot \theta}{2} \, d\theta \). Recognize that the integrand can be expressed as \( \frac{1}{2} \csc \theta \cot \theta \) which is the derivative of \( \csc \theta \).

Step 2: Apply the Reverse Differentiation Rule

Since the derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \), we can express the integral as: \[ \int \frac{1}{2} \csc \theta \cot \theta \, d\theta = -\frac{1}{2} \csc \theta + C, \] where \( C \) is the constant of integration.

Step 3: Verify by Differentiation

Differentiate \(-\frac{1}{2} \csc \theta + C \) to ensure it matches the original integrand. The derivative of \( -\frac{1}{2} \csc \theta \) is indeed \( \frac{1}{2} \csc \theta \cot \theta \), confirming that our solution is correct.

Key Concepts

Understanding the Antiderivative

An antiderivative, commonly known as an indefinite integral, is a function whose derivative is the original function given in the integrand. In this exercise, finding the antiderivative of the expression involves recognizing that we must reverse the process of differentiation.
A key part of solving this problem was realizing that the form, \( \frac{\csc \theta \cot \theta}{2} \), is familiar as it resembles the derivative of \( \csc \theta \). By identifying this pattern, we can directly deduce its antiderivative.
This process of reversing differentiation helps in constructing a broader understanding, as there may be multiple antiderivatives due to the constant of integration \( C \), which represents all possible vertical shifts of the function.

Introduction to Trigonometric Integration

Trigonometric integration is a technique used when the integrand is a trigonometric function. Such functions can often be simplified by recognizing standard forms or identities.
The integrand \( \csc \theta \cot \theta \) in the exercise is a perfect example. It's essential to recognize derivatives of basic trigonometric functions because they often reappear as integrands.
Common trigonometric identities and derivatives, like that of \( \csc \theta \), help streamline the process of integration. Familiarity with these can make the task significantly easier, especially when dealing with trigonometric functions such as sine, cosine, or secant.

Differentiation as a Verification Tool

Differentiation serves as a powerful tool to verify the correctness of an indefinite integral solution. Once you determine the antiderivative, differentiating the result should lead you back to the original function.
In our problem, once we found the antiderivative \( -\frac{1}{2} \csc \theta + C \), we differentiated it to check its original form. The process confirmed that the derivative is indeed \( \frac{1}{2} \csc \theta \cot \theta \), matching the integrand.
Thus, differentiation acts like a double-check mechanism, confirming accuracy and understanding each step of integration.

A Peek Into Integration by Parts

While not directly applicable in the provided solution, integration by parts is a vital technique for more complex integrations. It's especially useful when dealing with products of functions.
This technique is based on the product rule for derivatives and involves selecting parts of the integrand to represent \( u \) and \( dv \). You then use the formula:

• \( \int u \, dv = uv - \int v \, du \)

While our example was straightforward, it's vital to be prepared to use integration by parts for intricate expressions or where simple antiderivative recognition isn’t feasible.
Understanding when and how to use integration by parts is crucial for tackling a wide variety of integration problems in calculus.