Problem 44 - 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. $$\int \frac{1}{2}\left(\csc ^{2} x-\csc x \cot x\right) d x$$

Step-by-Step Solution

Answer

\(-\frac{1}{2} \cot x - \frac{1}{2} \csc x + C\) is the general antiderivative.

Step 1: Understand the Expression

The given expression is the indefinite integral: \( \int \frac{1}{2}\left(\csc ^{2} x-\csc x \cot x\right) d x \). Our task is to find the antiderivative for this expression.

Step 2: Break Down the Integral

First, separate the integral based on the terms given: \( \int \frac{1}{2} \csc^2 x \, dx - \int \frac{1}{2} \csc x \cot x \, dx \). This allows us to focus on each term individually.

Step 3: Find the Antiderivative of the First Term

For the first integral \( \int \frac{1}{2} \csc^2 x \, dx \), we know from standard antiderivative rules that \( \int \csc^2 x \, dx = -\cot x \). Thus, the antiderivative is \( -\frac{1}{2} \cot x + C_1 \), where \( C_1 \) is a constant of integration.

Step 4: Find the Antiderivative of the Second Term

For the second integral \( \int \frac{1}{2} \csc x \cot x \, dx \), the known rule \( \int \csc x \cot x \, dx = -\csc x \) is applied. Hence, the antiderivative is \( -\frac{1}{2} \csc x + C_2 \), with \( C_2 \) being another constant of integration.

Step 5: Combine the Antiderivatives

Combine the results from Steps 3 and 4 to form the general antiderivative: \( -\frac{1}{2} \cot x - \frac{1}{2} \csc x + C \), where \( C = C_1 + C_2 \) is the overall constant of integration.

Step 6: Verify by Differentiation

Differentiate the combined antiderivative: \( \frac{d}{dx} \left( -\frac{1}{2} \cot x - \frac{1}{2} \csc x + C \right) \). Ensure it equals the original function inside the integral: \( \frac{1}{2} \csc^2 x - \frac{1}{2} \csc x \cot x \). Verification confirms correctness.

Key Concepts

Antiderivative

The concept of an antiderivative is fundamental in calculus. An antiderivative of a function is a function whose derivative is the original function. In other words, if you have a function \( f(x) \), an antiderivative \( F(x) \) is such that \( F'(x) = f(x) \).

• Antiderivatives are vital for calculating indefinite integrals.
• Indefinite integrals do not have limits of integration, hence they yield a family of functions, differing only by a constant.
• The process of finding antiderivatives is called integration.

In the exercise, we find the antiderivative of the expression \( \int \frac{1}{2}(\csc^2 x - \csc x \cot x) \, dx \) by solving each part of the integral separately. Using known antiderivative forms helps in solving standard integral problems efficiently.

Cosecant

The cosecant function, denoted as \( \csc x \), is the reciprocal of the sine function, meaning \( \csc x = \frac{1}{\sin x} \). The cosecant function is less commonly used than sine or cosine, but it plays a crucial role in certain trigonometric integrals and antiderivatives.

• Like other trigonometric functions, \( \csc x \) is periodic, with a period of \( 2\pi \).
• The function has vertical asymptotes at every integer multiple of \( \pi \), where sine is zero.
• In calculus, the integral \( \int \csc^2 x \, dx \) and \( \int \csc x \cot x \, dx \) are standard forms, useful for evaluating integrals quickly.

While \( \csc^2 x \) and \( \csc x \cot x \) might seem complex, utilizing known integral rules can simplify these terms significantly, as seen in the exercise.

Integration Constant

In indefinite integrals, the integration constant is a crucial element. When you calculate an indefinite integral, you are essentially finding a family of functions. Each function in this family differs only by a constant. This constant is known as the integration constant and is represented by \( C \).

• This constant accounts for the fact that many functions can have the same derivative.
• Without it, the solution to an integral would be incomplete.
• The exact value of \( C \) can be determined if an initial condition is provided.

In the exercise solution, the integration constant is combined from two constants, \( C_1 \) and \( C_2 \), achieved during the integration of separate terms. This illustrates how each antiderivative has its own constant, combined for the final general solution. It reminds us to always include \( C \) for any indefinite integral.