Problem 44 - Thomas Calculus in SI Units

Question

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

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

Step 1: Simplify the Integral

The given integral to solve is \( \int \frac{1}{2}(\csc^2 x - \csc x \cot x) \, dx \). We can split this expression into two separate integrals: \( \int \frac{1}{2} \csc^2 x \, dx - \int \frac{1}{2} \csc x \cot x \, dx \). Introduce the constant \( \frac{1}{2} \) factor outside each integral for simplicity.

Step 2: Integrate Each Term

To solve these integrals, we use known results from calculus: 1. \( \int \csc^2 x \, dx = -\cot x + C_1 \) 2. \( \int \csc x \cot x \, dx = -\csc x + C_2 \).Applying these gives: \( \frac{1}{2} (-\cot x + C_1) - \frac{1}{2} (-\csc x + C_2) \).

Step 3: Combine the Expressions

Simplify the expression by distributing the \( \frac{1}{2} \) across each term: \( -\frac{1}{2} \cot x + \frac{1}{2} \csc x + C \), where \( C = \frac{1}{2}(C_1 + C_2) \) is the general constant of integration.

Step 4: Verify by Differentiation

Differentiate the expression \( -\frac{1}{2} \cot x + \frac{1}{2} \csc x + C \) to check if we return to the original integrand. The derivative of \( -\cot x \) is \( \csc^2 x \), and the derivative of \( \csc x \) is \( -\csc x \cot x \). Thus, the derivative is indeed \( \frac{1}{2}(\csc^2 x - \csc x \cot x) \), confirming our integral is correct.

Key Concepts

Antiderivative

An antiderivative, often called an indefinite integral, is a vital concept in calculus. It represents a function whose derivative is the given function. Think of it as reversing the process of differentiation.

For instance, if you're given a function and asked for its antiderivative, you're essentially asked to determine what function, when differentiated, will yield the given function.

• In this exercise, the original expression is integrated to find its antiderivative formula.
• We integrate the expression \( \frac{1}{2}(\csc^2 x - \csc x \cot x) \) to find its most general antiderivative.
• The integral contains both constants and variable expressions, indicating potential complexity.

Understanding and calculating antiderivatives are crucial for solving many calculus problems, notably defined integrals and differential equations.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and their reciprocals, like cosecant (\( \csc \)), are foundational in calculus. They often appear in integration and differentiation exercises.

Cosecant (\( \csc x \)) is the reciprocal of sine, while cotangent (\( \cot x \)) is the reciprocal of tangent. In this exercise, these functions play a central role:

• \( \csc^2 x \) and \( \csc x \cot x \) are common in calculus problems due to their straightforward derivatives and integrals.
• Knowledge of trigonometric identities and transformations can simplify these expressions, facilitating easier integration.

Mastering these functions not only aids in solving integrals but also enhances your understanding of oscillatory behavior in real-world phenomena.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. In our exercise, differentiation is used to verify the antiderivative solution.

Key points about differentiation in this context:

• To verify an integral, differentiate the result and ensure it matches the original integrand.
• For \( \cot x \), the derivative is \( -\csc^2 x \), while for \( \csc x \), the derivative is \( -\csc x \cot x \).
• The exercise confirms the integration solution by differentiating \( -\frac{1}{2} \cot x + \frac{1}{2} \csc x + C \) to recover the original integrand.

Differentiation is a powerful tool for checking your work and ensuring accuracy in calculus problems.

Calculus Techniques

Utilizing calculus techniques can simplify complex problems, especially in integration and differentiation. In this instance, several calculus techniques were applied:

Firstly, the expression is split into two simpler integrals, each with a constant coefficient. This method is often used to handle multiple terms more comfortably.

• Handling constants strategically, like factoring \( \frac{1}{2} \) out, simplifies the integration process.
• Employing known integral results, such as those for \( \csc^2 x \) and \( \csc x \cot x \), provides immediate solutions to the integrals without detailed working.

By combining all these techniques, the problem becomes more approachable, ultimately leading to a successful solution and verification.