Problem 15 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\csc x \cot x \quad\) b. \(-\csc 5 x \cot 5 x \quad\) c. \(-\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2}\)

Step-by-Step Solution

Answer

a. \\(-\csc x + C\\), b. \\(-\frac{1}{5} \csc 5x + C\\), c. \\(-2 \csc \frac{\pi x}{2} + C\\)."

Step 1: Understand the Integral of csc(x)cot(x)

To find an antiderivative of the function \(\csc x \cot x\), recognize that the derivative of \(\csc x\) is \(-\csc x \cot x\). Therefore, the antiderivative of \(\csc x \cot x\) is \(-\csc x\) plus a constant \(C\).

Step 2: Solve Part a

Given \(\csc x \cot x\), we know that an antiderivative is \(-\csc x + C\), where \(C\) is the constant of integration.

Step 3: Handle Negative Coefficient in Part b

For the function \(-\csc 5x \cot 5x\), observe that by chain rule adjustment, we need to factor in \(\frac{1}{5}\) for the inner function derivative. Thus, the antiderivative is \(-\frac{1}{5} \csc 5x + C\).

Step 4: Process Pi Coefficient in Part c

For \(-\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2}\), note that the antiderivative involves using the chain rule on \(\frac{\pi x}{2}\). The inner derivative is \(\frac{\pi}{2}\). Therefore, the antiderivative is \(\frac{2 \pi}{\pi} \cdot \left( -\csc \frac{\pi x}{2} \right) + C\), simplifying to \(-2 \csc \frac{\pi x}{2} + C\).

Step 5: Verify by Differentiation

Differentiate each derived antiderivative to check correctness. For example, differentiating \(-\csc x\) gives \(-\csc x \cot x\), matching the original function, confirming correctness.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc x \), is the reciprocal of the sine function: \( \csc x = \frac{1}{\sin x} \). This means it expresses how many times larger 1 is than the sine of \( x \). Knowing its derivative is essential in finding its antiderivative. The derivative of \( \csc x \) is \(-\csc x \cot x \), which forms the basis for solving antiderivatives involving the cosecant function.

The cotangent is another trigonometric function, \( \cot x = \frac{\cos x}{\sin x} \), that pairs with the cosecant in these integrals. When combined, \( \csc x \cot x \) results as part of differentiation involving \( \csc x \). Understanding these relationships is key to solving integrals involving trigonometric functions.

Integration Techniques

Integration is the reverse process of differentiation and involves determining the antiderivative of a function. Different strategies apply depending on function types. Here, we focus on techniques for trigonometric integrals, particularly those involving \( \csc x \cot x \).

Knowing the derivatives of common trigonometric functions allows for reverse calculation (integration). In this exercise:

• Recognize patterns between functions and their derivatives.
• Use substitution where necessary to simplify expressions.

Using these techniques, the integration becomes more accessible and less daunting, even without detailed calculations. This highlights the mental benefit of practice and familiarity with trigonometric behaviors.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It plays a crucial role when integrating expressions where an outside function's derivative depends on an inside function. Often, it demands adjustment to integrate properly.

In part (b) with \(-\csc 5x \cot 5x\), the chain rule appears as we need to account for the derivative of \(5x\) (the inner function), which is \(5\). Thus, for the antiderivative, we adjust by multiplying by \(\frac{1}{5}\), leading to:

• Inner derivative = \(5\)
• Antiderivative: \(-\frac{1}{5} \csc 5x + C\)

The chain rule ensures accurate adjustment of coefficients, maintaining the function's structure during integration.

Constant of Integration

In integration, the constant of integration, usually denoted as \(C\), represents an indefinite yet integral part of the antiderivative. Because differentiation of a constant yields zero, every indefinite integral needs \(C\) to encompass all possible functions that differentiate back to the original function.

When finding an antiderivative, always include \(C\) to represent the infinite set of parallel curves that could fit. Without this constant, the solution lacks generality, limiting interpretation to a single, specific curve rather than a family of solutions, crucial in indefinite integrals.