Problem 14 - Thomas Calculus in SI Units

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\csc ^{2} x\) b. \(-\frac{3}{2} \csc ^{2} \frac{3 x}{2}\) b. \(-\frac{3}{2} \csc ^{2} \frac{3 x}{2}\)

Step-by-Step Solution

Answer

a. \(-\cot x + C\), b. \(\cot \left( \frac{3x}{2} \right) + C\).

Step 1: Understand the Function to Integrate

Identify the functions given in the exercise for which we need to find antiderivatives. We have two functions: (a) \( \csc^2 x \) and (b) \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \).

Step 2: Recall Basic Antiderivatives

Recall that the antiderivative of \( \csc^2 u \) is \( -\cot u \). This is because \( \frac{d}{dx}(-\cot x) = \csc^2 x \).

Step 3: Find the Antiderivative for Part A

Using the formula from Step 2, the antiderivative for \( \csc^2 x \) is \( -\cot x + C \), where \( C \) is the integration constant.

Step 4: Consider the Chain Rule for Part B

For the function \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \), let \( u = \frac{3x}{2} \), so \( du = \frac{3}{2} dx \). This means the integral becomes \( -\frac{3}{2} \int \csc^2(u) \, du \).

Step 5: Antiderivative for Part B

Apply the antiderivative formula: \( -\cot u \) as \( u = \frac{3x}{2} \). Thus, the antiderivative is \( \cot \left( \frac{3x}{2} \right) + C \).

Step 6: Differentiate to Verify the Answer

Differentiate \( -\cot x \) to verify part a: \( \csc^2 x \), and differentiate \( \cot \left( \frac{3x}{2} \right) \) for part b to check that the differentiation returns \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \). This confirms the antiderivatives are correct.

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point. It is a fundamental concept in calculus and is used to calculate things like velocity, slope, and rate of change. For example, if we have a function \( y = f(x) \), the derivative of \( y \) with respect to \( x \) is denoted by \( \frac{dy}{dx} \).

In the context of antiderivatives, differentiation is used to verify our solution. If we have correctly found an antiderivative of a function, differentiating that antiderivative should give us back the original function.

For instance:

• Given the function \( \csc^2 x \), we know its derivative must be \( -\cot x \) because \( \frac{d}{dx}(-\cot x) = \csc^2 x \).
• Similarly, checking an antiderivative involves ensuring that applying the differentiation rules returns the initial given function.

Chain Rule

The Chain Rule is a formula for computing the derivative of the composition of two or more functions. It's a vital tool in calculus when dealing with complex functions. The rule states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Mathematically, if we have two functions: \( f(g(x)) \), the chain rule states that \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \).

In the exercise provided, we use the chain rule to find the antiderivative of a function following substitution. For the function \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \):

• We set \( u = \frac{3x}{2} \), making \( \csc^2(u) \) part of the inner function.
• \( du = \frac{3}{2} dx \), which helps us in identifying the integration process. This substitution simplifies integration.

This results in an antiderivative where we can verify the correctness by applying our understanding of the chain rule during differentiation.

Integration Constants

Integration constants are an essential part of finding antiderivatives. Whenever we find the antiderivative of a function, we must include an arbitrary constant known as the constant of integration, denoted as \( C \).

This constant emerges because the process of differentiation eliminates constants. Differentiating a constant results in zero, losing any information about its original value before differentiation. Therefore, when we work backward to find an antiderivative, we have a "family" of solutions differing by a constant.

For any antiderivative, we usually express it as:

• \( f(x) + C \), where other solutions are simply vertical shifts along the \( y \)-axis.

In our example, the antiderivative of \( \csc^2 x \) is expressed as \( -\cot x + C \). Similarly, for \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \), the solution becomes \( \cot \left( \frac{3x}{2} \right) + C \). This "\( C \)" ensures every possible initial condition is covered in our solution.