Chapter 4: Problem 14
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. \(-\frac{3}{2} \csc ^{2} \frac{3 x}{2}\)
Short Answer
Expert verified
a. \(-\cot x + C\), b. \(\cot \left( \frac{3x}{2} \right) + C\).
Step by step solution
01
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) \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \).
02
Recall Basic Antiderivatives
Recall that the antiderivative of \( \csc^2 u \) is \( -\cot u \). This is because \( \frac{d}{dx}(-\cot x) = \csc^2 x \).
03
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.
04
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 \( -\frac{3}{2} \int \csc^2(u) \, du \).
05
Antiderivative for Part B
Apply the antiderivative formula: \( -\cot u \) as \( u = \frac{3x}{2} \). Thus, the antiderivative is \( \cot \left( \frac{3x}{2} \right) + C \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \( -\frac{3}{2} \csc^2 \left( \frac{3x}{2} \right) \):
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.
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:
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.