Chapter 1: Problem 46
For the following exercises, find the antiderivatives for the functions.\(\int \frac{d x}{4-x^{2}}\)
Short Answer
Expert verified
\( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \).
Step by step solution
01
Recognize the Integral Form
The given integral is \( \int \frac{dx}{4-x^2} \). Notice that this is similar to the form \( \int \frac{dx}{a^2-x^2} \) where \( a = 2 \). This form suggests the use of an inverse trigonometric function.
02
Identify the Trigonometric Substitution
Recognize that \( \int \frac{dx}{a^2 - x^2} \) can be solved using the formula \( \int \frac{dx}{a^2 - x^2} = \frac{1}{a} \arcsin\left(\frac{x}{a}\right) + C \). Here, \( a = 2 \), so we'll substitute into the formula.
03
Substitute into the Formula
Using the identified formula, substitute \( a = 2 \) into \( \frac{1}{a} \arcsin\left(\frac{x}{a}\right) + C \). This gives us \( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \).
04
Write the Final Antiderivative
The antiderivative of the given function \( \int \frac{dx}{4-x^2} \) is \( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \), where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivatives
In calculus, an antiderivative of a function is essentially the reverse process of differentiation. When you find the antiderivative of a function, you are trying to determine what function, when differentiated, would produce the original function. This is akin to finding the parent function from its derivative.
An effective way to think about it is like retracing your steps. If you know the rate of change (derivative), finding the antiderivative tells you the original "position" function.
The process of finding antiderivatives is also known as integration, and it is fundamental in solving many calculus problems, such as calculating areas under curves and solving differential equations. In the given problem, we're finding the antiderivative of the function \( rac{1}{4-x^2}\) which requires the recognition of special integration techniques.
An effective way to think about it is like retracing your steps. If you know the rate of change (derivative), finding the antiderivative tells you the original "position" function.
The process of finding antiderivatives is also known as integration, and it is fundamental in solving many calculus problems, such as calculating areas under curves and solving differential equations. In the given problem, we're finding the antiderivative of the function \( rac{1}{4-x^2}\) which requires the recognition of special integration techniques.
Inverse trigonometric functions
Inverse trigonometric functions are the inverse of the standard trigonometric functions, like sine, cosine, and tangent. They are used to find the angle when the value of the trigonometric function is known.
In calculus, these functions often appear in integration problems, especially when dealing with the square roots of expressions involving quadratic forms.
For example, the integral \( \int \frac{1}{4-x^2} \ dx \) utilizes the arcsine, which is the inverse function of sine. The general formula used here is \( \int \frac{1}{a^2-x^2} \ dx = \frac{1}{a} \arcsin\left(\frac{x}{a}\right) + C \), where \( C \) is the constant of integration.
Understanding these inverse functions is crucial because they often simplify the process of integrating complex expressions.
In calculus, these functions often appear in integration problems, especially when dealing with the square roots of expressions involving quadratic forms.
For example, the integral \( \int \frac{1}{4-x^2} \ dx \) utilizes the arcsine, which is the inverse function of sine. The general formula used here is \( \int \frac{1}{a^2-x^2} \ dx = \frac{1}{a} \arcsin\left(\frac{x}{a}\right) + C \), where \( C \) is the constant of integration.
Understanding these inverse functions is crucial because they often simplify the process of integrating complex expressions.
Trigonometric substitution
Trigonometric substitution is a technique in integration that aids in evaluating integrals involving square roots. It often involves replacing variables with trigonometric functions to simplify the integrals.
In the problem \( \int \frac{dx}{4-x^2} \), recognizing that the denominator \( 4-x^2 \) fits the form \( a^2-x^2 \) is key.
The use of the arcsine function comes into play due to this substitution, transforming the integral into a simpler inverse trigonometric form. By substituting \( x = a\sin(\theta) \), other terms in the expression are also converted, which eventually allows the integral to simplify to an easily calculable form.
Keep in mind that these substitutions also often lead to a constant factor outside the integral, which is calculated based on the substitution used.
In the problem \( \int \frac{dx}{4-x^2} \), recognizing that the denominator \( 4-x^2 \) fits the form \( a^2-x^2 \) is key.
The use of the arcsine function comes into play due to this substitution, transforming the integral into a simpler inverse trigonometric form. By substituting \( x = a\sin(\theta) \), other terms in the expression are also converted, which eventually allows the integral to simplify to an easily calculable form.
Keep in mind that these substitutions also often lead to a constant factor outside the integral, which is calculated based on the substitution used.
Constant of integration
When computing indefinite integrals, the constant of integration \( C \) is an essential part of the solution. It represents an arbitrary constant, as indefinite integrals describe a family of functions as opposed to a single function.
The constant of integration emerges because the process of differentiation loses information about the original function's vertical position on a graph. Thus, integrating brings us back to a function that could be shifted up or down without affecting its derivative.
In expressing our final answer for the given problem with \( \int \frac{1}{4-x^2} dx \), the antiderivative is \( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \). Here, \( C \) covers all potential vertical shifts of the arcsine function solution.
Without \( C \), we wouldn’t account for all possible functions that fit this integration scenario.
The constant of integration emerges because the process of differentiation loses information about the original function's vertical position on a graph. Thus, integrating brings us back to a function that could be shifted up or down without affecting its derivative.
In expressing our final answer for the given problem with \( \int \frac{1}{4-x^2} dx \), the antiderivative is \( \frac{1}{2} \arcsin\left(\frac{x}{2}\right) + C \). Here, \( C \) covers all potential vertical shifts of the arcsine function solution.
Without \( C \), we wouldn’t account for all possible functions that fit this integration scenario.
