Problem 77
Question
Compute the indefinite integrals. $$ \int \frac{1}{\sqrt{1-x^{2}}} d x $$
Step-by-Step Solution
Verified Answer
\( \int \frac{1}{\sqrt{1-x^{2}}} dx = \sin^{-1}(x) + C \).
1Step 1: Recognize the Integral Form
The integral \( \int \frac{1}{\sqrt{1-x^{2}}} dx \) is recognized as a standard form that corresponds to the derivative of the inverse trigonometric function \( \sin^{-1}(x) \). This is based on the derivative \( \frac{d}{dx}\left(\sin^{-1}(x)\right) = \frac{1}{\sqrt{1-x^2}} \).
2Step 2: Apply the Antiderivative Formula
Since we have identified the integral as the derivation of \( \sin^{-1}(x) \), we can write the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) as \( \sin^{-1}(x) \). Integrating yields the function \( \sin^{-1}(x) \) plus a constant \( C \).
3Step 3: Write the Final Answer
The indefinite integral \( \int \frac{1}{\sqrt{1-x^{2}}} dx \) evaluates to \( \sin^{-1}(x) + C \), where \( C \) is the constant of integration.
Key Concepts
Inverse Trigonometric FunctionsAntiderivativeCalculus Problem Solving
Inverse Trigonometric Functions
Inverse trigonometric functions are essential in calculus, often used to simplify and evaluate integrals.
They are the inverses of standard trigonometric functions like sine, cosine, and tangent. For example, the inverse of the sine function is known as arcsine, denoted as \( \sin^{-1}(x) \). This means if \( y = \sin(x) \), then \( x = \sin^{-1}(y) \).
With inverse trigonometric functions, the key point is that they allow us to undo the effect of the original trigonometric function over its specific domain. This feature proves particularly useful when dealing with integrals such as \( \int \frac{1}{\sqrt{1-x^2}} dx \), which directly corresponds to the inverse sine function's derivative.
They are the inverses of standard trigonometric functions like sine, cosine, and tangent. For example, the inverse of the sine function is known as arcsine, denoted as \( \sin^{-1}(x) \). This means if \( y = \sin(x) \), then \( x = \sin^{-1}(y) \).
With inverse trigonometric functions, the key point is that they allow us to undo the effect of the original trigonometric function over its specific domain. This feature proves particularly useful when dealing with integrals such as \( \int \frac{1}{\sqrt{1-x^2}} dx \), which directly corresponds to the inverse sine function's derivative.
Antiderivative
The antiderivative is a primary concept in calculus, essentially reverse-engineering the process of differentiation. If you have a function \( f(x) \), its antiderivative could be any function \( F(x) \) such that \( F'(x) = f(x) \). This concept is used to find the original function that was differentiated to get a particular derivative.
In the context of the exercise, the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) can be determined because it matches the derivative of the inverse sine function, \( \sin^{-1}(x) \). So when we integrate \( \frac{1}{\sqrt{1-x^2}} \), the antiderivative is \( \sin^{-1}(x) \). Don't forget to add the constant of integration \( C \), since antiderivatives are not unique and can include any constant term.
In the context of the exercise, the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) can be determined because it matches the derivative of the inverse sine function, \( \sin^{-1}(x) \). So when we integrate \( \frac{1}{\sqrt{1-x^2}} \), the antiderivative is \( \sin^{-1}(x) \). Don't forget to add the constant of integration \( C \), since antiderivatives are not unique and can include any constant term.
Calculus Problem Solving
Solving calculus problems often requires a blend of recognizing patterns and applying integration rules effectively.
When faced with an integral, the first task is to identify its form. Different forms relate to different integration techniques, such as substitution, integration by parts, or applying known derivatives of trigonometric functions.
In this particular problem, recognizing the integral directly reflects the derivative of an inverse trigonometric function. Once it's identified, you apply the integration knowledge that it results in the inverse sine function, plus an arbitrary constant. This approach minimizes unnecessary steps and focuses on rapidly identifying solutions using established calculus rules.
When faced with an integral, the first task is to identify its form. Different forms relate to different integration techniques, such as substitution, integration by parts, or applying known derivatives of trigonometric functions.
In this particular problem, recognizing the integral directly reflects the derivative of an inverse trigonometric function. Once it's identified, you apply the integration knowledge that it results in the inverse sine function, plus an arbitrary constant. This approach minimizes unnecessary steps and focuses on rapidly identifying solutions using established calculus rules.
Other exercises in this chapter
Problem 76
Compute the indefinite integrals. $$ \int\left(1-\frac{x^{2}}{1+x^{2}}\right) d x $$
View solution Problem 76
Verify each inequality without evaluating the integrals. $$ \int_{1}^{2} x d x \leq \int_{1}^{2} x^{2} d x $$
View solution Problem 77
Verify each inequality without evaluating the integrals. $$ 0 \leq \int_{0}^{4} \sqrt{x} d x \leq 8 $$
View solution Problem 78
Compute the indefinite integrals. $$ \int \frac{5}{\sqrt{1-x^{2}}} d x $$
View solution