Chapter 2: Problem 28
Find the integral by using the simplest method. Not all problems require integration by parts. $$ \int \sin ^{-1} x d x $$
Short Answer
Expert verified
\(x\sin^{-1}x + \sqrt{1-x^2} + C\).
Step by step solution
01
Understand the Formula to Apply
Recognize that the integral of the inverse sine function \(\int \sin^{-1} x \, dx\) requires integration by parts, where we use \(u = \sin^{-1} x\) and \(dv = dx\). This is because the function inside the integral does not have a straightforward antiderivative.
02
Identify u and dv
Choose \(u = \sin^{-1} x\) (since the derivative of inverse functions can simplify the problem) and \(dv = dx\). Thus, we will later need to compute \(du\) and \(v\).
03
Compute du and v
Calculate the derivatives: \(du\) is the derivative of \(u\), given by \(du = \frac{1}{\sqrt{1 - x^2}} \, dx\), and \(v\) is the antiderivative of \(dv\), given by \(v = x\).
04
Apply Integration by Parts
Use the integration by parts formula \(\int u \, dv = uv - \int v \, du\) with our previous results:\(\int \sin^{-1} x \, dx = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx\).
05
Simplify and Integrate Remaining Integral
To solve \(\int \frac{x}{\sqrt{1 - x^2}} \, dx\), recognize it as a simple substitution integral, let \(w = 1-x^2\), thus \(dw = -2x \, dx\). Solve for x \(dx\) using this substitution.
06
Finalize the Computation
Substituting back and changing the limits (if definite), compute the integral as shown and simplify to find:\(\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\sqrt{1-x^2} + C\).Putting it all together:\(\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1-x^2} + C\).
07
Verify and Present the Final Solution
Ensure the solution is simplified and rechecked: the final answer being \(x \sin^{-1} x + \sqrt{1-x^2} + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Functions
Inverse trigonometric functions, like \( \sin^{-1}(x) \), are the reverse of the usual trigonometric functions. They let us find angles when the sine, cosine, or tangent values are known. This is especially useful in integration problems where the antiderivatives of these inverse functions need to be determined.
Unlike the basic trigonometric functions, inverse trigonometric functions cannot be integrated directly because they don't have simple antiderivatives. Instead, we use other methods like integration by parts to help us solve the integral. This process involves choosing appropriate parts to derive and integrate, making it easier to handle the complexity of inverse functions.
One key point about inverse trigonometric functions is that they are only defined for certain domains. For example, \( \sin^{-1}(x) \) is defined for \( x \in [-1, 1] \). Remembering these domains can help avoid errors in solving the problems involving them.
To sum up, understanding how inverse trigonometric functions work and how to apply them in the context of integration is essential. This prepares you to tackle integrals that may otherwise seem daunting.
Unlike the basic trigonometric functions, inverse trigonometric functions cannot be integrated directly because they don't have simple antiderivatives. Instead, we use other methods like integration by parts to help us solve the integral. This process involves choosing appropriate parts to derive and integrate, making it easier to handle the complexity of inverse functions.
One key point about inverse trigonometric functions is that they are only defined for certain domains. For example, \( \sin^{-1}(x) \) is defined for \( x \in [-1, 1] \). Remembering these domains can help avoid errors in solving the problems involving them.
To sum up, understanding how inverse trigonometric functions work and how to apply them in the context of integration is essential. This prepares you to tackle integrals that may otherwise seem daunting.
Antiderivatives
Antiderivatives are essentially the reverse process of derivatives. While a derivative shows how a function changes, an antiderivative provides a way to recover the original function given its derivative. This is directly linked to the concept of integration, where we seek the antiderivative to solve problems.
In cases where functions do not have straightforward antiderivatives, like inverse trigonometric functions, we employ methods such as integration by parts. This helps partition the function into manageable parts where we can handle both differentiation and integration easily.
When working with antiderivatives, it's crucial to remember to add the constant of integration, \( + C \). This represents the infinite set of possible original functions that could have resulted in the given derivative.
Understanding how to find antiderivatives efficiently gives you a powerful tool in solving integrals. Whether you are working with simple polynomials or complex inverse functions, mastering antiderivatives is a fundamental skill in calculus.
In cases where functions do not have straightforward antiderivatives, like inverse trigonometric functions, we employ methods such as integration by parts. This helps partition the function into manageable parts where we can handle both differentiation and integration easily.
When working with antiderivatives, it's crucial to remember to add the constant of integration, \( + C \). This represents the infinite set of possible original functions that could have resulted in the given derivative.
Understanding how to find antiderivatives efficiently gives you a powerful tool in solving integrals. Whether you are working with simple polynomials or complex inverse functions, mastering antiderivatives is a fundamental skill in calculus.
Substitution Method
The substitution method, often known as "u-substitution," is a powerful technique in integration. It simplifies an integral by substituting a part of the integral with a single variable, which makes the problem easier to solve.
The logic behind substitution is to transform a complicated integral into a simpler form that is much easier to tackle. By choosing an appropriate substitution, like letting \( w = 1 - x^2 \) in our example, you can reframe the problem to deal with polynomials or basic functions, which you know well how to integrate.
A crucial part of substitution is transforming the differential \( dx \) in terms of the new variable. This often involves calculating \( dw \) or another related term to align with the substitution you made. Always remember to substitute back the original variable if the integral ends up in terms of the new variable unless specified otherwise.
The method of substitution not only aids in solving integrals of inverse trigonometric functions but also many other types. It brings a broad application, providing a simpler path to obtain a solution that would otherwise seem complex.
The logic behind substitution is to transform a complicated integral into a simpler form that is much easier to tackle. By choosing an appropriate substitution, like letting \( w = 1 - x^2 \) in our example, you can reframe the problem to deal with polynomials or basic functions, which you know well how to integrate.
A crucial part of substitution is transforming the differential \( dx \) in terms of the new variable. This often involves calculating \( dw \) or another related term to align with the substitution you made. Always remember to substitute back the original variable if the integral ends up in terms of the new variable unless specified otherwise.
The method of substitution not only aids in solving integrals of inverse trigonometric functions but also many other types. It brings a broad application, providing a simpler path to obtain a solution that would otherwise seem complex.
