Chapter 29: Problem 14
$$ \text { Determine the following: } $$ $$ \int \frac{\sin ^{-1} x}{\sqrt{\left.1-x^{2}\right)}} d x $$
Short Answer
Expert verified
The integral is \( \frac{(\sin^{-1} x)^2}{2} + C \).
Step by step solution
01
Recognize the integration technique
Observe the integrand \( \frac{\sin^{-1}x}{\sqrt{1-x^2}} \). This integrates to a form where it can be simplified using substitution. Identifying such forms is crucial for integral evaluation.
02
Substitute the trigonometric function
Let \( x = \sin t \), then \( dx = \cos t \, dt \) and \( \sqrt{1-x^2} = \cos t \). The integral becomes \( \int \frac{t}{\cos t} \cdot \cos t \, dt = \int t \, dt \).
03
Evaluate the simpler definite integral
The integral \( \int t \, dt \) is straightforward, yielding \( \frac{t^2}{2} + C \), where \( C \) is the constant of integration.
04
Substitute back to the original variable
Since \( x = \sin t \), we know \( t = \sin^{-1} x \). Thus, substituting back gives us \( \frac{(\sin^{-1} x)^2}{2} + C \) as the solution to the original integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a fundamental technique in integration that simplifies complex integrals. It involves changing the variable of integration to transform a problem into a more manageable form. This often utilizes a substitution that aligns with the trigonometric identity or simplifies the integrand.
For instance, in the given exercise, we used the substitution \( x = \sin t \). This step converts the expression under the integral into terms of \( t \), making it simpler to integrate. This works well when the integrand contains expressions like \( \sqrt{1-x^2} \), which are recognizable as derivatives or related forms of trigonometric functions.
Here are some steps to effectively use the substitution method:
For instance, in the given exercise, we used the substitution \( x = \sin t \). This step converts the expression under the integral into terms of \( t \), making it simpler to integrate. This works well when the integrand contains expressions like \( \sqrt{1-x^2} \), which are recognizable as derivatives or related forms of trigonometric functions.
Here are some steps to effectively use the substitution method:
- Identify a substitution that might simplify the integral.
- Express the differential \( dx \) in terms of the new variable \( dt \).
- Substitute according to the chosen variable into the integral.
- Simplify and evaluate the integral with respect to the new variable.
- Finally, don't forget to substitute back to the original variable to get the result in terms of \( x \).
Trigonometric Integration
Trigonometric integration is a technique employed to solve integrals involving trigonometric functions like sine, cosine, tangent, etc. In this exercise, the integral involves \( \sin^{-1} x \), which is an inverse trigonometric function. Recognizing how these functions interact with one another can simplify seemingly complex problems.
When substituting \( x = \sin t \), you effectively leverage the identity \( \sin^2 t + \cos^2 t = 1 \) to rewrite the integrand expressions. Notice how \( \sqrt{1-x^2} = \cos t \), following directly from the Pythagorean identity once \( x = \sin t \) is substituted. This substitution aligns the integral components to reveal a straightforward integration path.
Benefits of trigonometric integration include:
When substituting \( x = \sin t \), you effectively leverage the identity \( \sin^2 t + \cos^2 t = 1 \) to rewrite the integrand expressions. Notice how \( \sqrt{1-x^2} = \cos t \), following directly from the Pythagorean identity once \( x = \sin t \) is substituted. This substitution aligns the integral components to reveal a straightforward integration path.
Benefits of trigonometric integration include:
- Utilizing known identities to simplify complex functions into basic forms.
- Removing the square roots and complex terms using identities.
- Creating a direct path to integration by reducing complex forms into familiar trigonometric terms.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \sin^{-1} x \), \( \cos^{-1} x \), and \( \tan^{-1} x \), often appear in integration problems. They represent angles whose trigonometric function equals a given number. In our exercise, the function \( \sin^{-1} x \) is integrated, requiring understanding both its definition and derivatives.
In mathematical terms, for \( y = \sin^{-1} x \), the relationship is \( x = \sin y \), implying that \( y \) is the angle whose sine is \( x \). This understanding is crucial when differentiating or integrating these functions.
Here are some key points about inverse trigonometric functions to remember:
In mathematical terms, for \( y = \sin^{-1} x \), the relationship is \( x = \sin y \), implying that \( y \) is the angle whose sine is \( x \). This understanding is crucial when differentiating or integrating these functions.
Here are some key points about inverse trigonometric functions to remember:
- They have specific derivatives which become useful during integration and differentiation tasks, e.g., \( \frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}} \).
- They translate angles measured as ratios into values within specific ranges: for \( \sin^{-1} x \), the range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
- Integrals involving these functions typically benefit from initial substitutions and expressions that match their derivative forms.