Problem 411
Question
In the following exercises, compute the antiderivative using appropriate substitutions. $$\int \frac{\sin ^{-1} t d t}{\sqrt{1-t^{2}}}$$
Step-by-Step Solution
Verified Answer
The antiderivative is \( \frac{(\sin^{-1} t)^2}{2} + C \).
1Step 1: Identify the substitution
For the integral \( \int \frac{\sin^{-1} t \, dt}{\sqrt{1-t^{2}}} \), we notice that the derivative of \( \sin^{-1} t \) is \( \frac{1}{\sqrt{1-t^2}} \). This suggests using a substitution that simplifies the expression. Set \( u = \sin^{-1} t \). Hence, \( du = \frac{1}{\sqrt{1-t^2}} \, dt \). This substitution implies that \( dt = \sqrt{1-t^2} \, du \).
2Step 2: Substitute into the integral
Replace \( \sin^{-1} t \) with \( u \) and \( dt \) with \( \sqrt{1-t^2} \, du \) from Step 1. The integral becomes:\[ \int u \, du. \]
3Step 3: Integrate with respect to u
Now integrate \( \int u \, du \). The formula for integrating \( u^n \) is \( \frac{u^{n+1}}{n+1} + C \). Thus, \( \int u \, du = \frac{u^2}{2} + C \).
4Step 4: Back-substitute to find the antiderivative
Replace \( u \) with \( \sin^{-1} t \) to return the integral to terms of \( t \). Hence, the antiderivative is:\[ \frac{(\sin^{-1} t)^2}{2} + C. \]
Key Concepts
Substitution MethodIntegration TechniquesInverse Trigonometric Functions
Substitution Method
The substitution method is an essential technique for solving integrals that are otherwise difficult to handle. It involves changing the variable of integration to simplify the integral's structure.
A good substitution typically aligns with parts of the integrand that have a clear derivative. For the given example, we have the integral \( \int \frac{\sin^{-1} t \, dt}{\sqrt{1-t^{2}}} \). The expression \( \sin^{-1} t \) has a derivative of \( \frac{1}{\sqrt{1-t^2}} \), which is present in the integral itself.
So, we set \( u = \sin^{-1} t \), meaning \( du = \frac{1}{\sqrt{1-t^2}} \, dt \). This transformation drastically simplifies how we integrate, turning the problem into \( \int u \, du \), which is much easier to solve.
A good substitution typically aligns with parts of the integrand that have a clear derivative. For the given example, we have the integral \( \int \frac{\sin^{-1} t \, dt}{\sqrt{1-t^{2}}} \). The expression \( \sin^{-1} t \) has a derivative of \( \frac{1}{\sqrt{1-t^2}} \), which is present in the integral itself.
So, we set \( u = \sin^{-1} t \), meaning \( du = \frac{1}{\sqrt{1-t^2}} \, dt \). This transformation drastically simplifies how we integrate, turning the problem into \( \int u \, du \), which is much easier to solve.
Integration Techniques
Integration can be tackled through a variety of methods depending on the function form. Fundamental techniques include basic rules, substitution, integration by parts, and the use of tables of integrals.
For example, once the substitution method has simplified an integral to \( \int u \, du \), it then requires basic integration techniques to solve it. Here the power rule applies.
The integral \( \int u \, du \) results in \( \frac{u^2}{2} + C \). This step taps into the power formula where \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Recognizing which technique to employ is crucial for efficient problem-solving.
For example, once the substitution method has simplified an integral to \( \int u \, du \), it then requires basic integration techniques to solve it. Here the power rule applies.
The integral \( \int u \, du \) results in \( \frac{u^2}{2} + C \). This step taps into the power formula where \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Recognizing which technique to employ is crucial for efficient problem-solving.
Inverse Trigonometric Functions
Inverse trigonometric functions link angles back to their ratio values. The arcsine function, \( \sin^{-1} t \), returns the angle whose sine is \( t \).
These functions frequently appear in integrals, especially when arcsine, arccosine, and arctangent involve derivatives. Importantly, the derivative of \( \sin^{-1} t \) is \( \frac{1}{\sqrt{1-t^2}} \), often essential for substitution.
Our exercise involved this function because the integral required handling \( \sin^{-1} t \). Understanding inverse trigonometric properties helps recognize patterns and simplifies the integration process, especially when paired with derivatives present in the integrand itself.
These functions frequently appear in integrals, especially when arcsine, arccosine, and arctangent involve derivatives. Importantly, the derivative of \( \sin^{-1} t \) is \( \frac{1}{\sqrt{1-t^2}} \), often essential for substitution.
Our exercise involved this function because the integral required handling \( \sin^{-1} t \). Understanding inverse trigonometric properties helps recognize patterns and simplifies the integration process, especially when paired with derivatives present in the integrand itself.
Other exercises in this chapter
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