Problem 11
Question
Find the indefinite integrels in \(1-20\). \(\int d t / \sqrt{1-t^{2}}\)
Step-by-Step Solution
Verified Answer
\( \int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C \)
1Step 1: Recognize the Integral Form
The integral \( \int \frac{dt}{\sqrt{1-t^2}} \) is in a standard form that corresponds to the inverse trigonometric integration formulas.
2Step 2: Recall the Standard Formula
The standard integral for \( \int \frac{dt}{\sqrt{1-t^2}} \) is known as the inverse sine function. It can be directly integrated as \( \sin^{-1}(t) + C \), where \( C \) is the constant of integration.
3Step 3: Apply the Formula
By applying the standard formula, we integrate \( \int \frac{dt}{\sqrt{1-t^2}} \) to get the result \( \sin^{-1}(t) + C \).
Key Concepts
Inverse Trigonometric FunctionsIntegration TechniquesConstant of Integration
Inverse Trigonometric Functions
Inverse trigonometric functions are the functions that reverse the action of the standard trigonometric functions like sine, cosine, and tangent. When we say "inverse," we mean finding the angle that corresponds to a given trigonometric value.
For example, the inverse sine function, denoted as \( \sin^{-1}(x) \) or arcsine, gives us an angle whose sine is \( x \). Therefore, if \( \sin(\theta) = x \), then \( \theta = \sin^{-1}(x) \).
Inverse trigonometric functions are not as straightforward as their non-inverse counterparts. They have specific domain and range restrictions to make them true functions (meaning they give exactly one output for each input). When integrating functions that resemble expressions involving inverse trigonometric forms, recognizing these patterns is crucial in finding an indefinite integral.
Key inverse trig functions in integration include:
For example, the inverse sine function, denoted as \( \sin^{-1}(x) \) or arcsine, gives us an angle whose sine is \( x \). Therefore, if \( \sin(\theta) = x \), then \( \theta = \sin^{-1}(x) \).
Inverse trigonometric functions are not as straightforward as their non-inverse counterparts. They have specific domain and range restrictions to make them true functions (meaning they give exactly one output for each input). When integrating functions that resemble expressions involving inverse trigonometric forms, recognizing these patterns is crucial in finding an indefinite integral.
Key inverse trig functions in integration include:
- The inverse sine (\( \sin^{-1}(x) \))
- The inverse cosine (\( \cos^{-1}(x) \))
- The inverse tangent (\( \tan^{-1}(x) \))
Integration Techniques
Integration techniques are methods used to solve integrals, which can range from basic to quite complex. Knowing which technique applies is essential to efficiently solve an integral.
In this scenario, we need an integral technique that involves recognizing patterns related to inverse trigonometric functions. The integral \( \int \frac{dt}{\sqrt{1-t^2}} \) specifically requires this approach. We achieve this by identifying it as similar to the derivative of an inverse trigonometric function.
Key techniques for such integrals involve:
In this scenario, we need an integral technique that involves recognizing patterns related to inverse trigonometric functions. The integral \( \int \frac{dt}{\sqrt{1-t^2}} \) specifically requires this approach. We achieve this by identifying it as similar to the derivative of an inverse trigonometric function.
Key techniques for such integrals involve:
- Recognizing the pattern that aligns with an inverse trig function.
- Recalling standard formulas associated with inverse trigonometric integrals, such as \( \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C \).
- Applying these formulas to calculate the integral.
Constant of Integration
When dealing with indefinite integrals, the result is not a single function but a family of functions. These integrals always include a "constant of integration," represented by \( C \).
The constant of integration exists because antiderivatives are not unique. Given a function, there are infinitely many ways to define its antiderivative by adding any constant. For instance, integrating \( f(x) = 3 \) can result in functions like \( 3x + 1 \), \( 3x - 4 \), or simply \( 3x \), each differing by a constant.
In our case of \( \sin^{-1}(t) \), the constant \( C \) is crucial because it acknowledges this infinite set of solutions. The function \( \sin^{-1}(t) + C \) tells us that we're considering all possible functions that could differentiate back to our original integrand.
The constant of integration exists because antiderivatives are not unique. Given a function, there are infinitely many ways to define its antiderivative by adding any constant. For instance, integrating \( f(x) = 3 \) can result in functions like \( 3x + 1 \), \( 3x - 4 \), or simply \( 3x \), each differing by a constant.
In our case of \( \sin^{-1}(t) \), the constant \( C \) is crucial because it acknowledges this infinite set of solutions. The function \( \sin^{-1}(t) + C \) tells us that we're considering all possible functions that could differentiate back to our original integrand.
- It is conventionally written as \( + C \) and signals the generality of indefinite integrals.
- In practical scenarios, initial or boundary conditions are used to find the value of \( C \), transforming the general solution into a specific one.
Other exercises in this chapter
Problem 10
Find the indefinite integrels in \(1-20\). \(\int \cos ^{3} x \sin 2 x d x\)
View solution Problem 11
Find the derivatives of the following functions \(F(x)\). $$ \int_{0}^{x}\left[\int_{0}^{t} v(u) d u\right] d t $$
View solution Problem 12
Find an antiderivative \(f(x)\) for \(v(x)\). Then compute the definite integral \(\int_{0}^{1} r(x) d x=f(1)-f(0)\). $$ \sin ^{2} x \cos x $$
View solution Problem 12
Find the indefinite integrels in \(1-20\). \(\int t \sqrt{1-t^{2}} d t\)
View solution