Problem 27 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{2} \frac{d s}{\sqrt{4-s^{2}}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} \).

Step 1: Recognize the Integral Type

The integral \( \int \frac{d s}{\sqrt{4-s^{2}}} \) is a standard form that matches the arcsine function integration formula. Specifically, it takes the form of \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \).

Step 2: Identify \( a \) and \( x \)

In the given integral \( \int_{0}^{2} \frac{d s}{\sqrt{4-s^{2}}} \), you can identify \( a^2 = 4 \), which means \( a = 2 \), and \( x = s \).

Step 3: Apply the Arcsine Integration Formula

Using the formula, the integral evaluates to \( \sin^{-1}\left(\frac{s}{2}\right) + C \). However, since this is a definite integral from \(0\) to \(2\), we apply the limits instead of adding \( C \).

Step 4: Evaluate the Definite Integral

Substitute the bounds into the result of the integration:\[\int_{0}^{2} \frac{d s}{\sqrt{4-s^{2}}} = \left[ \sin^{-1}\left(\frac{s}{2}\right) \right]_{0}^{2} = \sin^{-1}\left(1\right) - \sin^{-1}\left(0\right)\]Evaluate these arcsine values: \(\sin^{-1}(1) = \frac{\pi}{2}\) and \(\sin^{-1}(0) = 0\).

Step 5: Find the Final Result

Subtract the arcsine values obtained: \[ \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Thus, the final result of the integral is \( \frac{\pi}{2} \).

Key Concepts

Arcsine Function

The arcsine function, often denoted as \( \sin^{-1}(x) \), is the inverse of the sine function on its domain. This means it returns the angle whose sine is \( x \). The range of \( \sin^{-1} \) is between \(-\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

• The arcsine function is commonly used in calculus, especially in integration problems.
• It is particularly useful for integrals of the form \( \int \frac{dx}{\sqrt{a^2 - x^2}} \).
• Recognizing this form helps in applying the formula correctly.

This specific integral form leads directly to arcsine because the substitution \( u = \frac{x}{a} \) simplifies the problem. Once the integral \( \int \frac{dx}{\sqrt{4-s^2}} \) is recognized as such, we can directly apply the integration formula to get the expression \( \sin^{-1}\left(\frac{s}{2}\right) \). Remembering how to manipulate arcsine results is important to find the correct trigonometric values on specific intervals.

Definite Integral

A definite integral, like our example \( \int_{0}^{2} \frac{ds}{\sqrt{4-s^2}} \), involves evaluating the integral between two specific bounds. In contrast to indefinite integrals, which include a constant \( C \), definite integrals provide a specific numerical result.

• Definite integrals calculate the net area under a curve and are typically evaluated at the boundaries given by the limits.
• They are important for solving real-world problems, including physics and engineering challenges where total values are needed.
• In a definite integral, once the antiderivative is found, substitute the limits in sequence, subtracting the lower limit from the upper limit.

In our problem, substituting the bounds into the antiderivative results in \( \sin^{-1}\left(1\right) - \sin^{-1}\left(0\right) \), simplifying directly to \( \frac{\pi}{2} - 0 \), illustrating how calculus techniques can offer precise solutions.

Integration Techniques

Integration techniques provide methods for finding the antiderivatives of functions. Recognizing and matching integrals to known formulas, like the arcsine formula in our problem, is a key part of these techniques.

• The recognition technique involves identifying familiar integral forms and applying corresponding formulas or identities.
• Substitution is another method, where we simplify the integral by substituting variables to match known integrable forms.
• Other techniques include integration by parts, partial fraction decomposition, and trigonometric substitution.

Here, after identifying the integrand as \( \frac{1}{\sqrt{4-s^2}} \), we used the arcsine rule, applying the conversion directly to obtain \( \sin^{-1}\left(\frac{s}{2}\right) \). This is one of the simpler integration techniques often encountered early in calculus learning. Understanding when and how to use these techniques simplifies calculating integrals significantly.