Problem 53 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(47-70\) $$ \int_{0}^{1} \frac{4 d s}{\sqrt{4-s^{2}}} $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{2\pi}{3} \).

Step 1: Identify the Integral Form

Recognize that the integral \( \int \frac{a \, dx}{\sqrt{a^2 - x^2}} \) can be solved by using the formula \( \sin^{-1}\left(\frac{x}{a}\right) + C \), where \( a \) is a constant. In this case, \( a = 2 \).

Step 2: Apply Integration Formula

Apply the integration formula to solve the integral. For the given integral, \( \int \frac{4 \, ds}{\sqrt{4 - s^2}} \), substitute into the formula to find the antiderivative: \( 4 \cdot \sin^{-1}\left(\frac{s}{2}\right) \).

Step 3: Evaluate the Definite Integral

Evaluate the definite integral \( 4 \cdot \sin^{-1}\left(\frac{s}{2}\right) \) from \( s = 0 \) to \( s = 1 \). Substitute the limits into the antiderivative function: \( 4 \cdot \left[\sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0)\right] \).

Step 4: Simplify the Result

Calculate the result using the fundamental trigonometric values: \( \sin^{-1}\left(\frac{1}{2}\right) \) is \(\frac{\pi}{6}\) and \( \sin^{-1}(0) \) is 0. Therefore, the integral evaluates to \( 4 \cdot \frac{\pi}{6} = \frac{2\pi}{3} \).

Key Concepts

Integration Techniques

Understanding integration techniques is crucial for solving integrals, especially when dealing with complex functions. One important technique is recognizing integrals that resemble standard forms. These standard forms often have direct formulas we can use, which simplifies the work significantly. In our example, we have an integral of the form \( \int \frac{4 \, ds}{\sqrt{4 - s^2}} \). To solve this, it's essential to see it as an inverse trigonometric function form.

• Look for structures in the integrand that match known patterns, such as \( \sqrt{a^2 - x^2} \).
• Identify any constants or variables and compare them to the standard integral forms.
• Apply the known integral formula corresponding to the structure identified.

Practicing these integration techniques helps to quickly solve integrals by exploiting these forms, enhancing both speed and accuracy in calculations.

Inverse Trigonometric Functions

Inverse trigonometric functions play a pivotal role in integration, especially when dealing with square roots and specific forms that match trigonometric identities. The integral form for inverse sine, \( \int \frac{a \ dx}{\sqrt{a^2 - x^2}} \), is represented by \( \sin^{-1}\left(\frac{x}{a}\right) + C \).

• Recognize the structure: the form \( \sqrt{a^2 - x^2} \) is crucial for identifying potential inverse trigonometric integration.
• Remember the key relationship \( x = a \sin(\theta) \) which helps to see why this inverse function emerges.
• Substitute appropriately: Use the corresponding inverse trigonometric formula to directly integrate.

For our example, replacing \( s = 1 \) and \( s = 0 \) into \( \sin^{-1}\left(\frac{s}{2}\right) \) yields the definite integral solution. Utilizing these trigonometric relationships simplifies both the integration process and evaluation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge between antiderivatives and definite integrals. This powerful theorem allows us to evaluate definite integrals using antiderivatives, providing a thorough understanding of the net area under a curve. In our problem, once we find the antiderivative \( 4 \cdot \sin^{-1}\left(\frac{s}{2}\right) \), the theorem directs us to compute

• the difference of the antiderivative evaluated at the upper and lower limits.
• \[ 4 \cdot \left[\sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0)\right] \]
• Use known values of inverse sine: \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \) and \( \sin^{-1}(0) = 0 \).

Finally, multiplying by the constant leads to a neat result, \( \frac{2\pi}{3} \). Embracing this theorem ensures precise integration, as it uses the power of antiderivatives to solve definite integrals efficiently.