Problem 5 - Thomas Calculus

Question

Evaluate the integrals. $$\int_{0}^{3 / 2} \frac{d x}{\sqrt{9-x^{2}}}$$

Step-by-Step Solution

Answer

\(\frac{\pi}{6}\)

Step 1: Identify the Integral Type

The given integral \( \int_{0}^{3/2} \frac{d x}{\sqrt{9-x^{2}}} \) is an inverse trigonometric function integral. It resembles the form for the integral of the arcsine function, \( \int \frac{d x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C \).

Step 2: Determine \(a\)

In this integral, \(a = 3\) because the expression under the square root is \(9 - x^2\), where \(9\) is \(3^2\).

Step 3: Apply the Arcsine Formula

Use the formula for the integral of the arcsine function: \( \int \frac{d x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C \). Substitute \(a = 3\) into the formula to obtain \(\arcsin\left(\frac{x}{3}\right) + C\).

Step 4: Evaluate the Definite Integral

Evaluate the integral from \(0\) to \(\frac{3}{2}\):\[ \int_{0}^{3/2} \frac{d x}{\sqrt{9-x^{2}}} = \left[ \arcsin\left(\frac{x}{3}\right) \right]_{0}^{3/2} \]

Step 5: Calculate the Upper Limit

Calculate \(\arcsin\left(\frac{3/2}{3}\right) = \arcsin\left(\frac{1}{2}\right)\). The arcsine of \(\frac{1}{2}\) is \(\frac{\pi}{6}\).

Step 6: Calculate the Lower Limit

Calculate \(\arcsin\left(\frac{0}{3}\right) = \arcsin(0)\). The arcsine of \(0\) is \(0\).

Step 7: Subtract Lower Limit from Upper Limit

Subtract the lower limit result from the upper limit result:\[ \frac{\pi}{6} - 0 = \frac{\pi}{6} \]

Key Concepts

Integrals

Integrals are a fundamental concept in calculus, known for calculating the accumulated area under a curve. Think of them as a tool to add up tiny slices of a graph to find totals, such as area, volume, or even a distance. Integrals come in two main types:

• Indefinite Integrals: These represent a family of functions and include a constant of integration, "C." They are used to find the original function given its derivative.
• Definite Integrals: These have specific limits, such as \( \int_{a}^{b} f(x) \, dx \), which computes the total area under a curve from point \(a\) to point \(b\). The result is a numerical value representing this total area.

In the given exercise, we deal with a definite integral from 0 to \( \frac{3}{2} \). This involves finding the integral between two bounds, a lower limit and an upper limit. By using specific rules and formulas, we can determine the total area accurately. This often involves simplifying the expression or transforming it into a known form.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus for finding angles when the values of trigonometric functions are known. They reverse regular trigonometric roles:

• The arcsine function is the inverse of sine, denoted as \( \arcsin(x) \), which helps find an angle whose sine is \(x\).
• It is important to note that inverse trigonometric functions are defined within specific ranges. For arcsine, the range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\) because sine angles only rise from \(-1\) to \(1\).

In calculus, these functions assist in solving integrals that involve roots and are expressed in forms that resemble normal trigonometric functions. In the solution, recognizing the presence of an inverse trigonometric function helps transform the integral into a simpler, well-known expression.

Arcsine Function

The arcsine function, commonly written as \(\arcsin(x)\), gives the angle whose sine value is \(x\). It's crucial in integrating functions containing roots like \( \sqrt{a^2-x^2} \).

The classic integral form for arcsine is:\[\int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin\left(\frac{x}{a}\right) + C\]
This formula helps evaluate integrals involving square roots, typical in the various problems found in calculus.

In this exercise, recognizing the expression \( \sqrt{9-x^2} \) enabled the application of the arcsine formula. Substituting directly into this equation, it allowed us to simplify the calculation drastically, yielding the final values by evaluating at given bounds. The power of understanding and applying the arcsine form simplifies many potentially difficult integrals.