Problem 44 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d x}{\sqrt{1-4 x^{2}}}$$

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{1}{2} \arcsin(2x) + C \).

Step 1: Identify the Form of the Integral

The integral \( \int \frac{dx}{\sqrt{1-4x^2}} \) resembles the standard integral form \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C \). By comparing the integral to this form, we can identify that \( a^2 = 1 \) and \( 4x^2 \) matches with \( x^2 \).

Step 2: Substitute and Simplify

To match the integral exactly to the standard form, recognize that \( 4x^2 = (2x)^2 \). Therefore, we can rewrite the integral as: \( \int \frac{dx}{\sqrt{1 - (2x)^2}} \). Identify \( a = 1 \), meaning the substitution needed is \( u = 2x \), hence \( du = 2dx \) so \( dx = \frac{du}{2} \).

Step 3: Change Variables

Substitute \( u = 2x \) and \( dx = \frac{du}{2} \) into the integral: \[ \int \frac{dx}{\sqrt{1-4x^2}} = \int \frac{1}{2} \cdot \frac{du}{\sqrt{1-u^2}}. \] This simplifies to \( \frac{1}{2} \int \frac{du}{\sqrt{1-u^2}} \).

Step 4: Integrate Using Standard Form

The integral \( \int \frac{du}{\sqrt{1-u^2}} \) is the arcsine function, i.e., \( \arcsin(u) + C \). Therefore, \[ \frac{1}{2} \int \frac{du}{\sqrt{1-u^2}} = \frac{1}{2} \arcsin(u) + C. \]

Step 5: Back Substitute the Original Variable

Substitute back \( u = 2x \) into the integrated expression: \[ \frac{1}{2} \arcsin(2x) + C. \]This is the evaluated integral in terms of \( x \).

Key Concepts

Definite Integrals

Definite integrals provide a way to calculate the area under a curve between two specific points. They are represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. The value of a definite integral gives the net area between the function and the x-axis over the interval \([a, b]\).
To solve definite integrals, the Fundamental Theorem of Calculus is essential. It states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a). \]This powerful theorem connects the concept of differentiation with integration, allowing us to evaluate integrals by finding antiderivatives.
Evaluating definite integrals often involves identifying the function, finding its antiderivative, and then computing the difference between its values at the endpoints \( b \) and \( a \).

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of differentiation. They play a crucial role in solving integrals. The antiderivative of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \).
When we seek to integrate a function, we're essentially looking for its antiderivative. In notation, we write it as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration. This constant reflects that there are infinitely many antiderivatives for a function, differing only by a constant value.
To determine the antiderivative, we often rely on patterns and rules similar to differentiation. In particular, recognizing familiar forms such as power functions, exponential functions, and trigonometric functions helps efficiently find antiderivatives.

Integration Techniques

Integration techniques are essential tools that allow us to solve a wide variety of integrals. Sometimes, finding antiderivatives requires modes of solving beyond direct calculation. Here are some common techniques:

• Substitution: Often used when an integral contains a function and its derivative, substitution simplifies integration by changing variables.
• Integration by Parts: This technique is applied based on the product rule and is useful for integrating products of functions.
• Partial Fraction Decomposition: Used when dealing with rational functions, allowing complex fractions to be broken down into simpler parts.

In the given exercise, substitution was effectively employed by substituting \( u = 2x \) and transforming the integral into a formula that is easily recognized and integrated. The key to mastering integration techniques is practice and familiarity with a variety of functions.

Trigonometric Substitution

Trigonometric substitution is a method used to evaluate integrals involving expressions that resemble certain trigonometric identities. It simplifies the integration process by introducing a trigonometric function that transforms the integrand into a more manageable form.
In integrals like \( \int \frac{dx}{\sqrt{1-4x^2}} \), which matches the form \( \int \frac{dx}{\sqrt{a^2-x^2}} \), trigonometric substitution involves using identities like \( x = a\sin(\theta) \). This aligns the integrand with the arcsin function, allowing straightforward integration.
For example, substituting \( u = 2x \) effectively simplifies the original integral to a form that can be integrated directly as an arcsine function. Trigonometric substitution is particularly valuable when dealing with radicals containing squares, as it leverages the natural properties of trigonometric functions linked to unit circle identities.