Problem 30 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{2 d x}{x \sqrt{1-4 \ln ^{2} x}} $$

Step-by-Step Solution

Answer

The evaluated integral is \( \arcsin(2 \ln x) + C \).

Step 1: Identify the substitution

Consider the expression under the square root, \(1 - 4 (\ln x)^2\). Let \( u = \ln x \), which implies \( du = \frac{1}{x} dx \). This substitution simplifies the integrand.

Step 2: Substitute and Change Variables

With the substitution \( u = \ln x \), the differential \( dx \) changes to \( du \cdot x = x\cdot du = x \cdot du \). Since \( u = \ln x \), then \( x = e^u \) and \( dx = x \cdot du = e^u \cdot du \). Substitute these into the integral to get \( \int \frac{2}{x \sqrt{1 - 4u^2}} dx = \int \frac{2}{\sqrt{1 - 4u^2}} du \).

Step 3: Simplify the Integral

The integral is now in the form \( \int \frac{2}{\sqrt{1 - 4u^2}} du \). Factor out the constant: \[ 2 \int \frac{1}{\sqrt{1 - 4u^2}} du \]. The integral is now a standard form that relates to an inverse trigonometric function.

Step 4: Recognize the Standard Integral Form

The integral \( \int \frac{1}{\sqrt{1 - 4u^2}} du \) resembles the form \( \int \frac{1}{\sqrt{1 - a^2 x^2}} dx = \frac{1}{a} \arcsin(ax) + C \). Here, use \( a = 2 \), as the integral is \( \int \frac{1}{\sqrt{1 - (2u)^2}} du \). Thus, the integral becomes \( \frac{1}{2} \arcsin(2u) + C \).

Step 5: Solve the Integral

Apply the form: \( 2 \left( \frac{1}{2} \arcsin(2u) \right) = \arcsin(2u) + C \).

Step 6: Back-Substitute to Original Variable

Since \( u = \ln x \), replace \( u \) back with \( \ln x \), obtaining the final result as \( \arcsin(2 \ln x) + C \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arcsin \), \( \arccos \), and \( \arctan \), are the inverse operations of the primary trigonometric functions. They allow us to find angles when their trigonometric values are known. For example, if \( \sin(\theta) = x \), then \( \theta = \arcsin(x) \).
\( \arcsin \), specifically, is used to find the angle whose sine is a given number. It is defined for inputs ranging from -1 to 1, with outputs in the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
In integration, identifying the form that results in an inverse trigonometric function helps simplify the integral. Recognizing these forms is crucial for effectively evaluating integrals. In our exercise, the integral resembles the standard \( \arcsin \) form, allowing us to solve for the antiderivative quickly.Understanding how these functions operate provides a clear path to handling integrals that involve square roots and quadratic expressions. Recognizing standard integral forms and their corresponding inverse trigonometric functions is a powerful tool that simplifies the integration process.

Substitution Method

The substitution method is a fundamental technique in calculus used to simplify integrals. The main idea is to change variables in the integral to make it more manageable.
In our example, we identified \( u = \ln x \) as the substitution. This step translates the integral to be in terms of \( u \), removing complicated expressions like \( \ln^2 x \) which appear in the original form. Once the substitution is complete, the integral often transforms into a recognizable form that is easier to evaluate.
The substitution method follows a systematic approach:

• Identify and choose a substitution that simplifies the integrand.
• Replace all instances of the original variable and its differential in the integral.
• Evaluate the simplified integral.
• Switch back the substitute variable to the original one.

Mastering this skill allows solving a wide range of integrals effectively, especially when dealing with composite functions and tricky mathematical expressions. It is a valuable technique for reducing complex integrals into simple standard forms, including inverse trigonometric integrals.

Definite and Indefinite Integrals

Integration can be either definite or indefinite, serving different purposes in calculus.
Indefinite integrals, as illustrated in the original exercise, seek to find the most general form of the antiderivative. This is expressed with a constant \( C \) added to the function, representing any possible vertical shift. For example, the result \( \arcsin(2 \ln x) + C \) includes the constant to cover all potential antiderivatives.

Definite integrals, on the other hand, calculate the net area under a curve within a specific range on the x-axis. This gives a numerical value rather than a function, reflecting the accumulation of quantities between two bounds.Understanding both concepts:

• Indefinite integrals involve finding the antiderivative without limits, resulting in a function plus a constant \( C \).
• Definite integrals involve an evaluation from one limit to another, resulting in a specific numerical value.

Knowing when to use definite or indefinite integrals is crucial in solving real-world problems and working through complex calculus exercises. Each type serves unique purposes within mathematics, providing tools for both theoretical calculations and practical applications.