Problem 64 - Calculus Early Transcendentals

Question

Miscellaneous integrals Evaluate the following integrals. \(\int \frac{\sin (\ln x)}{4 x} d x\)

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \frac{\sin (\ln x)}{4 x} dx\). Answer: The integral evaluates to \(-\frac{1}{4}\cos(\ln x) + C\), where \(C\) is the constant of integration.

Step 1: Perform substitution

Let \(u = \ln x\). To find the differential \(du\), differentiate \(u\) with respect to \(x\): $$\frac{d u}{d x} = \frac{1}{x}.$$ Then, multiply by \(dx\) on both sides: $$du = \frac{1}{x} dx.$$ Now, substitute \(u = \ln x\) and \(du = \frac{1}{x}dx\) in the given integral: $$\int \frac{\sin (\ln x)}{4 x} d x = \int \frac{\sin (u)}{4} du.$$

Step 2: Compute the integral with the simplified expression

Evaluate the integral in terms of \(u\): $$\int \frac{\sin (u)}{4} du = \frac{1}{4} \int \sin(u) du.$$ The integral of \(\sin(u)\) with respect to \(u\) is \(-\cos(u)\), so we have: $$\frac{1}{4} \int \sin(u) du = -\frac{1}{4}\cos(u) + C,$$ where \(C\) is the constant of integration.

Step 3: Substitute the original variable

Now substitute back the original variable, \(u = \ln x\), into the solution: $$-\frac{1}{4}\cos(u) + C = -\frac{1}{4}\cos(\ln x) + C.$$ Therefore, the evaluated integral is: $$\int \frac{\sin (\ln x)}{4 x} d x = -\frac{1}{4}\cos(\ln x) + C.$$

Key Concepts

Indefinite Integral

An indefinite integral represents the antiderivative of a function and is a fundamental concept in calculus. In simple terms, it's the reverse of differentiation. When you integrate a function, you're finding all possible original functions (antiderivatives) that could have been differentiated to give you the original equation. The result of an indefinite integral includes a constant of integration, symbolized by the letter 'C', which denotes that there are infinitely many antiderivatives, each differing by a constant value.

For example, considering the function \(f(x) = x^2\), its indefinite integral is expressed as \( \int f(x) dx = \int x^2 dx \), which results in \( \frac{1}{3}x^3 + C \) - where 'C' is the constant of integration.

U-Substitution

U-substitution, also known as the substitution method, is a technique used in integration that simplifies complex integrals. It works by substituting a part of the integration with a new variable 'u' to make the integration easier to solve. The process involves three main steps:

• Choosing a substitution for 'u' that simplifies the integral.
• Finding the differential 'du' and replacing it in the integral.
• Computing the integral in terms of 'u' and then substituting back the terms with the original variable.

This method can turn a challenging integral into a more manageable one, often resulting in a basic form that's easier to integrate, such as the substitution used in the provided exercise where \( u = \ln x \) and \( du = \frac{1}{x}dx \).

Integration Techniques

There are various techniques to solve integrals in calculus, especially when tackling more complex functions. Some of the most commonly used methods include:

• Substitution: Simplifying integrals by changing variables.
• Integration by Parts: Based on the product rule of differentiation, useful for integrals that are products of two functions.
• Partial Fractions: Breaking down complex rational functions into simpler fractions that are easier to integrate.
• Trigonometric Substitution: Applying trigonometric identities to simplify integrals containing square roots.

Knowing when and how to apply these techniques is essential for solving integrals successfully. Often, a combination of methods is needed to find a solution.

Trigonometric Integration

Trigonometric integration involves integrating functions with trigonometric expressions, which is a common scenario in calculus. This technique uses trigonometric identities to simplify the integral before finding the antiderivative. Examples of trigonometric identities include \( \sin^2(x) + \cos^2(x) = 1 \) and \( 1 + \tan^2(x) = \sec^2(x) \).

For instance, integrating \( \sin(x) \) by itself, like in the example provided, is straightforward: \( \int \sin(x) dx = -\cos(x) + C \). However, more complex trigonometric integrals may require additional identities and substitutions for a solution. In the given exercise, the substitution \( u = \ln x \) turned the original integral into one involving a trigonometric function, making it a perfect candidate for trigonometric integration.