Problem 24 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the indefinite integral. \(\int \frac{\sin (\ln x)}{x} d x\)

Step-by-Step Solution

Answer

The integral is \(-\cos(\ln x) + C\).

Step 1: Identify the Form of the Integral

Look at the integral \( \int \frac{\sin(\ln x)}{x} \, dx \). This is a good candidate for a substitution method due to the composite function \( \ln x \).

Step 2: Perform Substitution

Let \( u = \ln x \). Then \( du = \frac{1}{x} \, dx \), which implies \( dx = x \, du \). Since \( \frac{1}{x} \) is included in the integral, it simplifies to \( \int \sin(u) \, du \).

Step 3: Integrate the Simplified Expression

Now integrate \( \int \sin(u) \, du \). The integral of \( \sin(u) \) is \( -\cos(u) + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back the Original Variable

Replace \( u \) with \( \ln x \) in the solution. So, \( -\cos(\ln x) + C \).

Step 5: Write the Final Solution

The indefinite integral \( \int \frac{\sin(\ln x)}{x} \, dx \) is \( -\cos(\ln x) + C \).

Key Concepts

Substitution Method

The substitution method is a popular technique used to simplify the process of integration, especially when dealing with complex expressions containing composite functions. When you encounter an integral that seems difficult to tackle directly, substitution can help by transforming it into a more familiar form. In our exercise, the presence of the function \( \ln x \) within \( \sin(\ln x) \) suggests the use of substitution.

To apply the substitution method, start by choosing a part of the integrand to replace with a new variable \( u \). In this case, letting \( u = \ln x \) transforms the integral into a solvable form. Remember to also convert \( dx \) into terms of \( du \), often with the help of the derivative of your substitution \( \frac{d}{dx}(\ln x) = \frac{1}{x} \), simplifying the integral to \( \int \sin(u) \, du \).

Identify a substitution that simplifies the integrand

Express \( dx \) in terms of \( du \)

Transform the integral accordingly

Use substitution to handle complicated functions more easily.

Composite Function

Composite functions are functions made by combining two or more functions, often through substitution. They are common in calculus problems, particularly in integration, because they frequently result in nested expressions or layers of operations.

In the context of our indefinite integral \( \int \frac{\sin(\ln x)}{x} \, dx \), we deal with a composite function where \( \sin(u) \) is constructed from the \( \ln x \) applied to \( x \). Here, \( \ln x \) is inside the sine function, creating a complexity that the substitution method can manage. The potential difficulty in evaluating an integral involving a composite function is identifying and isolating the inner function.

Recognize the inner function within the composite expression

Use substitution to simplify the complex layers

Evaluate the integral after transformation

Understanding composite functions is crucial for grasping how substitution helps unravel these intertwined layers.

Integration Techniques

Integration, a core concept in calculus, may involve various techniques to solve indefinite integrals. Depending on the form of the function to be integrated, different approaches can be suitable. The problem \( \int \frac{\sin(\ln x)}{x} \, dx \) is a perfect demonstration where basic integration techniques like direct integration, substitution, or parts may be appropriate.

Substitution is one of the most powerful tools in integration, especially useful when confronted with composite functions, as seen here. By converting the integral into \( \int \sin(u) \, du \), it tapped into simpler foundation techniques.

Direct Integration: Used when integrand forms are straightforward

Substitution: Suitable for composite functions

Integration by Parts: Applies when products of functions are present

Partial Fraction Decomposition: Useful for rational functions

By mastering these techniques, tackling otherwise complex integrals becomes manageable. Each method has its own application scenarios, adding flexibility to the toolbox of solving integrals.