Problem 27 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int \sin (\ln x) d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{2} (x \sin(\ln x) - x \cos(\ln x)) + C \).

Step 1: Identify the Parts

For integration by parts, we need to choose functions for \( u \) and \( dv \). In this integral, we set \( u = \sin(\ln x) \) and \( dv = dx \). This choice is typical because \( dv \) should be something easy to integrate.

Step 2: Differentiate and Integrate

Differentiate \( u \) and integrate \( dv \): \[ du = \cos(\ln x) \cdot \frac{1}{x} \, dx \] and \[ v = x \].

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting the expressions found in the previous steps, we get: \[ \int \sin(\ln x) \, dx = x \cdot \sin(\ln x) - \int x \cdot \left(\cos(\ln x) \cdot \frac{1}{x}\right) \, dx \].

Step 4: Simplify the Integral

Simplify the remaining integral: \[ \int \cos(\ln x) \, dx \].

Step 5: Second Integration by Parts

Again, apply integration by parts to \( \int \cos(\ln x) \, dx \). Set \( u = \cos(\ln x) \) and \( dv = dx \). Then \( du = -\sin(\ln x) \cdot \frac{1}{x} \, dx \) and \( v = x \). By applying the formula again: \[ \int \cos(\ln x) \, dx = x \cdot \cos(\ln x) + \int x \cdot \left(\sin(\ln x) \cdot \frac{1}{x}\right) \, dx \].

Step 6: Combine the Expansions

After substitution, we have: \[ \int \sin(\ln x) \, dx = x \cdot \sin(\ln x) - \left( x \cdot \cos(\ln x) + \int \sin(\ln x) \, dx \right) \].

Step 7: Solve for the Integral

Reorganizing, we find:\[ \int \sin(\ln x) \, dx = x \cdot \sin(\ln x) - x \cdot \cos(\ln x) - \int \sin(\ln x) \, dx \] Thus, \[ 2 \int \sin(\ln x) \, dx = x \cdot \sin(\ln x) - x \cdot \cos(\ln x) \], which means \[ \int \sin(\ln x) \, dx = \frac{1}{2} \left( x \cdot \sin(\ln x) - x \cdot \cos(\ln x) \right) + C \] where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics, dealing with the concepts of integration and accumulation of quantities. Integration is essentially the reverse process of differentiation. It allows us to find the total accumulation of quantities, such as areas under curves and volumes of solids. Understanding integration theory is essential for solving complex calculus problems.

There are two main components of integral calculus:

• Definite Integrals: These calculate the net area under a curve over a particular interval. They give a numerical value.
• Indefinite Integrals: These represent families of functions and include a constant of integration (often denoted as \(C\)). They help us understand the antiderivative.

In the original exercise, we deal with an indefinite integral of a trigonometric-logarithmic function, specifically, the integration of \(\sin(\ln x)\) with respect to \(x\). The fundamental theorem of calculus combines differentiation and integration, ensuring they function inversely to each other. This is a core principle that underscores the entirety of integral calculus and empowers many modern-day applications, from calculating distances to advanced physics models.

Logarithmic Functions

Logarithmic functions, often represented as \(\ln(x)\) or \(\log(x)\), are the inverse of exponential functions. They serve as a powerful tool for solving problems where exponential behavior is observed. In the context of calculus, logarithmic differentiation and integration involve elements that contain \(\ln(x)\) or \(e^x\), the base of natural logarithms.

Key features of logarithmic functions include:

• They grow slower than any linear, quadratic, or polynomial expression; hence, they are useful in understanding asymptotic behavior.
• The derivative of \(\ln(x)\) is straightforward: it is \(\frac{1}{x}\), which is a small detail of why logarithmic properties simplify calculus operations.

In the integration by parts exercise, the function \(\sin(\ln x)\) involves a logarithmic function inside a trigonometric function. The logarithmic function modifies the rate of change of the trigonometric function, revealing how powerful these combined operations can be in manipulating and understanding complex functional relationships.

Integration Techniques

Integration by Parts is a powerful technique in integral calculus, analogous to the product rule of differentiation. It is designed to simplify integrals that are products of two functions, one of which becomes simpler upon differentiation, and the other simpler upon integration.

Here is a simplified overview of how Integration by Parts works:

• Identify parts: Select \(u\) and \(dv\) from the integral \(\int u\ dv\). Typically, \(u\) is chosen to be the function that simplifies upon differentiation.
• Differentiation and Integration: Differentiate \(u\) yielding \(du\), and integrate \(dv\) yielding \(v\).
• Apply the Formula: Use the equation \(\int u\ dv = uv - \int v\ du\). This allows the original integral to break into simpler components.

In our exercise, integration by parts is applied twice due to the nature of the function \(\sin(\ln x)\). The initial integration simplifies only part of the expression, necessitating a second round of the technique to achieve a fully integrated form.

This method is especially handy when dealing with logarithmic or exponential functions, transforming challenges into more manageable expressions. Mastery of integration techniques like this broaden one's problem-solving toolkit immensely.