Problem 30 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int \cos (\ln x) d x $$

Step-by-Step Solution

Answer

\( \int \cos(\ln x) \, dx = \frac{1}{2}(x \cos(\ln x) + x \sin(\ln x)) + C \).

Step 1: Identify Functions for Integration by Parts

For integration by parts, we identify two functions from the integrand. Here, we can let \( u = \cos(\ln x) \) and \( dv = dx \).

Step 2: Differentiate and Integrate the Chosen Functions

Differentiate \( u = \cos(\ln x) \) to find \( du \). We have \( du = -\sin(\ln x) \cdot \frac{1}{x} \, dx \). Integrate \( dv = dx \) to get \( v = x \).

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute the identified \( u, v, du \): \[ \int \cos(\ln x) \, dx = x \cos(\ln x) - \int x \left( -\sin(\ln x) \cdot \frac{1}{x} \right) \, dx \].

Step 4: Simplify the Integral

Simplify the expression for the integral. The expression \( x \left( -\sin(\ln x) \cdot \frac{1}{x} \right) \) simplifies to \( -\sin(\ln x) \). Thus, \[ \int \cos(\ln x) \, dx = x \cos(\ln x) + \int \sin(\ln x) \, dx \].

Step 5: Apply Integration by Parts to Remaining Integral

To solve \( \int \sin(\ln x) \, dx \), use integration by parts again. Let \( u = \sin(\ln x) \) and \( dv = dx \). Differentiate \( u \) to get \( du = \cos(\ln x) \cdot \frac{1}{x} \, dx \), and integrate \( dv = dx \) to get \( v = x \).

Step 6: Substitute and Simplify Again

Substitute using integration by parts formulation: \[ \int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \left( \cos(\ln x) \cdot \frac{1}{x} \right) \, dx \]. This simplifies further to \[ \int \sin(\ln x) \, dx = x \sin(\ln x) - \int \cos(\ln x) \, dx \].

Step 7: Solving the System of Equations

Substitute Step 6 result \( \int \sin(\ln x) \, dx = x \sin(\ln x) - \int \cos(\ln x) \, dx \) into Step 4: \[ \int \cos(\ln x) \, dx = x \cos(\ln x) + x \sin(\ln x) - \int \cos(\ln x) \, dx \]. Let \( I = \int \cos(\ln x) \, dx \):\[ I = x \cos(\ln x) + x \sin(\ln x) - I \].

Step 8: Solve for the Integral

Rearrange terms to isolate \( I \):\[ 2I = x \cos(\ln x) + x \sin(\ln x) \]. Divide by 2: \[ I = \frac{1}{2}(x \cos(\ln x) + x \sin(\ln x)) \]. Thus, \[ \int \cos(\ln x) \, dx = \frac{1}{2}(x \cos(\ln x) + x \sin(\ln x)) + C \], where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is all about finding the area under a curve or, put simply, solving integrals. It is a branch of calculus concerned with the idea of accumulation and change. In the context of the given exercise, we're looking at a method known as **integration by parts**, which is a technique derived from the product rule for differentiation. This method is helpful when you have a product of two functions that you want to integrate. The basic formula is:

• \( \int u \, dv = uv - \int v \, du \).

In our exercise, integration by parts was used twice to solve the integral \( \int \cos(\ln x) \, dx \). First, we decomposed the integral into components that can be individually solved or simplified. Then, by smartly choosing our functions \( u \) and \( dv \), we were able to unravel and simplify the integral step by step. This shows how integration by parts can successfully tackle what seems like a complicated problem by breaking it down into more manageable parts.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus and play a crucial role in integration problems. In the provided integral \( \int \cos(\ln x) \, dx \), we encounter both the cosine and sine functions in tandem with the logarithmic function. These functions are cyclic and can transform based on their input values.

• Cosine function, \( \cos(\ln x) \), transforms based on the natural logarithm of \( x \), essentially creating a composite function.
• This requires strategic manipulation in integration, as shown in the solution by switching between cosine and sine through differentiation.
• The process involves using identities and derivatives of these trigonometric functions to simplify the integrals further.

Ultimately, understanding the properties of trigonometric functions allows us to handle the oscillatory behavior they introduce, especially when embedded within logarithmic inputs as seen here.

Logarithmic Functions

Logarithmic functions, such as the natural logarithm \( \ln(x) \), are the inverse of exponential functions and appear frequently in calculus. Their composition within other functions can make integrals challenging, as seen with \( \cos(\ln x) \). Logarithms transform multiplicative processes into additive ones, which affects how functions behave when you differentiate or integrate them.

• When integrating a function like \( \cos(\ln x) \), the presence of \( \ln x \) changes how we approach setting our \( u \) and \( dv \) in integration by parts.
• Differentiating \( \ln(x) \) results in \( \frac{1}{x} \), a simplification that plays a significant role in adjusting the integral components.
• This property helps to transform integrals like \( \int x \left( \cos(\ln x) \cdot \frac{1}{x} \right) \) into simpler forms, ultimately enabling precise simplification.

As we see in the solution steps, logarithmic functions can act as a pivot point to regroup and continuously refine the problem, leading to a balanced final result in calculus.