Problem 30 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Solve by integrating by parts twice, resulting in a solution in terms of original integral.

Step 1: Identify u and dv

For integration by parts, we use the formula: \[ \int u \, dv = uv - \int v \, du \]Let \( u = \cos(\ln x) \), leading us to consider the substitution \( dv = dx \).

Step 2: Differentiate and Integrate

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

Step 3: Apply Integration by Parts Formula

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

Step 4: Simplify the Integration

Simplify the integral:\[ x \cos(\ln x) + \int \sin(\ln x) \, dx \] Now, let \( v \) (from integration by parts) be simplified inside the context of the formula, leading to easier addressing of the resulting integral.

Step 5: Address the Next Integral Using Rotation Techniques

We observed alternating scenarios and may need further repeat techniques until final integral naturally resolves via identification mechanism:\[ = x \cos(\ln x) + x\sin(\ln x) - \int x \left(\cos(\ln x)\right) \frac{1}{x} dx \] Potentially may necessitate cycle method prolongs back visible initial state simplified formula contexts.

Key Concepts

Integral Calculus

Integral calculus is all about finding the total amount, accumulation, or the area under a curve. In the context of evaluating \( \int \cos(\ln x) \, dx \), we are interested in finding a function that represents this accumulation effectively.
In calculus, the average method to solve these types of problems is often through anti-differentiation. This is the inverse process of differentiation and is essential for calculating integrals.
Integration by parts is a key method in integral calculus where you break an integral into easier parts to manage and integrate.

• The formula used is \( \int u \, dv = uv - \int v \, du \), where we aim to choose \( u \) and \( dv \) wisely.
• This method can simplify complex integrals by transforming them into simpler forms, allowing you to integrate more challenging expressions accurately.

Trigonometric Integrals

Trigonometric integrals involve integrals comprising trigonometric functions like sine, cosine, tangent, etc.
In our specific problem, \( \int \cos(\ln x) \, dx \), the trick was to choose \( u = \cos(\ln x) \).
The differential \( du \) introduces sine, transforming our integral in a manner hinged on the interrelation between these two primary trigonometric functions.

• Trigonometric integrals often require identity transformations for simplification.
• They sometimes involve cycles or repeating processes leveraging known trigonometric identities to resolve.
• For example, knowing that \( \sin^2(x) + \cos^2(x) = 1 \) can help in simplifying or breaking down complex integrals.

These integrals, especially those involving natural logarithmic elements, may require special attention and inventive substitution strategies.

Differentiation and Integration Techniques

This concept combines differentiation and integration strategies to solve complex calculus issues. In our exercise, we used Integration by Parts which heavily relies on selecting which part of the integral to differentiate and which to integrate.
Let's break it down:

• Selection of \( u \) and \( dv \): Picking these parts accurately is crucial. Generally, we choose \( u \) to be a function that becomes simpler when differentiated, and \( dv \) to be easily integrable.
• Differentiation: Calculating \( du \) involves understanding how functions like \( \cos(\ln x) \) behave when differentiated, which becomes \( -\sin(\ln x) \cdot \frac{1}{x} \, dx \).
• Integration: Finding \( v \) from \( dv \) by integration. Here, since \( dv = dx \), it simply becomes \( v = x \).
• Combining Results: Substituting into the integration by parts formula simplifies the overall function and breaks it down further, transforming the initial complex problem into manageable smaller parts.

These techniques aid in navigating through intricate calculus problems by providing systematic approaches to unravel and compute them efficiently.