Problem 16 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x \csc ^{2} x d x $$

Step-by-Step Solution

Answer

The integral \( \int x \csc^2 x \, dx = -x \cot x + \ln |\sin x| + C \).

Step 1: Identify Functions for Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). Here, we need to identify parts of the integral: let \( u = x \) and \( dv = \csc^2 x \, dx \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) to find \( du \), which gives \( du = dx \). Now, integrate \( dv \) to find \( v \), giving \( v = -\cot x \) because \( \int \csc^2 x \, dx = -\cot x \).

Step 3: Apply the Integration by Parts Formula

Substitute into the formula: \( \int x \csc^2 x \, dx = uv - \int v \, du \). So we have: \( \int x \csc^2 x \, dx = x(-\cot x) - \int (-\cot x) \, dx \).

Step 4: Simplify the Expression

This simplifies to \( -x \cot x + \int \cot x \, dx \). The next task involves calculating \( \int \cot x \, dx \).

Step 5: Integrate \( \int \cot x \, dx \)

The integral of \( \cot x \) is \( \ln |\sin x| + C \). Thus, \( \int \cot x \, dx = \ln |\sin x| \).

Step 6: Combine the Results

Substitute back to obtain: \( -x \cot x + \ln |\sin x| + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

Integration by parts is a powerful tool when it comes to handling products of functions. The core idea here is to transform a challenging integral into parts that are easier to manage. The basic formula is \( \int u \, dv = uv - \int v \, du \). Here,

• Choose one part of the integrand as \( u \), which is typically a polynomial or function that simplifies when differentiated.
• The other part, \( dv \), should be something that is easy to integrate.

In our problem, we've chosen \( u = x \), as differentiating \( x \) yields a simple result. The other part \( dv \) is set to \( \csc^2 x \, dx \), a trigonometric function which integrates nicely into \( -\cot x \). This method often involves a little guesswork to pick \( u \) and \( dv \) effectively, but practice helps develop intuition. By transforming the original integral into a simpler one, integration by parts becomes an essential technique in calculus.

Calculus

Calculus is fundamental to understanding the change and motion in mathematics. It provides us with an in-depth approach to handle problems involving change rates and areas under curves, essential in various fields like engineering and physics.
There are two main branches: differentiation, focusing on the rate of change, and integration, concerning accumulation and area. In our example, integration by parts—a method derived from the product rule of differentiation—helps in finding the antiderivative of a product of functions.
As with our current integral, calculus often involves selecting appropriate methods to deconstruct and solve integrals, whether through substitution or more advanced techniques like integration by parts. By effectively applying these approaches, calculus enables us to tackle complex mathematical problems with precision and simplicity.

Trigonometric Integrals

Trigonometric integrals involve integrating functions with trigonometric expressions, a common occurrence in calculus problems. In this particular integral involving \(x\) and \(\csc^2 x\):

• The integral of \(\csc^2 x\) results in \(-\cot x\), based on basic integration rules of trigonometric functions.
• The integral of \(\cot x\) is \(\ln |\sin x|\), a result derived from simplifying the derivative of a logarithmic function.

Understanding these foundational integrals allows us to integrate more complex expressions involving trigonometric terms.
When computing trigonometric integrals, it's crucial to remember simpler forms and identities, as these often simplify our work significantly. This kind of problem-solving approach not only reinforces our understanding of trigonometric functions but also hones our skills in integration, making it an invaluable part of studying calculus.