Problem 16 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int x \csc ^{2} x d x $$

Step-by-Step Solution

Answer

The integral is \(-x \cot x + \ln |\sin x| + C\).

Step 1: Identify Parts for Integration by Parts

Integration by parts is given by the formula: \( \int u \, dv = uv - \int v \, du \).To apply this, we need to identify parts: let \( u = x \) and \( dv = \csc^2 x \, dx \). Then, we need to differentiate \( u \) and integrate \( dv \).

Step 2: Differentiate and Integrate

Differentiate \( u = x \), which gives \( du = dx \). Integrate \( dv = \csc^2 x \, dx \), knowing that \( \int \csc^2 x \, dx = -\cot x \). Therefore, \( v = -\cot x \).

Step 3: Apply Integration by Parts Formula

Plug \( u, du, v, \) and \( dv \) into the integration by parts formula:\[ \int x \csc^2 x \, dx = uv - \int v \, du = x (-\cot x) - \int (-\cot x) \, dx \]. Simplify this expression to \[ -x \cot x + \int \cot x \, dx \].

Step 4: Integrate Remaining Term

Now, find the integral \( \int \cot x \, dx \). This integral results in the natural logarithm of the sine function: \( \int \cot x \, dx = \ln |\sin x| + C \).

Step 5: Substitute Back and Simplify

Substitute the solution from Step 4 back into the equation from Step 3, resulting in:\[ -x \cot x + \ln |\sin x| + C \], where \( C \) is the constant of integration.

Key Concepts

Definite and Indefinite Integrals

Integrals are essential in calculus, helping us find areas under curves, among other things. Understanding definite and indefinite integrals is critical for solving various mathematical problems.

• Indefinite Integrals: These are used to find the antiderivative of a function. The result is a family of functions, each differing by a constant known as the constant of integration, often represented as C. When you see a problem like \( \int f(x) \, dx \), our aim is to determine the function whose derivative is f(x).
• Definite Integrals: Unlike indefinite integrals, these are used to calculate the net area between the graph of a function and the x-axis, bounded by specified limits \( a \) and \( b \). They are expressed as \( \int_{a}^{b} f(x) \, dx \). The answer is a number that represents the accumulation of f(x) over the interval from a to b, which is not just about finding antiderivatives but also evaluating them at these boundaries.

In our example, we focused on an indefinite integral, \( \int x \csc^2 x \, dx \). The solution provides an antiderivative, noting that results may vary according to the constant of integration. Understanding the nuances between these two can greatly improve problem-solving skills.

Trigonometric Functions

Trigonometric functions involving sine, cosine, tangent, and their reciprocals (cosecant, secant, cotangent) play a fundamental role in calculus. These functions describe angles and their relationships in a triangle or a unit circle, and they crop up often in various integrals:

• Understanding \(\csc^2 x\):
  • \( \csc^2 x \) is the square of the cosecant function, which is the reciprocal of sine. This means \( \csc x = 1/\sin x \), making it useful when dealing with functions that are difficult to integrate directly.
  • Knowing basic trigonometric relationships and derivatives, like the fact that \( \frac{d}{dx}(\cot x) = -\csc^2 x \), helps when identifying parts that can be integrated easily.
• Simplifying Integrals with Trigonometric Substitutions: Sometimes, integrals are simplified by substituting the trigonometric functions with more straightforward expressions. This is especially true in cases where regular functions don't easily integrate.

Understanding these basic properties can make calculus problems involving trigonometric functions more approachable and less intimidating.

Differentiation and Integration Techniques

Differentiation and integration are two sides of the same coin, crucial for solving calculus problems effectively. While differentiation breaks functions into their smallest rate of change, integration builds them back up to find total accumulation.When dealing with complex integrals, such as our \( \int x \csc^2 x \, dx \), leveraging various techniques is necessary:

• Integration by Parts: This technique is specifically for tackling integrals of products of functions. The formula \( \int u \, dv = uv - \int v \, du \) effectively turns the integral of a product into a simpler problem. Key to success is choosing u and dv wisely.
  • In our case, letting u = x helps to easily differentiate the term to \(du = dx \), making the integration straightforward after substituting \(\csc^2 x \) with its integral, \(-\cot x\).
  • It’s important to select functions that simplify under differentiation and integration, making the problem manageable.
• Pattern Recognition: Sometimes similar standard integral forms pop up, which can simplify problems quickly once recognized. Knowing integral formulas for common functions can expedite solving complex integrals.

Mastering these techniques enables you to tackle a broad range of problems, ensuring you can integrate with confidence and precision.