Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int x^{\alpha} \cos \beta x d x=\frac{x^{\alpha} \sin \beta x}{\beta}-\frac{\alpha}{\beta} \int x^{\alpha-1} \sin \beta x d x $$

Step-by-Step Solution

Answer

Use integration by parts with \(u = x^\alpha\) and \(dv = \cos \beta x \, dx\).

Step 1: Identify Functions for Integration by Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). For the given integral \( I = \int x^\alpha \cos \beta x \, dx \), choose \( u = x^\alpha \) and \( dv = \cos \beta x \, dx \).

Step 2: Differentiate and Integrate

Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \). We have \( du = \alpha x^{\alpha-1} \, dx \) and \( v = \int \cos \beta x \, dx = \frac{\sin \beta x}{\beta} \).

Step 3: Substitute into Integration by Parts Formula

Substitute \( u, du, v, \) and \( dv \) into the integration by parts formula: \[ I = \int x^\alpha \cos \beta x \, dx = \frac{x^\alpha \sin \beta x}{\beta} - \int \frac{\sin \beta x}{\beta} \alpha x^{\alpha-1} \, dx \].

Step 4: Simplify the Integral

Pull constants out of the integral: \[ I = \frac{x^\alpha \sin \beta x}{\beta} - \frac{\alpha}{\beta} \int x^{\alpha-1} \sin \beta x \, dx \].

Step 5: Confirm the Reduction Formula

The formula now matches the given reduction formula: \[ \int x^{\alpha} \cos \beta x \, dx = \frac{x^{\alpha} \sin \beta x}{\beta} - \frac{\alpha}{\beta} \int x^{\alpha-1} \sin \beta x \, dx \].

Key Concepts

Reduction Formula

A reduction formula is a valuable tool in calculus, especially when integrating complex expressions. It simplifies an integral into a form that is easier to evaluate, typically by reducing the power of a variable or transforming the integral into a more standard form. For instance, in the given exercise, the reduction formula takes:\[ \int x^{\alpha} \cos \beta x \, dx = \frac{x^{\alpha} \sin \beta x}{\beta} - \frac{\alpha}{\beta} \int x^{\alpha-1} \sin \beta x \, dx \]This approach helps in breaking down the original integral into simpler components, sometimes working recursively with a simpler integral, thus facilitating easier computation. In most cases, you would repeat the process of applying the reduction formula until the integral reaches a basic form you can evaluate directly, like a simple trigonometric or polynomial integral. This strategy effectively converts complex integration problems into more manageable steps.

Differentiation

Differentiation is a process that finds the rate at which a quantity changes, and it is the inverse operation of integration. In the context of integration by parts, differentiation is employed to determine the derivative of the chosen function. For our problem, if we choose \( u = x^{\alpha} \), then:

• The differential becomes \( du = \alpha x^{\alpha-1} \, dx \).

Differentiation here helps us 'break down' a polynomial term, reducing its degree, which is crucial for simplifying the integral. By determining \( du \), you're halfway to being able to express your integral in terms of parts that are much easier to manage. Combining these derivatives with integrals of complementary parts constructs a path to solve intricate integrals effectively.

Integration

Integration is a fundamental concept in calculus that computes the total accumulation or area under a curve, which essentially is the inverse process of differentiation. When handling integration by parts, identifying which part of the integrand should be integrated is vital.In our example:

• We set \( dv = \cos \beta x \, dx \), which upon integration gives us \( v = \frac{\sin \beta x}{\beta} \).

This integration allows us to construct the formula within integration by parts:\[ \int u \, dv = uv - \int v \, du \]Now we possess both the differentiated and integrated components needed to transform and simplify the original integral. Integration serves as a bridge, helping us convert differentiation results into areas under complex mathematical curves, crucially transforming problems into solvable steps.

Trigonometric Functions

Trigonometric functions like sine and cosine often appear in calculus problems due to their periodic nature and applications in modeling real-world scenarios. Understanding how to integrate and differentiate these functions is key to solving integrals involving them.In the context of the exercise problem, knowing the integral of \( \cos \beta x \) was essential:

• \( \int \cos \beta x \, dx = \frac{\sin \beta x}{\beta} \).

These functions, due to their oscillating properties, often result in repetitive patterns that can simplify integration by parts. As seen here, applying the correct integral form allows the control over other terms through the reduction formula, ultimately simplifying complex algebraic and trigonometric expressions into a manageable solution.