Problem 56

Question

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

Step-by-Step Solution

Verified

Answer

Integration by parts and simplification gives the reduction formula as provided.

1Step 1: Identify Functions for Integration by Parts

The integration by parts formula is $ \int u \, dv = uv - \int v \, du $. Let $ u = x^{\alpha} $ and $ dv = \sin(\beta x) \, dx $. Next, we need to find $ du $ and $ v $.

2Step 2: Differentiate and Integrate

Find $ du $ by differentiating $ u $: $ du = \alpha x^{\alpha-1} \, dx $. Integrate $ dv $ to get $ v $: $ v = -\frac{1}{\beta} \cos(\beta x) $.

3Step 3: Substitute into Integration by Parts Formula

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

4Step 4: Simplify the Expression

Simplify the expression: $-\frac{x^{\alpha} \cos(\beta x)}{\beta} + \frac{\alpha}{\beta} \int x^{\alpha-1} \cos(\beta x) \, dx $. This matches the given reduction formula.

Key Concepts

Integration by PartsDifferential CalculusIntegral Calculus

Integration by Parts

Integration by parts is a technique used in calculus for solving integrals. It helps us integrate the product of two functions.
The formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

This formula is derived from the product rule of differentiation. To apply this method correctly, we need to carefully choose which function to differentiate ($ u $) and which to integrate ($ dv $).
Sometimes, the choice can significantly simplify the problem. For example, in an integral $ \int x^\alpha \sin(\beta x) dx $, choosing $ u = x^\alpha $ makes $ du = \alpha x^{\alpha-1} dx $ straightforward, while choosing $ dv = \sin(\beta x) dx $ allows us to get $ v = -\frac{1}{\beta} \cos(\beta x) $ without much hassle. This technique is very helpful when dealing with products of polynomial and trigonometric functions.

Differential Calculus

Differential calculus focuses on understanding how functions change, primarily through derivatives.
In reduction formulas like the one in the given exercise, differentiation plays a key role. It helps us simplify steps in the reduction formula derivation.

Differentiating a function like $ u = x^\alpha $ involves applying the power rule. The derivative $ \frac{d}{dx}(x^\alpha) = \alpha x^{\alpha-1} $ is easy to compute and critical in solving integrals by parts.

Differential calculus helps us understand the behavior and slope of functions. It is foundational not only for integration by parts but also for other advanced calculus topics.

Integral Calculus

Integral calculus is concerned with finding functions given their rate of change, essentially reversing differentiation. It involves calculating areas under curves.
When tackling integrals using the reduction formula, integral calculus techniques guide us in integrating various components of the equation.

For instance, in the given formula, after applying integration by parts, we simplify and focus on the integral $ \int x^{\alpha-1} \cos(\beta x) dx $.
Integral calculus is not just about finding antiderivatives but also understanding the accumulation of quantities, such as the area or volume.

The strategic combination of differential and integral calculus empowers us to address complex integrals via reduction formulas, simplifying them to manageable expressions for further calculations.

Problem 56

Problem 57

Other exercises in this chapter

Problem 55

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

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Problem 56

Find the derivatives of the Fresnel functions (a) $\frac{d}{d x} S(x)$ (b) $\frac{d}{d x} C(x)$

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Problem 57

Evaluate $\int_{0}^{2 \pi} \frac{x|\sin x|}{1+\cos ^{2} x} d x$. Hint: Make the substitution $u=x-\pi$ in the definite integral and then use symmetry proper

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Problem 57

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}-\fra

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