Question

derive the given reduction formula using integration by parts. $$ \int \cos ^{\alpha} x d x=\frac{\cos ^{\alpha-1} x \sin x}{\alpha}+\frac{\alpha-1}{\alpha} \int \cos ^{\alpha-2} x d x $$

Short answer

Derived reduction formula is correct using integration by parts.

Step-by-step solution

Identify the Formula

The reduction formula to be derived is \( \int \cos^{\alpha} x \, dx = \frac{\cos^{\alpha-1} x \sin x}{\alpha} + \frac{\alpha-1}{\alpha} \int \cos^{\alpha-2} x \, dx \). We will use integration by parts, where \( u \) is a function of \( x \) and \( dv \) is a differential form.

Choose \( u \) and \( dv \)

Choose \( u = \cos^{\alpha - 1} x \) and \( dv = \cos x \, dx \). Thus, \( du = (\alpha - 1) \cos^{\alpha - 2} x (-\sin x) \, dx \) and \( v = \sin x \).

Apply Integration by Parts

Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \). First, calculate \( uv = \cos^{\alpha - 1} x \sin x \).

Compute \( \int v \, du \)

Evaluate the integral \( \int v \, du = \int \sin x \cdot (\alpha - 1) \cos^{\alpha - 2} x (-\sin x) \, dx = - (\alpha - 1) \int \sin^2 x \cos^{\alpha - 2} x \, dx \).

Simplify \( \sin^2 x \)

Use the trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) to simplify: \( \int \sin^2 x \cos^{\alpha - 2} x \, dx = \int (1 - \cos^2 x) \cos^{\alpha - 2} x \, dx \).

Split the Integral

Solve \( \int (1 - \cos^2 x) \cos^{\alpha - 2} x \, dx = \int \cos^{\alpha-2} x \, dx - \int \cos^{\alpha} x \, dx \).

Substitute and Arrange

Substitute back the result from Step 6 into the equation from Step 4. This gives \[ - (\alpha - 1) \left( \int \cos^{\alpha-2} x \, dx - \int \cos^{\alpha} x \, dx \right) \].

Combine All Parts

The integration by parts gives \( \cos^{\alpha - 1} x \sin x - (\alpha - 1) \left( \int \cos^{\alpha-2} x \, dx - \int \cos^{\alpha} x \, dx \right) \). Rearrange to isolate \( \int \cos^{\alpha} x \, dx \).

Finalize the Reduction Formula

Rearrange the terms to show that \[ \int \cos^{\alpha} x \, dx = \frac{\cos^{\alpha-1} x \sin x}{\alpha} + \frac{\alpha-1}{\alpha} \int \cos^{\alpha-2} x \, dx \]. This confirms the derived reduction formula.

Key concepts

Reduction Formula

Reduction formulas are handy tools in calculus that simplify complex integrations by breaking them into simpler parts. They involve transforming a complicated integral into an equivalent expression with a lower degree or a simpler form of the original integral. This technique is especially useful when dealing with integrals raised to a power or involving recursive expressions.

The formula presented is a reduction formula where the integral of \( \cos^{\alpha} x \) is rewritten in terms of \( \cos^{\alpha-2} x \) and other simpler terms. By doing so, it makes the process of evaluating the integral much more manageable.

Reduction formulas often build upon themselves, allowing repetitive use to solve otherwise arduous integrals by strategically reducing the degree of difficulty.

Trigonometric Identities

Trigonometric identities provide extensions and equivalencies for trigonometric functions. They act as foundational tools that can simplify expressions or help solve equations involving trigonometric functions.

In this exercise, the trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) is used to break down and simplify the integral. This identity is particularly useful in integration and differentiation processes because it transforms quadratic trigonometric terms into simpler linear or constant terms.

This application of a trigonometric identity illustrates how such identities can make complex calculus problems more tractable, streamlining the path to a solution. Furthermore, understanding and applying these identities are essential for solving many problems in calculus and higher-level mathematics.

Integration Techniques

Integration techniques are methods used to solve integrals, which may not be straightforward. Among them, integration by parts is a powerful technique derived from the product rule for differentiation.

When faced with the integral \( \int \cos^{\alpha} x \, dx \), applying integration by parts simplifies the integration process by strategically choosing parts of the function to differentiate and integrate. The formula is given by \( \int u \, dv = uv - \int v \, du \), where the functions \( u \) and \( dv \) are chosen to ease the integral computation.

Integration by parts often involves the choice of \( u \) and \( dv \) that simplifies \( \int v \, du \). It requires practice to master, as selecting the right \( u \) and \( dv \) is crucial to successfully applying this technique. It's particularly helpful when working with products of algebraic and trigonometric functions.

Calculus Problem Solving

Solving calculus problems like deriving reduction formulas requires a series of strategic steps and techniques. First, it's crucial to comprehend the problem, identify given elements, and understand what is being asked.

The step-by-step approach is a hallmark of calculus problem-solving. For example, successfully deriving the reduction formula involves breaking down each component of the integration process into manageable steps. This might include:

• Setting up the problem using known formulas
• Choosing appropriate functions and differentiating them
• Applying calculus techniques effectively

Each step builds on previous ones, requiring meticulous attention to detail and mathematical accuracy.

Developing a systematic approach to calculus problem-solving enhances not only proficiency in solving complex equations but also fosters a deeper understanding of the mathematical concepts involved.