How do you calculate a reduction formula?

A reduction formula expresses an integral in terms of a simpler integral of the same type. This allows you to iteratively simplify the integral until it can be directly evaluated. Calculating a reduction formula typically involves using integration by parts and algebraic manipulation. There isn't one single method; instead, you tailor your approach based on the specific integral.

Steps Involved in Calculating a Reduction Formula:

1. Identify the Integral and Parameter: Recognize the integral you want to reduce, and identify the parameter (often an exponent, n) that you want to reduce. For example, you might have an integral of the form ∫ f(x)^n dx, where you aim to express this in terms of an integral involving f(x)^(n-1), f(x)^(n-2), and so on.

2. Apply Integration by Parts: Choose u and dv strategically to reduce the power of the function being integrated. This is the most critical step. Consider what happens to the exponent when differentiating or integrating parts of the integrand. Remember the integration by parts formula: ∫ u dv = uv - ∫ v du

3. Perform the Integration by Parts: Carefully execute the integration by parts, obtaining an expression of the form uv - ∫ v du.

4. Algebraic Manipulation: This is often the most challenging part. The goal is to manipulate the resulting integral (∫ v du) to resemble the original integral, but with a reduced power of n. This may involve:

   • Adding zero in a clever way.
   • Using trigonometric identities (if trigonometric functions are involved).
   • Substituting back the original integral expression.
   • Factoring and simplifying.

5. Isolate the Reduction Formula: Rearrange the equation to isolate the original integral (∫ f(x)^n dx) on one side, expressed in terms of a constant term and an integral with a reduced power (e.g., ∫ f(x)^(n-1) dx or ∫ f(x)^(n-2) dx). This isolated expression is the reduction formula.

Examples:

Example 1: Reduction Formula for ∫ sin^n(x) dx

Let I_n = ∫ sinⁿ(x) dx

• Let u = sin^n-1(x) and dv = sin(x) dx
• Then du = (n-1)sin^n-2(x)cos(x) dx and v = -cos(x)

Applying integration by parts:

I_n = -cos(x)sin^n-1(x) + (n-1)∫ cos²(x)sin^n-2(x) dx
I_n = -cos(x)sin^n-1(x) + (n-1)∫ (1-sin²(x))sin^n-2(x) dx
I_n = -cos(x)sin^n-1(x) + (n-1)∫ sin^n-2(x) dx - (n-1)∫ sinⁿ(x) dx
I_n = -cos(x)sin^n-1(x) + (n-1)I_n-2 - (n-1)I_n

Now, isolate I_n:

I_n + (n-1)I_n = -cos(x)sin^n-1(x) + (n-1)I_n-2
nI_n = -cos(x)sin^n-1(x) + (n-1)I_n-2
I_n = (-1/n)cos(x)sin^n-1(x) + ((n-1)/n)I_n-2

Therefore, the reduction formula is:

∫ sinⁿ(x) dx = (-1/n)cos(x)sin^n-1(x) + ((n-1)/n)∫ sin^n-2(x) dx

Example 2: Reduction Formula for ∫ x^n e^x dx

Let I_n = ∫ xⁿe^x dx

• Let u = xⁿ and dv = e^x dx
• Then du = nx^n-1 dx and v = e^x

Applying integration by parts:

I_n = xⁿe^x - ∫ e^xnx^n-1 dx
I_n = xⁿe^x - n∫ x^n-1e^x dx
I_n = xⁿe^x - nI_n-1

Therefore, the reduction formula is:

∫ xⁿe^x dx = xⁿe^x - n∫ x^n-1e^x dx

Key Considerations:

• Choosing u and dv: This is crucial. A good choice will simplify the integral in the ∫ v du term. Often, you choose u to be a part of the integrand whose derivative reduces the power or complexity.
• Algebraic Skills: Proficiency in algebraic manipulation, trigonometric identities, and other mathematical techniques is necessary to successfully derive reduction formulas.
• Practice: Deriving reduction formulas requires practice. Working through different examples will build your intuition.

Reduction formulas provide a powerful technique for solving complex integrals by breaking them down into simpler forms that can eventually be evaluated. They rely on careful application of integration by parts and strategic algebraic manipulation.