Reduction Formulae Revision Notes for AQA A-Level Further Maths

Reduction Formulae

What is a reduction formula?

A reduction formula is a special type of equation that expresses an integral $I_n$ in terms of similar integrals with lower indices, such as $I_{n-1}$ or $I_{n-2}$ . This creates a recursive relationship that allows you to break down complex integrals into simpler ones.

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Definition: A reduction formula for $I_n$ is an equation that relates $I_n$ to $I_{n-1}$ and/or $I_{n-2}$ . It can be repeatedly applied to reduce the integral to one that does not require integration by parts.

Reduction formulae are particularly useful when you need to integrate expressions of the form $\int u \frac{dv}{dx} dx$ where repeated application of integration by parts would be extremely time-consuming. Instead of applying integration by parts multiple times manually, you can derive a formula once and then apply it repeatedly until you reach a simple integral you can evaluate directly.

Integration by parts formula

The foundation of reduction formulae is the integration by parts formula:

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$\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$

This formula is essential because deriving a reduction formula typically involves applying integration by parts to the integral $I_n$ and then recognising that the resulting expression contains $I_{n-1}$ or $I_{n-2}$ .

How reduction formulae work

When you have an integral such as $\int x^n e^{-x} dx$ , applying integration by parts once gives you another integral that looks similar but with a lower power of $x$ . The key insight is that this new integral is actually $I_{n-1}$ , creating a relationship:

$I_n = (\text{some expression}) + (\text{some constant}) \times I_{n-1}$

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You can then use this formula repeatedly, reducing $I_n$ to $I_{n-1}$ , then $I_{n-1}$ to $I_{n-2}$ , and so on, until you reach $I_0$ or $I_1$ , which are usually simple integrals you can evaluate directly.

Important: Reduction formulae can be used for both definite and indefinite integrals. For definite integrals, you substitute the limits at the appropriate stage after reaching the simplest form.

Deriving a reduction formula

To derive a reduction formula, follow these steps:

Define your integral as $I_n$ with the appropriate function involving $n$
Apply integration by parts with carefully chosen $u$ and $\frac{dv}{dx}$
Simplify the expression and identify where $I_{n-1}$ or $I_{n-2}$ appears
Rearrange to express $I_n$ explicitly in terms of lower-index integrals

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The choice of $u$ and $\frac{dv}{dx}$ is crucial. Generally, choose $u$ as the part containing the power $n$ , so that when you differentiate it, the power reduces.

Worked example 1: Logarithmic functions

Consider the integral $I_n = \int_0^1 (\ln x)^n dx$ . We need to prove that $I_n = -nI_{n-1}$ and use this to evaluate $I_4$ .

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Worked Example: Deriving and Using a Logarithmic Reduction Formula

Step 1: Apply integration by parts

Let $u = (\ln x)^n$ and $\frac{dv}{dx} = 1$ .

Then $\frac{du}{dx} = n(\ln x)^{n-1} \cdot \frac{1}{x} = \frac{n(\ln x)^{n-1}}{x}$ and $v = x$ .

Using the integration by parts formula:

$I_n = \int_0^1 (\ln x)^n dx = [x(\ln x)^n]_0^1 - \int_0^1 x \cdot \frac{n(\ln x)^{n-1}}{x} dx$

$= [x(\ln x)^n]_0^1 - n\int_0^1 (\ln x)^{n-1} dx$

Step 2: Evaluate the boundary term

The first term evaluates to 0 because when $x = 1$ , $\ln 1 = 0$ , and as $x \to 0^+$ , $x(\ln x)^n \to 0$ .

The remaining integral is $I_{n-1}$ , so:

$I_n = -nI_{n-1}$

This is our reduction formula.

Step 3: Apply the formula repeatedly to find $I_4$

$I_4 = -4I_3$
$= -4(-3I_2)$
$= 12I_2$
$= 12(-2I_1)$
$= -24I_1$

Step 4: Calculate $I_1$

Now we need $I_1 = \int_0^1 \ln x \, dx$ .

Using integration by parts again with $u = \ln x$ and $\frac{dv}{dx} = 1$ :

$I_1 = [x \ln x]_0^1 - \int_0^1 x \cdot \frac{1}{x} dx = 0 - [x]_0^1 = -1$

Step 5: Find the final answer

Therefore:

$I_4 = -24 \times (-1) = \boxed{-24}$

Worked example 2: Functions with square roots

Given that $I_n = \int_{-1}^0 x^n\sqrt{x+1} \, dx$ , show that $I_n = -\frac{2n}{3+2n}I_{n-1}$ and evaluate $\int_{-1}^0 x^3\sqrt{x+1} \, dx$ .

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Worked Example: Deriving a Square Root Reduction Formula

Step 1: Rewrite and apply integration by parts

First, rewrite the integrand as $x^n(x+1)^{1/2}$ .

Apply integration by parts with $u = x^n$ and $\frac{dv}{dx} = (x+1)^{1/2}$ .

Then $\frac{du}{dx} = nx^{n-1}$ and $v = \frac{2}{3}(x+1)^{3/2}$ .

Using integration by parts:

$I_n = \left[\frac{2}{3}x^n(x+1)^{3/2}\right]_{-1}^0 - \int_{-1}^0 \frac{2}{3}(x+1)^{3/2} \cdot nx^{n-1} dx$

The first term evaluates to 0 because both limits give $0$ .

$I_n = -\frac{2n}{3}\int_{-1}^0 x^{n-1}(x+1)^{3/2} dx$

Step 2: Manipulate to reveal $I_n$ and $I_{n-1}$

Now rewrite $(x+1)^{3/2} = (x+1)^{1/2}(x+1)$ :

$I_n = -\frac{2n}{3}\int_{-1}^0 x^{n-1}\sqrt{x+1}(x+1) dx$

$= -\frac{2n}{3}\int_{-1}^0 x^{n-1}\sqrt{x+1} \cdot x + x^{n-1}\sqrt{x+1} dx$

$= -\frac{2n}{3}\left(\int_{-1}^0 x^n\sqrt{x+1} dx + \int_{-1}^0 x^{n-1}\sqrt{x+1} dx\right)$

$= -\frac{2n}{3}(I_n + I_{n-1})$

Step 3: Rearrange to isolate $I_n$

$I_n = -\frac{2n}{3}I_n - \frac{2n}{3}I_{n-1}$

$I_n + \frac{2n}{3}I_n = -\frac{2n}{3}I_{n-1}$

$I_n\left(1 + \frac{2n}{3}\right) = -\frac{2n}{3}I_{n-1}$

$I_n = -\frac{2n}{3+2n}I_{n-1}$

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Worked Example: Evaluating $I_3$

Apply the formula repeatedly:

$I_3 = -\frac{6}{9}I_2 = -\frac{2}{3}I_2$

$= -\frac{2}{3} \times \left(-\frac{4}{7}\right)I_1 = \frac{8}{21}I_1$

$= \frac{8}{21} \times \left(-\frac{2}{5}\right)I_0 = -\frac{16}{105}I_0$

Calculate $I_0$ :

$I_0 = \int_{-1}^0 (x+1)^{1/2} dx = \left[\frac{2}{3}(x+1)^{3/2}\right]_{-1}^0 = \frac{2}{3}(1) - 0 = \frac{2}{3}$

Find the final answer:

$I_3 = -\frac{16}{105} \times \frac{2}{3} = \boxed{-\frac{32}{315}}$

Worked example 3: Trigonometric functions

Find a reduction formula for $I_n = \int_0^\pi \sin^n x \, dx$ and use it to find $\int_0^\pi \sin^8 x \, dx$ and $\int_0^\pi \sin^7 x \, dx$ .

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Sometimes the integrand is not initially in the form of a product of two functions. In such cases, you can split up the integral yourself. For example, write $\sin^n x$ as $\sin x \cdot \sin^{n-1} x$ .

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Worked Example: Deriving a Trigonometric Reduction Formula

Step 1: Split the function and apply integration by parts

Write $\sin^n x = \sin x \cdot \sin^{n-1} x$ and apply integration by parts.

Let $u = \sin^{n-1} x$ and $\frac{dv}{dx} = \sin x$ .

Then $\frac{du}{dx} = (n-1)\sin^{n-2} x \cos x$ and $v = -\cos x$ .

Using integration by parts:

$I_n = \left[-\cos x \sin^{n-1} x\right]_0^\pi - \int_0^\pi (-\cos x)(n-1)\sin^{n-2} x \cos x \, dx$

$= 0 + (n-1)\int_0^\pi \cos^2 x \sin^{n-2} x \, dx$

Step 2: Use the identity $\cos^2 x = 1 - \sin^2 x$

Replace $\cos^2 x$ with $1 - \sin^2 x$ :

$I_n = (n-1)\int_0^\pi (1 - \sin^2 x)\sin^{n-2} x \, dx$

$= (n-1)\int_0^\pi \sin^{n-2} x - \sin^n x \, dx$

$= (n-1)I_{n-2} - (n-1)I_n$

Step 3: Rearrange to find the reduction formula

$I_n + (n-1)I_n = (n-1)I_{n-2}$

$nI_n = (n-1)I_{n-2}$

$I_n = \frac{n-1}{n}I_{n-2}$

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Worked Example: Using the Formula to Find $I_8$

Apply the formula repeatedly:

$I_8 = \frac{7}{8}I_6$

$= \frac{7}{8} \times \frac{5}{6}I_4 = \frac{35}{48}I_4$

$= \frac{35}{48} \times \frac{3}{4}I_2 = \frac{105}{192}I_2$

$= \frac{105}{192} \times \frac{1}{2}I_0 = \frac{105}{384}I_0$

Now evaluate $I_0$ :

$I_0 = \int_0^\pi 1 \, dx = [x]_0^\pi = \pi$

Therefore:

$I_8 = \frac{105\pi}{384} = \boxed{\frac{35\pi}{128}}$

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Worked Example: Using the Formula to Find $I_7$

$I_7 = \frac{6}{7}I_5$

$= \frac{6}{7} \times \frac{4}{5}I_3 = \frac{24}{35}I_3$

$= \frac{24}{35} \times \frac{2}{3}I_1 = \frac{48}{105}I_1 = \frac{16}{35}I_1$

Now evaluate $I_1$ :

$I_1 = \int_0^\pi \sin x \, dx = [-\cos x]_0^\pi = -(-1) - (-1) = 2$

Therefore:

$I_7 = \frac{16}{35} \times 2 = \boxed{\frac{32}{35}}$

Strategy for finding and using reduction formulae

When tackling problems involving reduction formulae, follow this systematic approach:

Step 1: Split up the function if necessary and integrate

If the integrand is not already in the form of a product of two functions, rewrite it. For example, write $\sin^n x$ as $\sin x \cdot \sin^{n-1} x$ or $\sin^{n-2} x \cdot \sin^2 x$ , depending on which form will be most useful.

Step 2: Find an expression for $I_n$ in terms of $I_{n-1}$ and/or $I_{n-2}$

Apply integration by parts carefully, choosing $u$ and $\frac{dv}{dx}$ so that the power of $n$ reduces when you differentiate. After integration by parts, look for where $I_{n-1}$ or $I_{n-2}$ appears in your expression. You may need to factorise or use trigonometric identities to reveal the reduction formula.

Step 3: Use the formula repeatedly until you reach $I_0$ or $I_1$

Keep applying your reduction formula to reduce the index. Continue until you reach either $I_0$ or $I_1$ , which should be integrals you can evaluate directly without integration by parts.

Step 4: Calculate $I_0$ or $I_1$ and use with your formula to find $I_n$ for the value of $n$ required

Evaluate the simplest integral ( $I_0$ or $I_1$ ) directly. Then substitute this value back into your chain of reductions to find the final answer for the required value of $n$ .

Key exam tips

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Tip 1: Rearranging the integrand

After applying the integration by parts formula, you may need to rearrange the integrand so that it is clearly in the form of $I_{n-1}$ and/or $I_{n-2}$ . Look for opportunities to factorise expressions or use trigonometric identities like $\cos^2 x = 1 - \sin^2 x$ .

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Tip 2: Watch your limits

For definite integrals, be very careful when substituting limits. Terms involving $\ln 0$ or similar problematic expressions often evaluate to $0$ when you consider the limit properly. For example, $\lim_{x \to 0^+} x \ln x = 0$ .

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Tip 3: Know when you're done

Recognise when you've reached $I_0$ or $I_1$ . These should be integrals you can evaluate directly, such as $\int 1 \, dx$ , $\int e^x dx$ , or $\int \sin x \, dx$ .

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Tip 4: Check for algebraic errors

When applying the reduction formula repeatedly, it's easy to make arithmetic mistakes with fractions. Work carefully and consider simplifying at each step to keep expressions manageable.

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Tip 5: Even vs odd indices

In some reduction formulae (like the $\sin^n x$ example), notice that $I_n$ relates to $I_{n-2}$ , not $I_{n-1}$ . This means even powers of $n$ eventually reduce to $I_0$ , while odd powers reduce to $I_1$ . Be prepared to evaluate both.

Summary

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Key Points to Remember:

A reduction formula relates $I_n$ to $I_{n-1}$ and/or $I_{n-2}$ , allowing you to break down complex integrals systematically.
Derive reduction formulae using integration by parts, choosing $u$ and $\frac{dv}{dx}$ carefully so that the power reduces when you differentiate.
Apply the reduction formula repeatedly until you reach a simple integral ( $I_0$ or $I_1$ ) that you can evaluate directly.
Reduction formulae work for both definite and indefinite integrals – just remember to apply limits correctly for definite integrals.
When working with trigonometric functions, you may need to use identities like $\cos^2 x = 1 - \sin^2 x$ to reveal the reduction formula structure.

Reduction Formulae (AQA A-Level Further Maths): Revision Notes