Ultimate Guide to Integration Recurrence Relations (Reduction Formulas)

Introduction

What do you do if you are asked to evaluate $\int \sin^{10}(x) \, dx$ ? Writing out Integration by Parts ten times is completely unrealistic for an exam. In HSC Mathematics Extension 2, we solve this by creating Integration Recurrence Relations (often called Reduction Formulas). By generalizing the integral as $I_n$ , we can use Integration by Parts to create a formula that links $I_{10}$ to $I_8$ , $I_8$ to $I_6$ , and so on, stepping down until the integral collapses into something trivial.

Executive Summary

This guide covers advanced recurrence relations:

Defining $I_n$ : Setting up a general integral with an integer parameter $n$ .
Deriving the Relation: Using Integration by Parts to express $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$ .
Evaluating specific cases: "Stepping down" the formula to evaluate massive powers like $n=6$ .
Combining Techniques: Tackling theoretical problems that require substitution, partial fractions, and parts all at once.

What is this about?

A recurrence relation is a mathematical formula that defines a sequence of terms using the previous terms. In the context of calculus, a recurrence relation links the integral of a high power to the integral of a lower power.

For example, a typical HSC question will first ask you to prove that: $I_n = \frac{x^n e^x}{n} - \frac{1}{n} I_{n-1}$ Once you prove this, you can instantly evaluate $I_4$ just by plugging in $n=4$ and finding $I_3$ , which requires finding $I_2$ , which requires finding $I_1$ . It is a methodical, algorithmic approach that completely removes the need to do raw calculus on massive functions.

Main Content

1. Setting Up and Splitting for Parts

To derive a recurrence relation, you almost always use Integration by Parts: $\int u \, dv = uv - \int v \, du$ . The hardest step is deciding how to split your general integral $I_n$ into $u$ and $dv$ .

Rule of Thumb:

Algebraic / Exponentials: If $I_n = \int x^n e^x \, dx$ , let $u = x^n$ (so the power drops when you differentiate) and $dv = e^x \, dx$ .
Trigonometric powers: If $I_n = \int \sin^n x \, dx$ , you must "peel off" one power to create a $dv$ that you can actually integrate. Let $u = \sin^{n-1} x$ and $dv = \sin x \, dx$ .
Logarithms: If $I_n = \int (\ln x)^n \, dx$ , create a dummy $1$ . Let $u = (\ln x)^n$ and $dv = 1 \, dx$ .

2. The Algebraic Manipulation Step

After applying the Parts formula, you will get a new integral. This new integral rarely looks exactly like $I_{n-1}$ immediately. You will often have to use a trigonometric identity (like $\sin^2 x = 1 - \cos^2 x$ ) or algebraic manipulation to force the new integral to look like your original $I_n$ format.

3. Stepping Down to Evaluate

Once you have established the recurrence relation (e.g. $I_n = \frac{1}{n} + I_{n-2}$ ), the second part of the question will ask you to evaluate a specific integral, like $I_5$ .

Plug $n=5$ into the formula to find $I_5$ in terms of $I_3$ .
Plug $n=3$ into the formula to find $I_3$ in terms of $I_1$ .
The Base Case: You must manually integrate $I_1$ (which would just be $\int \sin^1 x \, dx$ ) using basic Extension 1 calculus.
Substitute $I_1$ back up the chain to get the final numerical answer.

4. Combining Multiple Techniques

The hardest Band 6 questions in the HSC will combine all the techniques you have learned. You might have to:

Use a trigonometric substitution ( $x = 2\sin\theta$ ).
Use Integration by Parts to create a recurrence relation on the result.
Use Partial Fractions to solve the base case.

When faced with a 4-mark theoretical integration problem, don't panic. Check if the integrand is a rational function (Partial fractions?), a product of different families (Parts?), or a high power (Recurrence relation?).

Simple Worked Example

Question: Let $I_n = \int_0^1 x^n e^x \, dx$ for $n \ge 0$ . (a) Show that $I_n = e - n I_{n-1}$ for $n \ge 1$ . (b) Hence, evaluate $I_3$ .

Solution: (a) Deriving the formula Step 1: Set up Parts. Let $u = x^n \implies du = nx^{n-1} \, dx$ Let $dv = e^x \, dx \implies v = e^x$

Step 2: Apply the Integration by Parts formula. $I_n = [uv]_0^1 - \int_0^1 v \, du$ $I_n = [x^n e^x]_0^1 - \int_0^1 e^x (nx^{n-1}) \, dx$

Step 3: Evaluate limits and extract constants. Evaluate the bracket: $(1^n e^1) - (0^n e^0) = e - 0 = e$ . Pull the constant $n$ out of the integral: $I_n = e - n \int_0^1 x^{n-1} e^x \, dx$

Notice that the remaining integral is exactly the definition of $I_n$ , just with $n-1$ instead of $n$ . Therefore: $I_n = e - n I_{n-1}$ (as required).

(b) Evaluating $I_3$ Step 1: Step down the formula. Using our formula: $I_3 = e - 3 I_2$ $I_2 = e - 2 I_1$ $I_1 = e - 1 I_0$

Step 2: Evaluate the base case ( $I_0$ ). What is $I_0$ ? Go back to the original definition and sub in $n=0$ : $I_0 = \int_0^1 x^0 e^x \, dx = \int_0^1 e^x \, dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ .

Step 3: Substitute back up the chain. $I_1 = e - 1(e - 1) = e - e + 1 = 1$ . $I_2 = e - 2(1) = e - 2$ . $I_3 = e - 3(e - 2) = e - 3e + 6 = 6 - 2e$ .

Final Answer: $I_3 = 6 - 2e$ .

mini-FAQ page

Q: Do I always step down to $I_0$ or $I_1$ ? A: It depends on the formula and the starting number. If your formula jumps by 2 (e.g. $I_n$ depends on $I_{n-2}$ ), then even numbers will step down to $I_0$ , and odd numbers will step down to $I_1$ . You must calculate the correct base case depending on the parity.

Q: Is it possible to have a recurrence relation that steps up instead of down? A: Yes! Sometimes questions ask you to express $I_{n-2}$ in terms of $I_n$ . You still derive the normal formula using parts, and then you just algebraically rearrange it at the end to make $I_{n-2}$ the subject.

Common mistakes to avoid

Forgetting to apply the limits to the $uv$ bracket: In a definite recurrence relation, the $uv$ part of the parts formula MUST be evaluated between the bounds immediately. Many students leave it as $[uv]_a^b$ and forget to actually calculate the numbers (like we got $e$ in the example above).
Misidentifying the base case: If the question states for $n \ge 1$ , then $I_1$ might be your base case, not $I_0$ . Always read the domain of $n$ carefully before evaluating the final integral.

Practice on Vu's Maths Hub

Recurrence relations are often the final, hardest question in the HSC exam.

Practice deriving complex trigonometric reduction formulas in the HSC Integration Booklet.
Combine all your skills in the ultimate Extension 2 test at Vu's Maths Hub.

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