Part 10: Recurrence Relations | Beginner’s Guide to Year 12 Ext 2 Maths

In this article, we discuss how to approach recurrence relations problems and provide practice questions soy ou don't have to deal with recurrently weak marks!

beginner's guide to year 12 math ext 2 recurrence relations banner

In this article, we will discuss how to deal with two types of recurrence relation problems: questions that require integration by parts and questions that don’t require integration by parts.


In this article, we’ll discuss:


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NESA Syllabus Outcomes

  • Further Integration: Students derive and use recurrence relations.



Assumed Knowledge

Students should be familiar with how to integrate a range of functions (power, exponential, logarithmic and trigonometric) as well as how to do integration by parts.



Introduction to Recurrence Relations

When you used integration by parts to evaluate integrals such as \( \int x^4 sin(x) \ dx\), you may have noticed that the ‘remaining integral’ obtained was almost identical to the original one.

Let’s have a look at the example we have here.

Applying integration by parts by letting \(u=x^4\) and \(dv=sin(x)\) yields:

\(\int x^4 sin(x) \ dx= -x^4cos(x)+4 \int x^3 cos(x) \ dx\)


Now, looking at \( \int x^3 cos(x) \ dx\), applying integration by parts on this again, and letting \(u=x^3\) and yields \(dv=cos(x)\):

\( \int x^3 cos(x) \ dx = x^3 sin(x) \ – 3 \int x^2 sin(x) \ dx \)


After applying integration by parts twice, you’ll notice that we have obtained a similar integral to the one we had originally but with a lower power for the \(x\) term.

We refer to relationships of this kind as recurrence relations.

In recurrence relations questions, we generally want to find \(I_n\) (the \(n^{th}\) power of the integral) and express it in terms of its \((n-1)^{th}, (n-2)^{th}, … etc.\) powers of the integral \((I_{n-1}, I_{n-2}, …)\).

We can then use these relationships to evaluate integrals where we are given a deterministic value of \(n\).



Without Integration by Parts:

As a rule of thumb, if the formula you are required to prove does not have a function of \(n\) in front of the \(I_{n-k}\) term, you are generally not required to use integration by parts.


Example 1:

Let \(I_n= \int_0^1 \frac{x^n}{x^2+1} \ dx\). Show that \(I_n= \frac{1}{n-1}-I_{n-2}\).



First, we notice that that there is no function of \(n\) in front of the \(I_{n-2}\) term, so it is likely we won’t need to use integration by parts here.

We proceed in this question by manipulating the integral algebraically.

I_n &= \int_0^1 \frac{x^n}{x^2+1} \ dx\\
&= \int_0^1 \frac{x^n+x^{n-2}-x^{n-2}}{x^2+1} \ dx\\
&= \int_0^1 \frac{x^{n-2}(x^2+1)-x^{n-2}}{x^2+1} \ dx\\
&= \int_0^1 \left( x^{n-2} – \frac{x^{n-2}}{x^2+1} \right)\\
&= \int_0^1 x^{n-2} \ dx \ – \int_0^1 \frac{x^{n-2}}{x^2+1} \ dx\\
&= \left[ \frac{x^{n-1}}{n-1} \right]_0^1 – I_{n-2}\\
&= \frac{1}{n-1}-I_{n-2}\\


As you can see here, we did not use any integration by parts but managed to derive the recurrence relation!

The difficult part about dealing with this type of recurrence relation is correctly manipulating the integral algebraically to obtain lower powers of the integral.

Another method of dealing with this question would be to rearrange the recurrence relation to try to prove that \(I_n+I_{n-2}= \frac{1}{n-1}\).


Using Integration by Parts:

If the recurrence formula you are required to prove has a function of n in front of the \(I_{n-k}\) term, you are most likely required to use integration by parts.

The difficult part of these types of questions is determining what to let your \(u\) and \(dv\) equal such that you can get a lower power of the integral or produce another \(I_n\) term.

Harder problems may require you to use integration by parts multiple times (as per our example in the introduction section).

Harder problems may also require you to algebraically manipulate the integral you obtain after integration by parts to produce \(I_n, \ I_{n-1}, \ I_{n-2} \ …\) etc. terms.


Example 1:

Let \(I_n= \int_0^1 x(1-x^3)^n \ dx\). Show that \(I_n = \frac{3n}{3n+2} I_{n-1}\)



First, we notice that there is a function of \(n\) in front of the \(I_{n-1}\) term, so it is likely we will need to use integration by parts.

We want to choose a \(u\) and \(dv\) such that we can obtain a lower power of the integral.

Here, notice if we let \(u=(1-x^3)^n, \ \frac{du}{dx}\), would equal \(-3nx^2(1-x^3)^{n-1}\).

This looks promising!


Here, we have a lower power for the \((1-x^3)\) term.

This is what we are trying to obtain with the \(I_{n-1}\) term. So, using \(u=(1-x^3)^n\) and \(dv=x\) and applying integration by parts yields:

I_n &= \int_0^1 x(1-x^3)^n \ dx\\
&= \left[ \frac{x^2}{2} (1-x^3)^n \right]_0^1 + \frac{3n}{2} \int_0^1 x^4(1-x^3)^{n-1} \ dx\\
&= \frac{3n}{2} \int_0^1 x^4(1-x^3)^{n-1} \ dx\\


Here, we want to make the integral we’ve just obtained look closer to the form of \(I_n\), so we isolate an \(x\) here:

\(I_n = \frac{3n}{2} \int_0^1 x(x^3)(1-x^3)^{n-1} \ dx \)


Manipulating this algebraically:

I_n &= \frac{3n}{2} \int_0^1 x[1-(1-x^3)](1-x^3)^{n-1} \ dx\\
&= \frac{3n}{2} \int_0^1 [x(1-x^3)^{n-1} -x(1-x^3)^n] \ dx\\


By manipulating the integral in this way, we have just produced lower powers of the integral and more \(I_n\) terms!

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



Concept Check Questions:

1. Let \(I_n= \int_0^{\frac{π}{4}} tan^n(x) \ dx\). Prove that \(I_n = \frac{1}{n-1} – I_{n-2}\), and hence, evaluate \( \int_0^{\frac{π}{4}} tan^3(x) \ dx\)


2. Let \(I_n= \int_0^1 x^n e^{kx} \ dx\) where \(k≠0\). Prove that \(I_n = \frac{e^k}{k} – \frac{n}{k} I_{n-1}\), and hence, evaluate \( \int_0^1 x^2 e^{3x} \ dx\)

Concept Check Solutions:


\text{Consider } I_n + I_{n-2} &= \int_0^{\frac{π}{4}} [tan^n(x) + tan^{n-2}(x)] \ dx\\
&=\int_0^{\frac{π}{4}} (tan^2(x)+1)tan^{n-2}(x) \ dx\\
&= \int_0^{\frac{π}{4}} sec^2(x)tan^{n-2}(x) \ dx\\
&= \left[ \frac{tan^{n-1}(x)}{n-1} \right]_0^{\frac{π}{4}}\\
&= \frac{1}{n-1}\\
∴ I_n &= \frac{1}{n-1} – I_{n-2}\\
\int_0^{\frac{π}{4}} tan^3(x) \ dx &= I_3
&= \frac{1}{2} – I_1\\
&= \frac{1}{2} – \int_0^{\frac{π}{4}} tan(x) \ dx\\
&= \frac{1}{2} – \int_0^{\frac{π}{4}} \frac{sin(x)}{cos(x)} \ dx\\
&= \frac{1}{2} + [ln|cos(x)|]_0^{\frac{π}{4}}\\
&= \frac{1}{2} + ln \left( \frac{1}{\sqrt{2}} \right)\\


2. Applying integration by parts with \(u=x^n\) and \(dv=e^{kx}\) yields:

\int_0^1 x^ne^{kx} \ dx &= \left[ \frac{x^ne^{kx}}{k} \right]_0^1 – \int_0^1 \frac{e^{kx}}{k} nx^{n-1} \ dx\\
&= \frac{e^k}{k} – \frac{n}{k} \int_0^1 e^{kx}x^{n-1} \ dx\\
&= \frac{e^k}{k} – \frac{n}{k} I_{n-1}\\

\( \int_0^1 x^2 e^{3x} \ dx\) is \(I_2\) with \(k=3\):

I_2 &= \frac{e^3}{3} – \frac{2}{3}I_1\\
&= \frac{e^3}{3} – \frac{2}{3} \left( \frac{e^3}{3} – \frac{1}{3}I_0 \right)\\
&= \frac{e^3}{9}+\frac{2}{9} \int_0^1 e^{3x} \ dx\\
&= \frac{e^3}{9} + \frac{2}{9} \left[ \frac{e^{3x}}{3} \right]_0^1\\
&= \frac{5e^3}{27} – \frac{2}{27}\\



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