Demystifying Integration By Recognition in VCE Maths Methods

If you’ve ever stared blankly at a weird-looking integral like ∫xcos(x)dx and asked, ‘When were we ever taught this?’ – you’re in good company.

Integration by recognition is one of the most misunderstood areas of VCE Maths Methods, despite being little more than an exercise in rearranging algebra. Don’t believe it? Read on as Vanguard’s Head of Methods walks you through the foolproof strategy our Raw 50 tutors use to solve these questions. Once you know what to look for, these ‘hard’ integrals become both predictable and easily doable.

Walkthrough: integration by recognition, solved by our Head of Methods

The fundamental aim of this topic

Integration by recognition is all about this key idea:

Sometimes, an integral you can’t solve directly happens to be embedded in the derivative of another known function – i.e. something you can integrate.  

That’s why many of these questions come in two parts:

1. Part (a): Differentiate something like xsin⁡(x)
2. Part (b): Use that result to integrate something like xcos⁡(x)
   

It’s basically reverse-engineering your differentiation – using what you know from Part (a) to help you ‘recognise’ the derivative/integral relationship hidden in Part (b). Put simply, recognition.

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Step 1: Carefully differentiate the initial function 

Watch the above video to follow along with our tutor’s solving process. We begin with this classic Methods example:

(a) Find the derivative of xsin⁡(x)

This is a standard product rule job:

d/dx[xsin⁡(x)] = sin⁡(x) + xcos⁡(x)

But remember: the question doesn’t end after this one differentiation – you’re doing it because the resulting expression – sin⁡(x) + xcos⁡(x) – is going to help you solve the next part of the question.

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Step 2: Set up the opposing integral equation

Now the question asks:

(b) Hence, find ∫xcos⁡(x) dx

This is where you may freeze – wondering if you’d ever been taught the rule for integrating this expression. Hint: you weren’t, but all the necessary info is right in front of you!

From (a) we just learned that:

1. d/dx[xsin⁡(x)] = sin⁡(x) + xcos⁡(x)

Conversely, that means:

2. ∫(sin⁡(x) + xcos⁡(x)) dx = xsin⁡(x) + c

See what we did there? You already know that differentiating and anti-differentiating are opposite processes with an inverse relationship. Therefore, if we integrate both sides of equation (1), we end up with equation (2): a relationship we can now use to answer part (b).

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Step 3: Isolate the part of the integral you actually want

Here’s the catch – we were only asked to find:

∫xcos⁡(x) dx

not 

∫(sin⁡(x) + xcos⁡(x)) dx

So we need to break down the larger expression using basic integral algebra:

∫(sin⁡(x)+xcos⁡(x)) dx = ∫sin⁡(x) dx + ∫xcos⁡(x) dx

Rearrange this equation to isolate the integral we want:

∫xcos⁡(x) dx = ∫(sin⁡(x)+xcos⁡(x)) dx − ∫sin⁡(x) dx

Now our right-hand side is made of integrals we can solve for! : ) Sub in what we know:

• From part (a): ∫(sin⁡(x) + xcos⁡(x)) dx = xsin⁡(x) + c
• From general knowledge: ∫sin⁡(x) dx = −cos⁡(x) + c

Final answer:

∫xcos⁡(x) dx 

= xsin⁡(x) − (−cos⁡(x) + c)

= xsin⁡(x) + cos⁡(x) + c 

And there you have it! An unfamiliar integral solved using nothing but product rule, basic integration and rearranging algebra.

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Recap: Integration by recognition in 3 simple steps

Here’s the process you can use every time you see one of these questions:

1. Differentiate carefully in Part (a)

Don’t rush it – make sure you apply product rule or chain rule properly, because the answer you get is the building block for Part (b).

2. Set up the integral relationship from Part (a)

Using the fact that differentiating and anti-differentating are opposites, write:

∫(your derivative from Part a) dx = your original function from Part a + c

3. Break up the integral, solve the parts you know and isolate the target

Use the fact that:

∫(a+b) dx = ∫a dx + ∫b dx

Then rearrange to isolate the integral you were asked to find.

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Recognising when to use this process

How do you know when a question wants integration by recognition?

Look for these signs:

• It comes in two parts, with the first one asking for a derivative
  
• The second part asks for an integral you can’t do directly
  
• You see products of functions (like xcos⁡(x), xln⁡(x) etc.)
  

If you see these flags, think: ‘Can I link this back to what I just differentiated?’ 

Spoiler: usually, yes.

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Don’t be intimidated by this ‘weird’ area of VCE Methods

Integration by recognition is examinable. It appears every year in Methods papers. And if you don’t know how to use it, you’ll waste time – or worse, skip mutli-part questions entirely.

The key is:

• Understand the connection between derivatives and anti-derivatives
  
• Use that to reverse-engineer relationships via integration
  
• Practice the rearrangement of these integral equations
  

Once you’ve done a few of these, the pattern becomes clear. They’re not hard – they’re just unfamiliar.

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