Techniques of Integration (DP IB Maths: AI HL)

Revision Note

Paul

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Paul

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Integrating Composite Functions (ax+b)

What is a composite function?

  • A composite function involves one function being applied after another
  • A composite function may be described as a “function of a function”
  • This Revision Note focuses on one of the functions being linear – i.e. of the formbold space bold italic a bold italic x bold plus bold italic b

How do I integrate linear (ax+b) functions?

  • A linear function (ofspace x) is of the formspace a x plus b
  • The special cases for trigonometric functions and exponential and logarithm functions are
    •  space integral sin left parenthesis a x plus b right parenthesis space straight d x equals negative 1 over a cos left parenthesis a x plus b right parenthesis plus c
    •  space integral cos left parenthesis a x plus b right parenthesis space straight d x equals 1 over a sin left parenthesis a x plus b right parenthesis plus c
    • space integral straight e to the power of a x plus b end exponent space straight d x equals 1 over a straight e to the power of a x plus b end exponent plus c
    • space integral fraction numerator 1 over denominator a x plus b end fraction space straight d x equals 1 over a ln open vertical bar a x plus b close vertical bar plus c
  • There is one more special case
    • space integral left parenthesis a x plus b right parenthesis to the power of n space straight d x equals fraction numerator 1 over denominator a left parenthesis n plus 1 right parenthesis end fraction left parenthesis a x plus b right parenthesis to the power of n plus 1 end exponent plus c where space n element of straight rational numbers comma space n not equal to negative 1
  • space c, in all cases, is the constant of integration
  • All the above can be deduced using reverse chain rule
    • However, spotting them can make solutions more efficient

Examiner Tip

  • Although the specific formulae in this revision note are NOT  in the formula booklet
    • almost all of the information you will need to apply reverse chain rule is provided
    • make sure you have the formula booklet open at the right page(s) and practice using it

Worked example

Find the following integrals

a)      space integral 3 left parenthesis 7 minus 2 x right parenthesis to the power of 5 over 3 end exponent space straight d x

5-4-2-ib-sl-aa-only-we1-soltn-a

b)      space integral 1 half cos left parenthesis 3 x minus 2 right parenthesis space straight d x

5-4-2-ib-sl-aa-only-we1-soltn-b

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Reverse Chain Rule

What is reverse chain rule?

  • The Chain Rule is a way of differentiating two (or more) functions
  • Reverse Chain Rule (RCR) refers to integrating by inspection
    • spotting that chain rule would be used in the reverse (differentiating) process

How do I know when to use reverse chain rule?

  • Reverse chain rule is used when we have the product of a composite function and the derivative of its secondary function
  • Integration is trickier than differentiation; many of the shortcuts do not work
    • For example, in general integral e to the power of f left parenthesis x right parenthesis end exponent space straight d x not equal to fraction numerator 1 over denominator f apostrophe left parenthesis x right parenthesis end fraction e to the power of f left parenthesis x right parenthesis end exponent
    • However, this result is true ifspace f left parenthesis x right parenthesis is linearspace left parenthesis a x plus b right parenthesis
  • Formally, in function notation, reverse chain rule is used for integrands of the form 

I equals integral g apostrophe open parentheses x close parentheses f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses space straight d x

    • this does not have to be strictly true, but ‘algebraically’ it should be
      • if coefficients do not match ‘adjust and compensate’ can be used
      • e.g. space 5 x squared is not quite the derivative ofspace 4 x cubed
        • the algebraic partspace left parenthesis x squared right parenthesis is 'correct'
        • but the coefficient 5 is ‘wrong’
        • use ‘adjust and compensate’ to ‘correct’ it
  • A particularly useful instance of reverse chain rule to recognise is

I equals integral fraction numerator f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space straight d x equals ln space vertical line f left parenthesis x right parenthesis vertical line plus c

    • i.e.  the numerator is (almost) the derivative of the denominator
    • 'adjust and compensate' may need to be used to deal with any coefficients
      • e.g.  I equals integral fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third integral 3 fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third integral fraction numerator 3 x squared plus 3 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third ln space vertical line x cubed plus 3 x vertical line plus c

How do I integrate using reverse chain rule?

  • If the product can be identified, the integration can be done “by inspection
    • there may be some “adjusting and compensating” to do
  • Notice a lot of the "adjust and compensate method” happens mentally
    • this is indicated in the steps below by quote marks 

STEP 1
Spot the ‘main’ function
e.g. space I equals integral x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space straight d x
"the main function isspace left parenthesis space... space right parenthesis to the power of 6 which would come fromspace left parenthesis space... space right parenthesis to the power of 7
 
STEP 2
Adjust’ and ‘compensate’ any coefficients required in the integral
e.g.  "space left parenthesis space... space right parenthesis to the power of 7 would differentiate to 7 left parenthesis space... space right parenthesis to the power of 6"
“chain rule says multiply by the derivative ofspace 5 x squared minus 2, which isspace 10 x
“there is no '7' or ‘10’ in the integrand so adjust and compensate”
space I equals 1 over 7 cross times 1 over 10 cross times integral 7 cross times 10 cross times x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space straight d x
 
STEP 3
Integrate and simplify
e.g. space I equals 1 over 7 cross times 1 over 10 cross times left parenthesis 5 x squared minus 2 right parenthesis to the power of 7 plus c
space I equals 1 over 70 left parenthesis 5 x squared minus 2 right parenthesis to the power of 7 plus c
 
  • Differentiation can be used as a means of checking the final answer
  • After some practice, you may find Step 2 is not needed
    • Do use it on more awkward questions (negatives and fractions!)
  • If the product cannot easily be identified, use substitution

Examiner Tip

  • Before the exam, practice this until you are confident with the pattern and do not need to worry about the formula or steps anymore
    • This will save time in the exam
  • You can always check your work by differentiating, if you have time

Worked example

A curve has the gradient functionspace f apostrophe left parenthesis x right parenthesis equals 5 x squared sin left parenthesis 2 x cubed right parenthesis.

Find an expression forspace straight f left parenthesis x right parenthesis.

iiq~htJ9_5-4-2-ib-sl-aa-only-we2-soltn

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Substitution: Reverse Chain Rule

What is integration by substitution?

  • When reverse chain rule is difficult to spot or awkward to use then integration by substitution can be used
    • substitution simplifies the integral by defining an alternative variable (usuallyspace u) in terms of the original variable (usuallyspace x)
    • everything (including “straight d x” and limits for definite integrals) is then substituted which makes the integration much easier

How do I integrate using substitution?

STEP 1
Identify the substitution to be used – it will be the secondary function in the composite function

Sospace g left parenthesis x right parenthesis inspace f left parenthesis g left parenthesis x right parenthesis right parenthesis andspace u equals g left parenthesis x right parenthesis

STEP 2
Differentiate the substitution and rearrange

fraction numerator straight d u over denominator straight d x end fractioncan be treated like a fraction
(i.e. “multiply byspace straight d x to get rid of fractions)

STEP 3
Replace all parts of the integral
Allspace x terms should be replaced with equivalentspace u terms, includingspace straight d x
If finding a definite integral change the limits fromspace x-values tospace u-values too

STEP 4
Integrate and either
substitutespace x back in
or
evaluate the definte integral using thespace u limits (either using a GDC or manually)

STEP 5
Findspace c, the constant of integration, if needed

  • For definite integrals, a GDC should be able to process the integral without the need for a substitution
    • be clear about whether working is required or not in a question

Examiner Tip

  • Use your GDC to check the value of a definite integral, even in cases where working needs to be shown

Worked example

a)
Find the integral

space integral fraction numerator 6 x plus 5 over denominator left parenthesis 3 x squared plus 5 x minus 1 right parenthesis cubed end fraction space straight d x

5-4-2-ib-sl-aa-only-we3-soltn-a

b)
Evaluate the integral

           space integral subscript 1 superscript 2 fraction numerator 6 x plus 5 over denominator left parenthesis 3 x squared plus 5 x minus 1 right parenthesis cubed end fraction space straight d x

giving your answer as an exact fraction in its simplest terms.

5-4-2-ib-sl-aa-only-we3-soltn-b

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Paul

Author: Paul

Expertise: Maths

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.