Techniques of Integration (DP IB Analysis & Approaches (AA)) : Revision Note

Did this video help you?

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 Tips and Tricks

  • 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

Did this video help you?

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 Tips and Tricks

  • 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

Did this video help you?

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 Tips and Tricks

  • 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
👀 You've read 1 of your 5 free revision notes this week
An illustration of students holding their exam resultsUnlock more revision notes. It's free!

By signing up you agree to our Terms and Privacy Policy.

Already have an account? Log in

Did this page help you?

Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

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.

Download notes on Techniques of Integration