Integration by Substitution (Edexcel A Level Further Maths: Core Pure)

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Integrating using Trigonometric Substitutions

The integrals covered in this revision note are based on the standard results

space integral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x equals arcsin stretchy left parenthesis x over a stretchy right parenthesis plus straight c

and

space integral fraction numerator 1 over denominator a squared plus x squared end fraction space straight d x equals 1 over a arc tan stretchy left parenthesis x over a stretchy right parenthesis plus straight c comma space space straight a greater than 0 comma space space vertical line x vertical line greater than straight a

These are given in the formulae booklet

How do I know when to use a trigonometric substitution in integration?

There are three main types of problem

  •  Type 1
    Showing the standard results using a substitution (a may have a value)
    The substitution will not be given in such cases
    e.g.  Use a suitable substitution to show that space integral fraction numerator 1 over denominator square root of a squared minus x squared space end root end fraction space space straight d x equals arcsin left parenthesis x over a right parenthesis plus c
    Let x equals a sin space u, so fraction numerator straight d x over denominator straight d u end fraction equals a cos space u and u equals arcsi n space left parenthesis x over a right parenthesis
    integral fraction numerator 1 over denominator square root of a squared minus x squared space end root end fraction space space straight d x equals integral fraction numerator 1 over denominator square root of a squared minus left parenthesis a sin space u right parenthesis squared end root end fraction space space left parenthesis a cos space u right parenthesis space straight d u equals integral fraction numerator a cos space u over denominator square root of a squared left parenthesis 1 minus sin squared u right parenthesis end root end fraction space space straight d u equals integral 1 space space straight d u equals u plus c
therefore integral fraction numerator 1 over denominator square root of a squared minus x squared space end root end fraction space space straight d x equals arcsin stretchy left parenthesis x over a stretchy right parenthesis plus c
    The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity , so it can be integrated using standard results or techniques

  • Type 2
    Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
    e.g.  Find integral fraction numerator 3 over denominator 9 squared plus left parenthesis 2 x right parenthesis squared end fraction space space straight d x
    integral fraction numerator 3 over denominator 9 squared plus left parenthesis 2 x right parenthesis squared end fraction space space straight d x equals 3 cross times 1 half integral fraction numerator 2 over denominator 9 squared plus left parenthesis 2 x right parenthesis squared end fraction space space straight d x equals 3 over 2 stretchy left parenthesis 1 over 9 arctan space left parenthesis fraction numerator 2 x over denominator 9 end fraction right parenthesis stretchy right parenthesis plus c equals 1 over 6 arctan left parenthesis fraction numerator 2 x over denominator 9 end fraction right parenthesis plus c

  • Type 3
    The denominator contains a three-term quadratic expression – i.e. there is an x term
    In such cases complete the square and use reverse chain rule
    e.g.  Find integral fraction numerator 1 over denominator 25 plus x squared minus 6 x end fraction space space straight d x
    integral fraction numerator 1 over denominator 25 plus x squared minus 6 x end fraction space space straight d x equals integral fraction numerator 1 over denominator 25 plus left parenthesis x minus 3 right parenthesis squared minus 9 end fraction space space straight d x equals integral fraction numerator 1 over denominator 4 squared plus left parenthesis x minus 3 right parenthesis squared end fraction space space straight d x equals 1 fourth arctan stretchy left parenthesis fraction numerator x minus 3 over denominator 4 end fraction stretchy right parenthesis plus c
    (This works since fraction numerator straight d space over denominator straight d x end fraction left square bracket x minus 3 right square bracket equals 1, so effectively there is no reverse chain rule involved) 
  • A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course

How do I use a trigonometric substitution to find integrals?

  • STEP 1
    Identify the type of problem and if a substitution is required
    Determine the substitution if needed

  • STEP 2
    For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
    For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
    For Type 3 problems complete the square

  • STEP 3
    Integrate using standard techniques and results, possibly from the formulae booklet
    For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation

  • STEP 4
    Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
    For indefinite integration, simplify where obvious and/or rearrange into a required format

Why is arccos x not involved in any of the integration results?

  • fraction numerator straight d space over denominator straight d x end fraction stretchy left square bracket arccos space stretchy left parenthesis x over a stretchy right parenthesis stretchy right square bracket equals negative fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction
    For integration the "-" at the start can be treated as the constant "-1" and so integrating would lead to "-arcsin ..."
    • i.e. space integral negative fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x equals negative arcsin space stretchy left parenthesis x over a stretchy right parenthesis plus c

Examiner Tip

  • The general form of the functions involving trigonometric and hyperbolic functions are very similar
  • Be clear about which form needs a trigonometric substitution and which form need a hyperbolic substitution
  • Always have a copy of the formula booklet to hand when practising these problems

Worked example

(a) Use an appropriate substitution to show that

integral fraction numerator 1 over denominator a squared plus x squared end fraction space space straight d x equals 1 over a arctan space stretchy left parenthesis x over a stretchy right parenthesis plus c

5-2-5-edex-fm--alevel-we1-trigsub-soltn-a

(b) Find

integral fraction numerator 3 over denominator 16 plus 9 x squared end fraction space space straight d x

5-2-5-edex-fm--alevel-we1-trigsub-soltn-b

Integrating using Hyperbolic Substitutions

The integrals covered in this revision note are based on the standard results

integral fraction numerator 1 over denominator square root of a squared plus x squared end root end fraction space space straight d x equals arsinh space stretchy left parenthesis x over a stretchy right parenthesis plus c

and

integral fraction numerator 1 over denominator square root of x squared minus a squared end root end fraction space space straight d x equals arcosh space stretchy left parenthesis x over a stretchy right parenthesis plus c comma space x greater than space a 

These are given in the formulae booklet

How do I know when to use a hyperbolic substitution in integration?

There are three main types of problem

  • Type 1
    Showing the standard results using a substitution (a may have a value)
    The substitution will not be given in such cases
    e.g.  Use a suitable substitution to show that integral fraction numerator 1 over denominator square root of x squared minus a squared end root end fraction space space straight d x equals ar cos text h end text space stretchy left parenthesis x over a stretchy right parenthesis plus c
    Let x equals a cosh space u, so fraction numerator straight d x over denominator straight d u end fraction equals a sinh space u and u equals arcosh space stretchy left parenthesis x over a stretchy right parenthesis
    integral fraction numerator 1 over denominator square root of x squared minus a squared end root end fraction space space straight d x equals integral fraction numerator 1 over denominator square root of left parenthesis a cosh space u right parenthesis squared minus a squared end root end fraction space left parenthesis a sinh space u right parenthesis space straight d u equals integral fraction numerator a sinh space u over denominator square root of a squared left parenthesis cosh squared space u minus 1 right parenthesis end root end fraction space straight d u equals integral 1 space straight d u equals u plus c
    therefore integral fraction numerator 1 over denominator square root of x squared minus a squared end root end fraction space space straight d x equals arcosh space stretchy left parenthesis x over a stretchy right parenthesis plus c
    The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity cosh squared space x minus sinh squared space x identical to 1, so it can be integrated using standard results or techniques

  • Type 2
    Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
    e.g.  Find integral fraction numerator 3 over denominator square root of 4 x squared plus 9 end root end fraction space space straight d x
    integral fraction numerator 3 over denominator square root of left parenthesis 2 x right parenthesis squared plus 3 squared end root end fraction space space straight d x equals 3 cross times 1 half integral fraction numerator 2 over denominator square root of left parenthesis 2 x right parenthesis squared plus 3 squared end root end fraction space space straight d x equals 3 over 2 stretchy left parenthesis arsinh stretchy left parenthesis fraction numerator 2 x over denominator 3 end fraction stretchy right parenthesis stretchy right parenthesis plus c equals 3 over 2 arsinh stretchy left parenthesis fraction numerator 2 x over denominator 3 end fraction stretchy right parenthesis plus c

  • Type 3
    The denominator contains a three-term quadratic expression – i.e. there is an x term
    In such cases complete the square and use reverse chain rule
    e.g.  Find integral fraction numerator 1 over denominator square root of x squared minus 6 x plus 25 end root end fraction space space straight d x
    integral fraction numerator 1 over denominator square root of x squared minus 6 x plus 25 end root end fraction space space straight d x equals integral fraction numerator 1 over denominator square root of left parenthesis x minus 3 right parenthesis squared minus 9 plus 25 end root end fraction space space straight d x equals integral fraction numerator 1 over denominator square root of left parenthesis x minus 3 right parenthesis squared plus 4 to the power of 2 end exponent end root end fraction space space straight d x equals arsinh space stretchy left parenthesis fraction numerator x minus 3 over denominator 4 end fraction stretchy right parenthesis plus c
    (This works since fraction numerator straight d space over denominator straight d x end fraction left square bracket x minus 3 right square bracket equals 1,so effectively there is no reverse chain rule involved

  • A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course, although hyperbolic functions are not

How do I use a hyperbolic substitution to find integrals?

  • STEP 1
    Identify the type of problem and if a substitution is required
    Determine the substitution if needed

  • STEP 2
    For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
    For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
    For Type 3 problems complete the square

  • STEP 3
    Integrate using standard techniques and results, possibly from the formulae booklet
    For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation

  • STEP 4
    Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
    For indefinite integration, simplify where obvious and/or rearrange into a required format

Is artanh x involved in integration?

  • The standard result, given in the formulae booklet is
    integral fraction numerator 1 over denominator a squared minus x squared end fraction space space straight d x equals 1 over a artanh space stretchy left parenthesis x over a stretchy right parenthesis plus c comma space vertical line x vertical line less than a
    with the alternative result fraction numerator 1 over denominator 2 a end fraction ln space stretchy vertical line fraction numerator a plus x over denominator a minus x end fraction stretchy vertical line plus c
     also given
  • Problems involving these often involve partial fractions (since a squared minus x squared is the difference of two squares) leading to the 'ln' result
  • If you happen to recognise the integral and can use the formulae booklet result involving "artanh" to solve a problem, then do so!

Examiner Tip

  • The general form of the functions involving trigonometric and hyperbolic functions are very similar
  • Be clear about which form needs a trigonometric substitution and which form need a hyperbolic substitution
  • Always have a copy of the formula booklet to hand when practising these problems

Worked example

(a) Use an appropriate substitution to show that

integral fraction numerator 1 over denominator square root of x squared plus a squared end root end fraction space space straight d x equals ar sinh space open parentheses x over a close parentheses plus c

5-2-5-edex-fm--alevel-we2-hypsub-soltn-a-correct

(b) Find

integral fraction numerator 5 over denominator square root of 9 x squared minus 25 end root end fraction space space straight d x

5-2-5-edex-fm--alevel-we2-hypsub-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.