DP IB Maths: AA HL

Revision Notes

5.9.2 Further Techniques of Integration

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Integration by Substitution

What is integration by substitution?

  • Integration by substitution is used when an integrand where reverse chain rule is either not obvious or is not spotted
    • in the latter case it is like a “back-up” method for reverse chain rule

How do I use integration by substitution?

  • For instances where the substitution is not obvious it will be given in a question
    • e.g.  Find integral subscript blank superscript blank cot space x space d x using the substitution u equals sin space x
  • Substitutions are usually of the form u equals g left parenthesis x right parenthesis
    • in some cases u squared equals g left parenthesis x right parenthesis and other variations are more convenient
      • as these would not be obvious, they would be given in a question
    • if need be, this can be rearranged to find x in terms of u
  • Integration by substitution then involves rewriting the integral, including “straight d x” in terms of u

STEP 1
Name the integral to save rewriting it later
Identify the given substitution u equals g left parenthesis x right parenthesis

STEP 2
Find fraction numerator straight d u over denominator straight d x end fraction and rearrange into the form f left parenthesis u right parenthesis space straight d u equals g left parenthesis x right parenthesis space straight d x such that (some of) the integral can be rewritten in terms of u

STEP 3
If limits are involved, use u equals g left parenthesis x right parenthesis to change them from x values to u values 

STEP 4
Rewrite the integral so everything is in terms of u rather than x
This is the step when it may become apparent that x is needed in terms of u

STEP 5
Integrate with respect to u and either rewrite in terms of x or apply the limits using their u values

  • For quotients the substitution usually involves the denominator
  • It may be necessary to use ‘adjust and compensate’ to deal with any coefficients in the integrand
  • Although fraction numerator straight d u over denominator straight d x end fraction can be treated like a fraction it should be appreciated that this is a ‘shortcut’ and the maths behind it is beyond the scope of the IB course

Exam Tip

  • If a substitution is not given in a question, it is usually because it is obvious
    • If you can't see anything obvious, or you find that your choice of substitution doesn't reduce the integrand to something easy to integrate, consider that it may not be a substitution question

Worked example

Use the substitution u equals open parentheses 1 plus 2 x close parentheses to evaluate integral subscript 0 superscript 1 x open parentheses 1 plus 2 x close parentheses to the power of 7 space end exponent d x.

5-9-2-ib-hl-aa-only-we1-soltn

Integration by Parts

What is integration by parts?

  • Integration by parts is generally used to integrate the product of two functions
    • however reverse chain rule and/or substitution should be considered first
      • e.g. integral subscript blank superscript blank 2 x cos open parentheses x squared close parentheses space d x can be solved using reverse chain rule or the substitution u equals x squared
    • Integration by parts is essentially ‘reverse product rule’
      • whilst every product can be differentiated, not every product can be integrated (analytically)

What is the formula for integration by parts?

  • integral subscript blank superscript blank u fraction numerator straight d v over denominator straight d x end fraction space d x equals u v minus integral subscript blank superscript blank v fraction numerator d u over denominator d x end fraction space d x
  • This is given in the formula booklet alongside its alternative form integral subscript blank superscript blank u space d v equals u v minus integral subscript blank superscript blank v space d u

How do I use integration by parts?

  • For a given integral u and fraction numerator d v over denominator d x end fraction (rather than u and v) are assigned functions of x
  • Generally, the function that becomes simpler when differentiated should be assigned to u
  • There are various stages of integrating in this method
    • only one overall constant of integration (“+c”) is required
    • put this in at the last stage of working
    • if it is a definite integral then “+c” is not required at all 

STEP 1
Name the integral if it doesn’t have one already!
This saves having to rewrite it several times – I is often used for this purpose.
e.g. I equals integral subscript blank superscript blank x sin space x space d x

STEP 2
Assign u and fraction numerator d v over denominator d x end fraction.
Differentiate u to find fraction numerator d u over denominator d x end fraction and integrate fraction numerator d v over denominator d x end fraction to find v
e.g. table row cell u equals x space space space end cell cell space space space space space v equals negative cos space x end cell row cell fraction numerator d u over denominator d x end fraction equals 1 end cell cell space space space fraction numerator d v over denominator d x end fraction equals sin space x end cell end table


STEP 3
Apply the integration by parts formula
e.g. I equals negative x cos space x minus integral subscript blank superscript blank minus cos space x space d x

STEP 4
Work out the second integral, integral subscript blank superscript blank v fraction numerator d u over denominator d x end fraction space d x
Now include a “+c” (unless definite integration) 
e.g. I equals negative x cos space x space plus sin space x plus c  

STEP 5
Simplify the answer if possible or apply the limits for definite integration
e.g. I equals sin space x minus x cos space x plus c

  • In trickier problems other rules of differentiation and integration may be needed
    • chain, product or quotient rule
    • reverse chain rule, substitution

Can integration by parts be used when there is only a single function?

  • Some single functions (non-products) are awkward to integrate directly
    • e.g. y equals ln space x, y equals arcsin space x, y equals arccos space x, y equals arctan space x
  • These can be integrated using parts however
    • rewrite as the product ‘1 cross times f left parenthesis x right parenthesis’ and choose u equals f left parenthesis x right parenthesis and fraction numerator d v over denominator d x end fraction equals 1
    • 1 is easy to integrate and the functions above have standard derivatives listed in the formula booklet

Exam Tip

  • If  ln space x  or one of the inverse trig functions are one of the functions involved in the product then these should be assigned to "u" when applying parts
    • They are (realtively) easy to differentiate (to find u apostrophe) but are awkward to integrate

Worked example

a)       Find integral subscript blank superscript blank 5 x straight e to the power of 3 x end exponent space d x.

5-9-2-ib-hl-aa-only-we2a-soltn

b)       Show that integral subscript blank superscript blank 8 x ln space x space d x equals 2 x squared open parentheses 1 plus ln space x squared close parentheses plus c.

5-9-2-ib-hl-aa-only-we2b-soltn

Repeated Integration by Parts

When will I have to repeat integration by parts?

  • In some problems, applying integration by parts still leaves the second integral as a product of two functions of x
    • integration by parts will need to be applied again to the second integral
  • This occurs when one of the functions takes more than one derivative to become simple enough to make the second integral straightforward
    • These functions usually have the form x squared g left parenthesis x right parenthesis

How do I apply integration by parts more than once?

STEP 1
Name the integral if it doesn’t have one already!

STEP 2
Assign u and fraction numerator d v over denominator d x end fraction.  Find fraction numerator d u over denominator d x end fraction and v

STEP 3
Apply the integration by parts formula

STEP 4
Repeat STEPS 2 and 3 for the second integral

STEP 5
Work out the second integral and include a “+c” if necessary

STEP 6
Simplify the answer or apply limits

What if neither function ever becomes simpler when differentiating?

  • It is possible that integration by parts will end up in a seemingly endless loop
    • consider the product straight e to the power of x sin space x
    • the derivative of straight e to the power of x is straight e to the power of x
      • no matter how many times a function involving straight e to the power of x is differentiated, it will still involve straight e to the power of x
    • the derivative of sin space x is cos space x
      • cos space x would then have derivative negative sin space x, and so on
      • no matter how many times a function involving sin space x or cos space x is differentiated, it will still involve sin space x or cos space x
  • This loop can be trapped by spotting when the second integral becomes identical to (or a multiple of) the original integral
    • naming the original integral (I) at the start helps
    • I then appears twice in integration by parts
      • e.g. I equals g left parenthesis x right parenthesis minus I
        where g left parenthesis x right parenthesis are parts of the integral not requiring further work
    • It is then straightforward to rearrange and solve the problem
      • e.g. 2 I equals g left parenthesis x right parenthesis plus c
        I equals 1 half g left parenthesis x right parenthesis plus c

                                                                                                                                     

Worked example

a)       Find integral subscript blank superscript blank x squared cos space x space d x.

5-9-2-ib-hl-aa-only-we3a-soltn

b)       Find integral subscript blank superscript blank straight e to the power of x sin space x space d x.

5-9-2-ib-hl-aa-only-we3b-soltn

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Paul

Author: Paul

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.