Integration by Parts (DP IB Analysis & Approaches (AA)): Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

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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’

    • Though note that 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, parts of the integrand are assigned to be u and fraction numerator d v over denominator d x end fraction (rather than u and v)

    • Generally, the function that becomes simpler when differentiated should be assigned to u

  • There are various stages of integrating using this method

    • However 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 to save rewriting it later (space I is often used)

    • e.g. space I equals integral x sin 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. space 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. table row cell space space I end cell equals cell negative x cos space x minus integral open parentheses negative cos space x close parentheses space d x equals negative x cos space x plus integral cos space x space d x end cell end table

  • STEP 4
    Work out the second integral, integral v fraction numerator d u over denominator d x end fraction space d x
    Now include a “+c” (unless definite integration) 

    • e.g. space 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

Examiner Tips and Tricks

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.

These functions are (relatively) easy to differentiate (to find u apostrophe), but are awkward to integrate.

Worked Example

a)  Find integral 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 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 differentiation to become simple enough to make the second integral straightforward

    • This often involves integrands of 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 to save rewriting it later (space I is often used)
     

  • 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 is parts of the original integral that have already been integrated

    • It is then straightforward to rearrange and solve the problem

      • e.g.  table row cell space space 2 I end cell equals cell g left parenthesis x right parenthesis plus c space space rightwards double arrow space space I equals 1 half g left parenthesis x right parenthesis plus c end cell end table

    • Don't forget to include a constant of integration ('+c') at the end

Examiner Tips and Tricks

'+c' just represents 'any constant', so it doesn't matter that we've divided by 2 between the final two steps above.

  • I.e. you don't need to worry about changing '+c' to '+1 halfc' in the final step there

  • But you do need the 1 half in front of g open parentheses x close parentheses for the answer to be correct!

Worked Example

a)       Find integral x squared cos space x space d x.

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

b)       Find integral 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

Expertise: Maths Content Creator

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.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.