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

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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 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

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