Integrals of Composite Functions (College Board AP® Calculus AB)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Integrating composite functions

What is meant by integrating composite functions?

  • Integrating composite functions refers to integrating 'by inspection'

    • by spotting that the chain rule would be used in the inverse (differentiation) process

    • This is sometimes referred to as 'reverse chain rule'

  • This method can be used to integrate the product of

    • a composite function

    • and the derivative of its secondary ('inside') function

  • In function notation, this method is to integrate integrals of the form integral f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses times g to the power of apostrophe open parentheses x close parentheses space straight d x

    • By the chain rule, fraction numerator d over denominator d x end fraction open square brackets f open parentheses g open parentheses x close parentheses close parentheses close square brackets equals f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses times g to the power of apostrophe open parentheses x close parentheses

    • And differentiation and integration are inverse operations, therefore

      • integral f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses times g to the power of apostrophe open parentheses x close parentheses space straight d x equals f open parentheses g open parentheses x close parentheses close parentheses plus C

  • If coefficients do not match exactly, ‘adjust and compensate’ can be used

    • E.g. space 5 x squared is not quite the derivative of g open parentheses x close parentheses equals 4 x cubed

      • the algebraic partspace left parenthesis x squared right parenthesis is 'correct'

      • but the coefficient 5 is ‘wrong’

        • g to the power of apostrophe open parentheses x close parentheses equals 12 x squared

      • use ‘adjust and compensate to ‘correct it

        • 5 x squared equals 5 over 12 open parentheses 12 x squared close parentheses equals 5 over 12 g to the power of apostrophe open parentheses x close parentheses

Special case: a function raised to a power

  • One common example is an integral involving a function raised to a power

  • In this case the general pattern becomes

    • integral f to the power of apostrophe open parentheses x close parentheses times open square brackets f open parentheses x close parentheses close square brackets to the power of n space straight d x equals fraction numerator 1 over denominator n plus 1 end fraction open square brackets f open parentheses x close parentheses close square brackets to the power of n plus 1 end exponent plus C

What are the steps for integrating composite functions?

  • STEP 1
    Spot the ‘main’ function

    • E.g. integral x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space d x

    • Think: "the main function is left parenthesis space... space right parenthesis to the power of 6 which would come from 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 times 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”

    • integral x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space d x 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 d x

  • STEP 3
    Integrate and simplify

    • E.g. 1 over 70 integral 70 x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space d x

      • Now 70 x open parentheses 5 x squared minus 6 close parentheses is the exact derivative of open parentheses 5 x squared minus 2 close parentheses to the power of 7

    • So space integral x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space d x equals 1 over 70 integral 70 x left parenthesis 5 x squared minus 2 right parenthesis to the power of 6 space d x equals 1 over 70 left parenthesis 5 x squared minus 2 right parenthesis to the power of 7 plus C

  • After some practice, you may find Step 2 is not needed (because you can do it in your head)

    • Do use it on more awkward questions (negatives and fractions!)

Examiner Tips and Tricks

Integrals of this form can also be integrated by substitution

  • See the 'Integration Using Substitution' study guide

You can always check your work by differentiating, if you have time

  • Differentiating your answer should turn it back into the function you were trying to integrate

Worked Example

Let f be a function whose derivative, f to the power of apostrophe, is given byspace f to the power of apostrophe left parenthesis x right parenthesis equals 5 x squared sin left parenthesis 2 x cubed right parenthesis.

Given that the graph of f passes through the point left parenthesis 0 comma space 1 right parenthesis, find an expression forf left parenthesis x right parenthesis.

Answer:

Use f open parentheses x close parentheses equals integral f to the power of apostrophe open parentheses x close parentheses space d x

Start by spotting the 'main' function, sin open parentheses horizontal ellipsis close parentheses

When differentiating, sin 'comes from' cos, so use the chain rule to find the derivative of cos open parentheses 2 x cubed close parentheses

fraction numerator d over denominator d x end fraction open parentheses cos open parentheses 2 x cubed close parentheses close parentheses equals negative sin open parentheses 2 x cubed close parentheses times 6 x squared

'Adjust and compensate' to get the inside of the integral to be equal to that

table row cell f open parentheses x close parentheses end cell equals cell integral 5 x squared sin left parenthesis 2 x cubed right parenthesis space d x end cell row blank equals cell negative 5 over 6 integral open parentheses negative sin open parentheses 2 x cubed close parentheses times 6 x squared close parentheses space d x end cell row blank equals cell negative 5 over 6 cos open parentheses 2 x cubed close parentheses plus C end cell end table

The graph of f goes through open parentheses 0 comma space 1 close parentheses, therefore f open parentheses 0 close parentheses equals 1

table row cell negative 5 over 6 cos open parentheses 0 close parentheses plus C end cell equals 1 row cell negative 5 over 6 plus C end cell equals 1 row C equals cell 11 over 6 end cell end table

f open parentheses x close parentheses equals negative 5 over 6 cos open parentheses 2 x cubed close parentheses plus 11 over 6

Integrating f'(x)/f(x)

How do I integrate f'(x)/f(x) ?

  • A particularly useful special case of integrating composite functions is

    • integral fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space d x equals ln vertical line f left parenthesis x right parenthesis vertical line plus C

      • I.e.  the numerator of a fraction being integrated is the derivative of the denominator

    • Make sure you recognise this pattern!

      • It speeds up and simplifies integrals of this sort

  • 'Adjust and compensate' may need to be used to deal with any coefficients

    • e.g.  integral fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space d x equals 1 third integral 3 fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space d x equals 1 third integral fraction numerator 3 x squared plus 3 over denominator x cubed plus 3 x end fraction space d x equals 1 third ln space vertical line x cubed plus 3 x vertical line plus C

Examiner Tips and Tricks

Don't forget the modulus sign in the answer when finding integrals of this form.

Worked Example

Find the indefinite integral integral fraction numerator 2 over denominator 5 x ln x end fraction space d x.

Answer:

It may not be obvious at first, but this is an example of integral fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space d x

Recall that the derivative of ln x is 1 over x, and note that fraction numerator 2 over denominator 5 x ln x end fraction equals 2 over 5 times fraction numerator open parentheses 1 over x close parentheses over denominator ln x end fraction

All that's left is to 'adjust and compensate', and then use integral fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space d x equals ln vertical line f left parenthesis x right parenthesis vertical line plus C

table row cell integral fraction numerator 2 over denominator 5 x ln x end fraction space d x end cell equals cell 2 over 5 integral fraction numerator open parentheses 1 over x close parentheses over denominator ln x end fraction space d x end cell row blank equals cell 2 over 5 ln open vertical bar ln x close vertical bar plus C end cell end table

table row cell integral fraction numerator 2 over denominator 5 x ln x end fraction space d x end cell equals cell 2 over 5 ln open vertical bar ln x close vertical bar plus C end cell end table

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths 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.