Integrals of Composite Functions (College Board AP® Calculus BC): Revision Note

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 $\int f^{\text{'}}\left(g\left(x\right)\right)·g^{\text{'}}\left(x\right)\mathrm{d}x$

  • By the chain rule, $\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f^{\text{'}}\left(g\left(x\right)\right)·g^{\text{'}}\left(x\right)$

  • And differentiation and integration are inverse operations, therefore

    • $\int f^{\text{'}}\left(g\left(x\right)\right)·g^{\text{'}}\left(x\right)\mathrm{d}x=f\left(g\left(x\right)\right)+C$

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

  • E.g. $5x^2$ is not quite the derivative of $g\left(x\right)=4x^3$

    • the algebraic part$\left(x^2\right)$ is 'correct'

    • but the coefficient 5 is ‘wrong’

      • $g^{\text{'}}\left(x\right)=12x^2$

    • use ‘adjust and compensate’ to ‘correct’ it

      • $5x^2=\frac{5}{12}\left(12x^2\right)=\frac{5}{12}{g}^{\text{'}}\left(x\right)$

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

  • $\int f^{\text{'}}\left(x\right)·{\left[f\left(x\right)\right]}^n\mathrm{d}x=\frac{1}{n+1}{\left[f\left(x\right)\right]}^{n+1}+C$

  • E.g. $\int \cos x{\sin }^{5}xdx=\frac{1}{6}{\sin }^{6}x+C$

What are the steps for integrating composite functions?

• STEP 1
  Spot the ‘main’ function

  • E.g. $\int {x(5x^2-2)}^{6}dx$

  • Think: "the main function is ${(...)}^6$ which would come from ${(...)}^7$”

• STEP 2
  ‘Adjust and compensate’ any coefficients required in the integral

  • E.g.  "${(...)}^7$ would differentiate to $7·{(...)}^6$"

    • “Chain rule says multiply by the derivative of$5x^2-2$, which is$10x$”

    • “There is no '7' or ‘10’ in the integrand so adjust and compensate”

  • $\int {x(5x^2-2)}^{6}dx=\frac{1}{7}\times \frac{1}{10}\times \int 7\times 10\times x{(5x^2-2)}^{6}dx$

• STEP 3
  Integrate and simplify

  • E.g. $\frac{1}{70}\int 70x{(5x^2-2)}^{6}dx$

    • Now $70x{\left(5x^2-2\right)}^6$ is the exact derivative of ${\left(5x^2-2\right)}^7$

  • So $\int {x(5x^2-2)}^{6}dx=\frac{1}{70}\int 70x{(5x^2-2)}^{6}dx=\frac{1}{70}{(5x^2-2)}^7+C$

• 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!)

Integrating f'(x)/f(x)

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

• A particularly useful special case of integrating composite functions is

  • $\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}dx=\ln \left|f\right(x\left)\right|+C$

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

  • Make sure you recognize 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.  $\int \frac{x^2+1}{x^3+3x}dx=\frac{1}{3}\int 3\frac{x^2+1}{x^3+3x}dx=\frac{1}{3}\int \frac{3x^2+3}{x^3+3x}dx=\frac{1}{3}\ln |x^3+3x|+C$