Reverse Chain Rule (DP IB Applications & Interpretation (AI): HL): Revision Note

Integrating composite functions (ax+b)

What is a composite function?

• A composite function involves one function being applied after another

• A composite function may be described as a “function of a function”

  • E.g. $\sin \left(ax+b\right)$

    • First the function $ax+b$ is applied to $x$

    • Then the function is applied to $ax+b$

• This Revision Note focuses on one of the functions being linear – i.e. of the form$\mathit{a}\mathit{x}\mathbf{+}\mathit{b}$

How do I integrate linear (ax+b) functions?

• A linear function (of$x$) is of the form$ax+b$, where $a$ and $b$ are constants

• The special cases for trigonometric functions and exponential and logarithm functions are

  •  $\int \sin (ax+b)\mathrm{d}x=-\frac{1}{a}\cos (ax+b)+c$

  •  $\int \cos (ax+b)\mathrm{d}x=\frac{1}{a}\sin (ax+b)+c$

  • $\int {\mathrm{e}}^{ax+b}\mathrm{d}x=\frac{1}{a}{\mathrm{e}}^{ax+b}+c$

  • $\int \frac{1}{ax+b}\mathrm{d}x=\frac{1}{a}\ln \left|ax+b\right|+c$

• There is one more special case you should know

  • $\int {(ax+b)}^n\mathrm{d}x=\frac{1}{a(n+1)}{(ax+b)}^{n+1}+c$ where $n\in \mathrm{\mathbb{Q}},n\ne -1$

• $c$, in all cases, is the constant of integration

• All the above can be deduced using reverse chain rule

  • However, recognising the patterns and knowing these results can make solutions more efficient

Reverse chain rule

What is reverse chain rule?

• The Chain Rule is a way of differentiating composite functions made up of two (or more) functions

• Reverse Chain Rule (RCR) refers to integrating by inspection

  • You do this by spotting that chain rule would be used in the reverse (differentiating) process

How do I know when to use reverse chain rule?

• Reverse chain rule is used when we have the product of a composite function and the derivative of its secondary function

  • E.g. $2x\cos \left(x^2\right)$

    • $\cos \left(x^2\right)$ is a composite function

      • is the primary (or main or 'outside') function

      • $x^2$ is the secondary (or 'inside') function

    • $2x$ is the derivative of the secondary function $x^2$

• Formally, in function notation, reverse chain rule can be written in the following way 

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

  • E.g. $\int 2x\cos \left(x^2\right)\mathrm{d}x$

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

    • And is the derivative of , so

    • Therefore $\int 2x\cos \left(x^2\right)\mathrm{d}x=\sin \left(x^2\right)+c$

• Be sure also to recognise this useful instance of reverse chain rule

  
  $\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}\mathrm{d}x=\ln \left|f\right(x\left)\right|+c$
  

  • I.e.  the numerator is the derivative of the denominator

  • E.g. $\int \frac{3x^2+1}{x^3+x}\mathrm{d}x$

    • $f\left(x\right)=x^3+x$

    • And the derivative of that is $f^{\text{'}}\left(x\right)=3x^2+1$

    • Therefore $\int \frac{3x^2+1}{x^3+x}\mathrm{d}x=\ln \left|x^3+x\right|+c$

How do I integrate using reverse chain rule?

• If you can spot the patterns, the integration can be done “by inspection”

  • Though there may be some “adjusting and compensating” to do

• A lot of the method happens mentally

  • This is indicated in the steps below by quote marks 

• STEP 1

  Spot the ‘main’ function

  • e.g. $\int {x(5x^2-2)}^6\mathrm{d}x$

  • "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”

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

• STEP 3

  Integrate and simplify

$\begin{array}{rcl}\int {x(5x^2-2)}^6\mathrm{d}x& =& \frac{1}{7}\times \frac{1}{10}\times {(5x^2-2)}^7+c\\ & =& \frac{1}{70}{(5x^2-2)}^7+c\end{array}$ 

• After some practice, you may find Step 2 is not needed

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