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

Revision Note

Author

Roger B

Expertise

Maths

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 $\int f^{'} (g (x)) \cdot g^{'} (x) d x$
- By the chain rule, $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$
- And differentiation and integration are inverse operations, therefore
  - $\int f^{'} (g (x)) \cdot g^{'} (x) d x = f (g (x)) + C$
If coefficients do not match exactly, ‘adjust and compensate’ can be used
- E.g. $5 x^{2}$ is not quite the derivative of $g (x) = 4 x^{3}$
  - the algebraic part $(x^{2})$ is 'correct'
  - but the coefficient 5 is ‘wrong’
    - $g^{'} (x) = 12 x^{2}$
  - use ‘adjust and compensate’ to ‘correct’ it
    - $5 x^{2} = \frac{5}{12} (12 x^{2}) = \frac{5}{12} g^{'} (x)$

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
- $\int f^{'} (x) \cdot {[f (x)]}^{n} d x = \frac{1}{n + 1} {[f (x)]}^{n + 1} + C$

What are the steps for integrating composite functions?

STEP 1
Spot the ‘main’ function
- E.g. $\int {x (5 x^{2} - 2)}^{6} d x$
- 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 $7 \cdot {(. . .)}^{6}$ "
  - “Chain rule says multiply by the derivative of $5 x^{2} - 2$ , which is $10 x$ ”
  - “There is no '7' or ‘10’ in the integrand so adjust and compensate”
- $\int {x (5 x^{2} - 2)}^{6} d x = \frac{1}{7} \times \frac{1}{10} \times \int 7 \times 10 \times x {(5 x^{2} - 2)}^{6} d x$
STEP 3
Integrate and simplify
- E.g. $\frac{1}{70} \int 70 x {(5 x^{2} - 2)}^{6} d x$
  - Now $70 x (5 x^{2} - 6)$ is the exact derivative of ${(5 x^{2} - 2)}^{7}$
- So $\int {x (5 x^{2} - 2)}^{6} d x = \frac{1}{70} \int 70 x {(5 x^{2} - 2)}^{6} d x = \frac{1}{70} {(5 x^{2} - 2)}^{7} + C$
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!)

Exam Tip

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, $f^{'}$ , is given by $f^{'} (x) = 5 x^{2} \sin (2 x^{3})$ .

Given that the graph of $f$ passes through the point $(0, 1)$ , find an expression for $f (x)$ .

Answer:

Use $f (x) = \int f^{'} (x) d x$

Start by spotting the 'main' function, $\sin (\dots)$

When differentiating, $\sin$ 'comes from' $\cos$ , so use the chain rule to find the derivative of $\cos (2 x^{3})$

$\frac{d}{d x} (\cos (2 x^{3})) = - \sin (2 x^{3}) \cdot 6 x^{2}$

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

$\begin{array}{rcl} f (x) & = & \int 5 x^{2} \sin (2 x^{3}) d x \\ = & - \frac{5}{6} \int (- \sin (2 x^{3}) \cdot 6 x^{2}) d x \\ = & - \frac{5}{6} \cos (2 x^{3}) + C \end{array}$

The graph of $f$ goes through $(0, 1)$ , therefore $f (0) = 1$

$\begin{array}{rcl} - \frac{5}{6} \cos (0) + C & = & 1 \\ - \frac{5}{6} + C & = & 1 \\ C & = & \frac{11}{6} \end{array}$

$f (x) = - \frac{5}{6} \cos (2 x^{3}) + \frac{11}{6}$

Integrating f'(x)/f(x)

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

A particularly useful special case of integrating composite functions is
- $\int \frac{f^{'} (x)}{f (x)} d x = \ln | f (x) | + C$
  - 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. $\int \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int 3 \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int \frac{3 x^{2} + 3}{x^{3} + 3 x} d x = \frac{1}{3} \ln | x^{3} + 3 x | + C$

Exam Tip

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

Worked Example

Find the indefinite integral $\int \frac{2}{5 x \ln x} d x$ .

Answer:

It may not be obvious at first, but this is an example of $\int \frac{f^{'} (x)}{f (x)} d x$

Recall that the derivative of $\ln x$ is $\frac{1}{x}$ , and note that $\frac{2}{5 x \ln x} = \frac{2}{5} \cdot \frac{(\frac{1}{x})}{\ln x}$

All that's left is to 'adjust and compensate', and then use $\int \frac{f^{'} (x)}{f (x)} d x = \ln | f (x) | + C$

$\begin{array}{rcl} \int \frac{2}{5 x \ln x} d x & = & \frac{2}{5} \int \frac{(\frac{1}{x})}{\ln x} d x \\ = & \frac{2}{5} \ln |\ln x| + C \end{array}$

$\begin{array}{rcl} \int \frac{2}{5 x \ln x} d x & = & \frac{2}{5} \ln |\ln x| + C \end{array}$

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

Revision Note

Integrating composite functions

What is meant by integrating composite functions?

Special case: a function raised to a power

What are the steps for integrating composite functions?

Exam Tip

Worked Example

Integrating f'(x)/f(x)

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

Exam Tip

Worked Example

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Author: Roger B