5.4.2 Techniques of Integration

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”
This Revision Note focuses on one of the functions being linear – i.e. of the form $a x + b$

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

A linear function (of $x$ ) is of the form $a x + b$
The special cases for trigonometric functions and exponential and logarithm functions are
- $\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c$
- $\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c$
- $\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$
- $\int \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + c$
There is one more special case
- $\int {(a x + b)}^{n} d x = \frac{1}{a (n + 1)} {(a x + b)}^{n + 1} + c$ where $n \in ℚ, n \neq - 1$
$c$ , in all cases, is the constant of integration
All the above can be deduced using reverse chain rule
- However, spotting them can make solutions more efficient

Exam Tip

Although the specific formulae in this revision note are NOT in the formula booklet
- almost all of the information you will need to apply reverse chain rule is provided
- make sure you have the formula booklet open at the right page(s) and practice using it

Worked Example

Find the following integrals

\int 3 {(7 - 2 x)}^{\frac{5}{3}} d x

5-4-2-ib-sl-aa-only-we1-soltn-a

\int \frac{1}{2} \cos (3 x - 2) d x

5-4-2-ib-sl-aa-only-we1-soltn-b

Reverse Chain Rule

What is reverse chain rule?

The Chain Rule is a way of differentiating two (or more) functions
Reverse Chain Rule (RCR) refers to integrating by inspection
- 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 second function
Integration is trickier than differentiation; many of the shortcuts do not work
- For example, in general $\int e^{f (x)} d x \neq \frac{1}{f' (x)} e^{f (x)}$
- However, this result is true if $f (x)$ is linear $(a x + b)$
Formally, in function notation, reverse chain rule is used for integrands of the form

$I = \int g' (x) f (g (x)) d x$

- this does not have to be strictly true, but ‘algebraically’ it should be
  - if coefficients do not match ‘adjust and compensate’ can be used
  - e.g. $5 x^{2}$ is not quite the derivative of $4 x^{3}$
    - the algebraic part $(x^{2})$ is 'correct'
    - but the coefficient 5 is ‘wrong’
    - use ‘adjust and compensate’ to ‘correct’ it
A particularly useful instance of reverse chain rule to recognise is

$I = \int \frac{f' (x)}{f (x)} d x = \ln | f (x) | + c$

- i.e. the numerator is (almost) the derivative of the denominator
- 'adjust and compensate' may need to be used to deal with any coefficients
  - e.g. $I = \int \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int 3 \frac{x^{2} + 1}{x^{3} + 3 x} d x = \frac{1}{3} \int \frac{3 x^{2} + 3}{x^{3} + 3 x} d x = \frac{1}{3} \ln | x^{3} + 3 x | + c$

How do I integrate using reverse chain rule?

If the product can be identified, the integration can be done “by inspection”
- there may be some “adjusting and compensating” to do
Notice a lot of the "adjust and compensate method” happens mentally
- this is indicated in the steps below by quote marks

STEP 1

Spot the ‘main’ function

e.g.

I = \int {x (5 x^{2} - 2)}^{6} 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

5 x^{2} - 2

, which is

10 x

”

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

I = \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.

I = \frac{1}{7} \times \frac{1}{10} \times {(5 x^{2} - 2)}^{7} + c

I = \frac{1}{70} {(5 x^{2} - 2)}^{7} + c

Differentiation can be used as a means of checking the final answer
After some practice, you may find Step 2 is not needed
- Do use it on more awkward questions (negatives and fractions!)
If the product cannot easily be identified, use substitution

Exam Tip

Before the exam, practice this until you are confident with the pattern and do not need to worry about the formula or steps anymore
- This will save time in the exam
You can always check your work by differentiating, if you have time

Worked Example

A curve has the gradient function $f' (x) = 5 x^{2} \sin (2 x^{3})$ .

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

iiq~htJ9_5-4-2-ib-sl-aa-only-we2-soltn

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 (usually $u$ ) in terms of the original variable (usually $x$ )
- everything (including “ $d x$ ” 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

So $g (x)$ in $f (g (x))$ and $u = g (x)$

STEP 2
Differentiate the substitution and rearrange

$\frac{d u}{d x}$ can be treated like a fraction
(i.e. “multiply by $d x$ ” to get rid of fractions)

STEP 3

Replace all parts of the integral

All

x

terms should be replaced with equivalent

u

terms, including

d x

If finding a definite integral change the limits from

x

-values to

u

-values too

STEP 4

Integrate and either

substitute

x

back in

evaluate the definte integral using the

u

limits (either using a GDC or manually)

STEP 5

Find

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

Exam Tip

Use your GDC to check the value of a definite integral, even in cases where working needs to be shown

Worked Example

Find the integral

$\int \frac{6 x + 5}{{(3 x^{2} + 5 x - 1)}^{3}} d x$

5-4-2-ib-sl-aa-only-we3-soltn-a

Evaluate the integral

$\int_{1}^{2} \frac{6 x + 5}{{(3 x^{2} + 5 x - 1)}^{3}} d x$

giving your answer as an exact fraction in its simplest terms.

5-4-2-ib-sl-aa-only-we3-soltn-b

DP IB Maths: AA SL

Revision Notes

5.4.2 Techniques of Integration

Integrating Composite Functions (ax+b)

What is a composite function?

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

Exam Tip

Worked Example

Reverse Chain Rule

What is reverse chain rule?

How do I know when to use reverse chain rule?

How do I integrate using reverse chain rule?

Exam Tip

Worked Example

Substitution: Reverse Chain Rule

What is integration by substitution?

How do I integrate using substitution?

Exam Tip

Worked Example

Author: Paul

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