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

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

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

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

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

Exam Tip

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

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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?**

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?

Worked Example

Substitution: Reverse Chain Rule

What is integration by substitution?

How do I integrate using substitution?

Worked Example

Exam Tip

Author: Paul

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