Integration by Substitution (Edexcel A Level Further Maths: Core Pure)

Revision Note

Test yourself

Author

Paul

Last updated

7 February 2024

Integrating using Trigonometric Substitutions

The integrals covered in this revision note are based on the standard results

$\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \arcsin (\frac{x}{a}) + c$

and

$\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} arc \tan (\frac{x}{a}) + c, a > 0, | x | > a$

These are given in the formulae booklet

How do I know when to use a trigonometric substitution in integration?

There are three main types of problem

Type 1
Showing the standard results using a substitution ( $a$ may have a value)
The substitution will not be given in such cases
e.g. Use a suitable substitution to show that $\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \arcsin (\frac{x}{a}) + c$
Let $x = a \sin u$ , so $\frac{d x}{d u} = a \cos u$ and $u = arcsi n (\frac{x}{a})$
$\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \int \frac{1}{\sqrt{a^{2} - {(a \sin u)}^{2}}} (a \cos u) d u = \int \frac{a \cos u}{\sqrt{a^{2} (1 - \sin^{2} u)}} d u = \int 1 d u = u + c ∴ \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \arcsin (\frac{x}{a}) + c$
The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity , so it can be integrated using standard results or techniques
Type 2
Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
e.g. Find $\int \frac{3}{9^{2} + {(2 x)}^{2}} d x$
$\int \frac{3}{9^{2} + {(2 x)}^{2}} d x = 3 \times \frac{1}{2} \int \frac{2}{9^{2} + {(2 x)}^{2}} d x = \frac{3}{2} (\frac{1}{9} \arctan (\frac{2 x}{9})) + c = \frac{1}{6} \arctan (\frac{2 x}{9}) + c$
Type 3
The denominator contains a three-term quadratic expression – i.e. there is an x term
In such cases complete the square and use reverse chain rule
e.g. Find $\int \frac{1}{25 + x^{2} - 6 x} d x$
$\int \frac{1}{25 + x^{2} - 6 x} d x = \int \frac{1}{25 + {(x - 3)}^{2} - 9} d x = \int \frac{1}{4^{2} + {(x - 3)}^{2}} d x = \frac{1}{4} \arctan (\frac{x - 3}{4}) + c$
(This works since $\frac{d}{d x} [x - 3] = 1$ , so effectively there is no reverse chain rule involved)

A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course

How do I use a trigonometric substitution to find integrals?

STEP 1
Identify the type of problem and if a substitution is required
Determine the substitution if needed

STEP 2
For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
For Type 3 problems complete the square

STEP 3
Integrate using standard techniques and results, possibly from the formulae booklet
For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation

STEP 4
Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
For indefinite integration, simplify where obvious and/or rearrange into a required format

Why is arccos x not involved in any of the integration results?

$\frac{d}{d x} [\arccos (\frac{x}{a})] = - \frac{1}{\sqrt{a^{2} - x^{2}}}$
For integration the "-" at the start can be treated as the constant "-1" and so integrating would lead to "-arcsin ..."
- i.e. $\int - \frac{1}{\sqrt{a^{2} - x^{2}}} d x = - \arcsin (\frac{x}{a}) + c$

Examiner Tip

The general form of the functions involving trigonometric and hyperbolic functions are very similar
Be clear about which form needs a trigonometric substitution and which form need a hyperbolic substitution
Always have a copy of the formula booklet to hand when practising these problems

Worked example

(a) Use an appropriate substitution to show that

$\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \arctan (\frac{x}{a}) + c$

5-2-5-edex-fm--alevel-we1-trigsub-soltn-a

(b) Find

$\int \frac{3}{16 + 9 x^{2}} d x$

5-2-5-edex-fm--alevel-we1-trigsub-soltn-b

Integrating using Hyperbolic Substitutions

The integrals covered in this revision note are based on the standard results

$\int \frac{1}{\sqrt{a^{2} + x^{2}}} d x = arsinh (\frac{x}{a}) + c$

and

$\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = arcosh (\frac{x}{a}) + c, x > a$

These are given in the formulae booklet

How do I know when to use a hyperbolic substitution in integration?

There are three main types of problem

Type 1
Showing the standard results using a substitution ( $a$ may have a value)
The substitution will not be given in such cases
e.g. Use a suitable substitution to show that $\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = ar \cos h (\frac{x}{a}) + c$
Let $x = a \cosh u$ , so $\frac{d x}{d u} = a \sinh u$ and $u = arcosh (\frac{x}{a})$
$\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \int \frac{1}{\sqrt{{(a \cosh u)}^{2} - a^{2}}} (a \sinh u) d u = \int \frac{a \sinh u}{\sqrt{a^{2} (\cosh^{2} u - 1)}} d u = \int 1 d u = u + c$
$∴ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = arcosh (\frac{x}{a}) + c$
The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity $\cosh^{2} x - \sinh^{2} x \equiv 1$ , so it can be integrated using standard results or techniques
Type 2
Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
e.g. Find $\int \frac{3}{\sqrt{4 x^{2} + 9}} d x$
$\int \frac{3}{\sqrt{{(2 x)}^{2} + 3^{2}}} d x = 3 \times \frac{1}{2} \int \frac{2}{\sqrt{{(2 x)}^{2} + 3^{2}}} d x = \frac{3}{2} (arsinh (\frac{2 x}{3})) + c = \frac{3}{2} arsinh (\frac{2 x}{3}) + c$
Type 3
The denominator contains a three-term quadratic expression – i.e. there is an $x$ term
In such cases complete the square and use reverse chain rule
e.g. Find $\int \frac{1}{\sqrt{x^{2} - 6 x + 25}} d x$
$\int \frac{1}{\sqrt{x^{2} - 6 x + 25}} d x = \int \frac{1}{\sqrt{{(x - 3)}^{2} - 9 + 25}} d x = \int \frac{1}{\sqrt{{(x - 3)}^{2} + 4^{2}}} d x = arsinh (\frac{x - 3}{4}) + c$
(This works since $\frac{d}{d x} [x - 3] = 1$ ,so effectively there is no reverse chain rule involved
A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course, although hyperbolic functions are not

How do I use a hyperbolic substitution to find integrals?

STEP 1
Identify the type of problem and if a substitution is required
Determine the substitution if needed
STEP 2
For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
For Type 3 problems complete the square
STEP 3
Integrate using standard techniques and results, possibly from the formulae booklet
For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation
STEP 4
Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
For indefinite integration, simplify where obvious and/or rearrange into a required format

Is artanh x involved in integration?

The standard result, given in the formulae booklet is
$\int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{a} artanh (\frac{x}{a}) + c, | x | < a$
with the alternative result $\frac{1}{2 a} \ln | \frac{a + x}{a - x} | + c$ also given
Problems involving these often involve partial fractions (since $a^{2} - x^{2}$ is the difference of two squares) leading to the 'ln' result
If you happen to recognise the integral and can use the formulae booklet result involving "artanh" to solve a problem, then do so!

Examiner Tip

The general form of the functions involving trigonometric and hyperbolic functions are very similar
Be clear about which form needs a trigonometric substitution and which form need a hyperbolic substitution
Always have a copy of the formula booklet to hand when practising these problems