Introduction
• This is a very big chapter which covers many
ways of Integrating more difficult expressions
• You are sometimes told which method to use
but in addition must be able to identify the
most appropriate way
• We also cover using Integration to work out
volumes as well as areas, as well as setting up
and using equations (similar to the chapter on
differentiation)
Integration
You need to be able to integrate standard
functions
You met the following in C3, in the differentiation chapter:
𝑦 = 𝑥𝑛
𝑦 = 𝑠𝑖𝑛𝑥
𝑦 = 𝑐𝑜𝑠𝑒𝑐𝑥
𝑑𝑦
= 𝑛𝑥 𝑛−1
𝑑𝑥
𝑑𝑦
= 𝑐𝑜𝑠𝑥
𝑑𝑥
𝑑𝑦
= −𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥
𝑑𝑥
𝑦 = 𝑒 𝑓(𝑥)
𝑦 = 𝑐𝑜𝑠𝑥
𝑦 = 𝑐𝑜𝑡𝑥
𝑑𝑦
= 𝑓′(𝑥)𝑒 𝑓(𝑥)
𝑑𝑥
𝑑𝑦
= −𝑠𝑖𝑛𝑥
𝑑𝑥
𝑑𝑦
= −𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
𝑦 = ln(𝑓 𝑥 )
𝑦 = 𝑡𝑎𝑛𝑥
𝑦 = 𝑠𝑒𝑐𝑥
𝑑𝑦 𝑓′(𝑥)
=
𝑑𝑥 𝑓(𝑥)
𝑑𝑦
= 𝑠𝑒𝑐 2 𝑥
𝑑𝑥
𝑑𝑦
= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
𝑑𝑥
6A
Integration
You need to be able to integrate standard
functions
Therefore, you already can deduce the following
𝑛+1
𝑥
+𝐶
𝑥𝑛 =
𝑛+1
𝑐𝑜𝑠𝑥 = 𝑠𝑖𝑛𝑥 + 𝐶
𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥 = −𝑐𝑜𝑠𝑒𝑐𝑥 + 𝐶
𝑒𝑥 = 𝑒𝑥 + 𝐶
𝑠𝑖𝑛𝑥 = −𝑐𝑜𝑠𝑥 + 𝐶
𝑐𝑜𝑠𝑒𝑐 2 𝑥 = −𝑐𝑜𝑡𝑥 + 𝐶
1
= ln |𝑥| + 𝐶
𝑥
𝑠𝑒𝑐 2 𝑥 = 𝑡𝑎𝑛𝑥 + 𝐶
𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 = 𝑠𝑒𝑐𝑥 + 𝐶
The modulus sign is used here to
avoid potential problems with
negative numbers…
6A
Integration
2𝑐𝑜𝑠𝑥 +
You need to be able to integrate
standard functions
Find the following integral:
3
2𝑐𝑜𝑠𝑥 + − 𝑥 𝑑𝑥
𝑥
2𝑐𝑜𝑠𝑥
= 2𝑠𝑖𝑛𝑥
As the terms are
separate you can
integrate them
separately
3
− 𝑥 𝑑𝑥
𝑥
3
𝑥
𝑥
= 3ln |𝑥|
=
1
𝑥2
Rewrite this
term as a
power
3
𝑥2
=
3
2
=
2 3
𝑥2
3
2 3
= 2𝑠𝑖𝑛𝑥 + 3 ln 𝑥 − 𝑥 2 + 𝐶
3
Remember
the + C!
Integration
𝑐𝑜𝑠𝑥
− 2𝑒 𝑥 𝑑𝑥
2
𝑠𝑖𝑛 𝑥
You need to be able to integrate
standard functions
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛2 𝑥
Find the following integral:
𝑐𝑜𝑠𝑥
− 2𝑒 𝑥 𝑑𝑥
2
𝑠𝑖𝑛 𝑥
Try to rewrite as
an integral you
‘know’
=
𝑐𝑜𝑠𝑥 1
𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥
=
𝑐𝑜𝑡𝑥𝑐𝑜𝑠𝑒𝑐𝑥
As the terms are
separate you can
integrate them
separately
2𝑒 𝑥
= 2𝑒 𝑥
= −𝑐𝑜𝑠𝑒𝑐𝑥
= −𝑐𝑜𝑠𝑒𝑐𝑥 − 2𝑒 𝑥 + 𝐶
Remember
the + C!
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
𝑦 = sin(2𝑥 + 3)
𝑑𝑦
= 2cos(2𝑥 + 3)
𝑑𝑥
Find the following integral:
cos 2𝑥 + 3 𝑑𝑥
cos 2𝑥 + 3 𝑑𝑥
=
1
sin 2𝑥 + 3 + 𝐶
2
Consider starting with
sin(2x + 3) and the answer
that would give
This is double what we
are wanting to integrate
 Therefore, we must
‘start’ with half the
amount…
Divide the original
‘guess’ by 2
This is a VERY common method of integration
– considering what we might start with that
would differentiate to our answer…
6B
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
Find the following integral:
𝑒 4𝑥+1 𝑑𝑥
𝑦 = 𝑒 4𝑥+1
𝑑𝑦
= 4𝑒 4𝑥+1
𝑑𝑥
Consider starting with e4x + 1
and the answer that would
give
This is four times what we are
wanting to integrate
 Therefore, we must ‘start’
with a quarter of the amount…
𝑒 4𝑥+1 𝑑𝑥
1
4
= 𝑒 4𝑥+1 + 𝐶
Divide the original
‘guess’ by 4
6B
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
Find the following integral:
𝑠𝑒𝑐 2 3𝑥 𝑑𝑥
𝑦 = 𝑡𝑎𝑛3𝑥
𝑑𝑦
= 3𝑠𝑒𝑐 2 3𝑥
𝑑𝑥
Consider starting with tan3x
and the answer that would
give
This is three times what we
are wanting to integrate
 Therefore, we must ‘start’
with a third of the amount…
𝑠𝑒𝑐 2 3𝑥 𝑑𝑥
1
3
= 𝑡𝑎𝑛3𝑥 + 𝐶
Divide the original
‘guess’ by 3
6B
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
=
cos 2𝑥 + 3 𝑑𝑥
𝑒 4𝑥+1 𝑑𝑥
1
sin 2𝑥 + 3 + 𝐶
2
= 𝑒 4𝑥+1 + 𝐶
1
4
𝑠𝑒𝑐 2 3𝑥 𝑑𝑥
1
3
= 𝑡𝑎𝑛3𝑥 + 𝐶
These three answers illustrate a
rule:
1
𝑓 ′ 𝑎𝑥 + 𝑏 = 𝑓 𝑎𝑥 + 𝑏 + 𝐶
𝑎
1) Integrate the function using what
you know from C3
2) Divide by the coefficient of x
3) Simplify if possible and add C
6B
𝑓 ′ 𝑎𝑥 + 𝑏 =
1
𝑓 𝑎𝑥 + 𝑏 + 𝐶
𝑎
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
1
𝑑𝑥
3𝑥 + 2
1
= ln 3𝑥 + 2 + 𝐶
3
1) Integrate the function using
what you know from C3
2) Divide by the coefficient of x
3) Simplify if possible and add C
Find the following integral:
1
𝑑𝑥
3𝑥 + 2
6B
𝑓 ′ 𝑎𝑥 + 𝑏 =
1
𝑓 𝑎𝑥 + 𝑏 + 𝐶
𝑎
Integration
You can integrate using the
reverse of the Chain rule
This technique will only work for
linear transformations of
functions such as f(ax + b)
𝑦 = (2𝑥 + 3)5
𝑑𝑦
= 5(2𝑥 + 3)4 (2)
𝑑𝑥
Consider a function that would
leave you with a power 4 and the
same bracket
Simplify after using the Chain rule
𝑑𝑦
= 10(2𝑥 + 3)4
𝑑𝑥
Find the following integral:
4
(2𝑥 + 3)
As this is 10 times what we want, we need to divide our
‘guess’ by 10
𝑑𝑥
(2𝑥 + 3)4 𝑑𝑥
=
1
(2𝑥 + 3)5 + 𝐶
10
6B
Integration
You can use Trigonometric Identities
in Integration
There are some trigonometric
expression you cannot integrate yet,
but a way to do so is to write them in
terms of ones you can integrate…
Trigonometric Identities are
invaluable in this!
Find:
𝑡𝑎𝑛2 𝑥 𝑑𝑥
=
Divide by
cos
Subtract
1
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 ≡ 1
𝑡𝑎𝑛2 𝑥 + 1 ≡ 𝑠𝑒𝑐 2 𝑥
𝑡𝑎𝑛2 𝑥 ≡ 𝑠𝑒𝑐 2 𝑥 − 1
Using the identity
above, replace tan2x
(𝑠𝑒𝑐 2 𝑥 − 1) 𝑑𝑥
Integrate each part
separately
𝑠𝑒𝑐 2 𝑥
= 𝑡𝑎𝑛𝑥
1
=𝑥
= 𝑡𝑎𝑛𝑥 − 𝑥 + 𝐶
𝑡𝑎𝑛2 𝑥 𝑑𝑥
6C
Integration
You can use Trigonometric Identities
in Integration
There are some trigonometric
expression you cannot integrate yet,
but a way to do so is to write them in
terms of ones you can integrate…
Trigonometric Identities are
invaluable in this!
𝑠𝑖𝑛2 𝑥 𝑑𝑥
1
2
=
Find:
2
𝑠𝑖𝑛 𝑥 𝑑𝑥
1 1
− 𝑐𝑜𝑠2𝑥
2 2
=
=
1
𝑥
2
𝑠𝑖𝑛2 𝑥 ≡
1 1
− 𝑐𝑜𝑠2𝑥
2 2
𝑑𝑥
1
𝑠𝑖𝑛2𝑥
4
1
1
𝑥 − 𝑠𝑖𝑛2𝑥 + 𝐶
2
4
Divide by
2
2𝑠𝑖𝑛2 𝑥 = 1 − 𝑐𝑜𝑠2𝑥
Using the identity
above, replace sin2x
1
𝑐𝑜𝑠2𝑥
2
=
Rearrange
𝑐𝑜𝑠2𝑥 = 1 − 2𝑠𝑖𝑛2 𝑥
Integrate each part
separately
Use the
‘guessing’
method
𝑦 = 𝑠𝑖𝑛2𝑥
𝑑𝑦
= 2𝑐𝑜𝑠2𝑥
𝑑𝑥
This is 4 times what we
want so divide the
‘guess’ by 4
6C
Integration
You can use Trigonometric Identities
in Integration
There are some trigonometric
expression you cannot integrate yet,
but a way to do so is to write them in
terms of ones you can integrate…
Trigonometric Identities are
invaluable in this!
Find:
Double angle
formula
Follow the
pattern…
Divide by 2
s𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
s𝑖𝑛4𝑥 = 2𝑠𝑖𝑛2𝑥𝑐𝑜𝑠2𝑥
s𝑖𝑛6𝑥 = 2𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥
1
𝑠𝑖𝑛6𝑥 = 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥
2
𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥
1
𝑠𝑖𝑛6𝑥 𝑑𝑥
2
𝑦 = 𝑐𝑜𝑠6𝑥
𝑑𝑦
= −6𝑠𝑖𝑛6𝑥
𝑑𝑥
Replace with the
above…
This will give us the sin 6x when
differentiating, but is negative
and 12 times too big!
 Make the guess negative and
divide by 12!
1
𝑠𝑖𝑛6𝑥 𝑑𝑥
2
𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥
=−
1
𝑐𝑜𝑠6𝑥 + 𝐶
12
6C
Integration
You can use Trigonometric Identities
in Integration
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 ≡ 1
Divide by
cos
𝑡𝑎𝑛2 𝑥 + 1 ≡ 𝑠𝑒𝑐 2 𝑥
Subtract
1
𝑡𝑎𝑛2 𝑥 ≡ 𝑠𝑒𝑐 2 𝑥 − 1
(𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)2 𝑑𝑥
Expand the
bracket
𝑠𝑒𝑐 2 𝑥 + 𝑡𝑎𝑛2 𝑥 + 2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥
There are some trigonometric
expression you cannot integrate yet,
but a way to do so is to write them in
terms of ones you can integrate…
Replace
tan2x
𝑠𝑒𝑐 2 𝑥 + (𝑠𝑒𝑐 2 𝑥 − 1) + 2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥
Simplify
2𝑠𝑒𝑐 2 𝑥 − 1 + 2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥
Integrate
separately
Trigonometric Identities are
invaluable in this!
2𝑠𝑒𝑐 2 𝑥
1
Find:
= 2𝑡𝑎𝑛𝑥
=𝑥
2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
= 2𝑠𝑒𝑐𝑥
(𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥)2 𝑑𝑥
= 2𝑡𝑎𝑛𝑥 − 𝑥 + 2𝑠𝑒𝑐𝑥 + 𝐶
6C
Integration
You can use partial fractions
to integrate expressions
This allows you to split a
fraction up – it can sometimes
be recombined after
integration…
𝐴
𝐵
𝑥−5
=
+
(𝑥 + 1) (𝑥 − 2)
(𝑥 + 1)(𝑥 − 2)
𝑥−5
𝐴(𝑥 − 2)
𝐵(𝑥 + 1)
=
+
(𝑥 + 1)(𝑥 − 2)
(𝑥 + 1)(𝑥 − 2) (𝑥 + 1)(𝑥 − 2)
𝑥−5
𝐴 𝑥 − 2 + 𝐵(𝑥 + 1)
=
(𝑥 + 1)(𝑥 − 2)
(𝑥 + 1)(𝑥 − 2)
Write as two
fractions and make
the denominators
equal
Combine
The numerators
must be equal
𝑥 − 5 = 𝐴 𝑥 − 2 + 𝐵(𝑥 + 1)
Find:
𝑥−5
(𝑥 + 1)(𝑥 − 2)
Let x = 2
−3 = 3𝐵
−1 = 𝐵
Let x = -1
Calculate A and B by
choosing appropriate
x values
−6 = −3𝐴
2 =𝐴
2
1
−
(𝑥 + 1) (𝑥 − 2)
𝐴
𝐵
𝑥−5
=
+
(𝑥 + 1) (𝑥 − 2)
(𝑥 + 1)(𝑥 − 2)
2
1
𝑥−5
=
−
(𝑥 + 1) (𝑥 − 2)
(𝑥 + 1)(𝑥 − 2)
Replace A and B
from the start
6D
Integration
You can use partial fractions
to integrate expressions
This allows you to split a
fraction up – it can sometimes
be recombined after
integration…
2
1
−
(𝑥 + 1) (𝑥 − 2)
Integrate
separately
2
(𝑥 + 1)
1
(𝑥 − 2)
= 2ln |𝑥 + 1|
= ln |𝑥 − 2|
= ln |(𝑥 + 1)2 |
Find:
𝑥−5
(𝑥 + 1)(𝑥 − 2)
= ln |(𝑥 + 1)2 | − ln 𝑥 − 2 + 𝐶
𝑥+1 2
= ln
+𝐶
𝑥−2
You can combine the
natural logarithms as a
division
2
1
−
(𝑥 + 1) (𝑥 − 2)
6D
Integration
You can use partial fractions
to integrate expressions
1
9𝑥 2 − 4
9𝑥 2 − 3𝑥 + 2
−4
9𝑥 2
−3𝑥 + 6
This allows you to split a
fraction up – it can sometimes
be recombined after
integration…
1) Divide the first term by
the highest power
−
2) Multiply the answer by
the whole expression you’re
dividing by
3) Subtract to find the
remainder
4) Remember to write the
remainder as a fraction of
the original expression
Find:
9𝑥 2 − 3𝑥 + 2
9𝑥 2 − 4
9𝑥 2 − 3𝑥 + 2
9𝑥 2 − 4
= 1 +
−3𝑥 + 6
9𝑥 2 − 4
= 1 +
6 − 3𝑥
9𝑥 2 − 4
Looks tidier!
6D
Integration
You can use partial fractions
to integrate expressions
This allows you to split a
fraction up – it can sometimes
be recombined after
integration…
9𝑥 2 − 3𝑥 + 2
9𝑥 2 − 4
We now need to write the
remainder as partial
fractions
6 − 3𝑥
6 − 3𝑥
=
9𝑥 2 − 4
(3𝑥 + 2)(3𝑥 − 2)
Find:
9𝑥 2 − 3𝑥 + 2
9𝑥 2 − 4
6 − 3𝑥
= 1 +
9𝑥 2 − 4
=
𝐴
𝐵
+
(3𝑥 + 2) (3𝑥 − 2)
=
𝐴 3𝑥 − 2 + 𝐵(3𝑥 + 2)
(3𝑥 − 2)(3𝑥 + 2)
𝐴 3𝑥 − 2 + 𝐵 3𝑥 + 2 = 6 − 3𝑥
Let x = 2/3
4𝐵 = 4
Set the numerators
equal and solve for A
and B
𝐵=1
Let x = -2/3
−4𝐴 = 8
𝐴 = −2
1 +
6 − 3𝑥
9𝑥 2 − 4
= 1−
2
1
+
3𝑥 + 2 3𝑥 − 2
Write the final answer
with the remainder
broken apart!
6D
Integration
You can use partial fractions
to integrate expressions
This allows you to split a
fraction up – it can sometimes
be recombined after
integration…
Find:
9𝑥 2 − 3𝑥 + 2
9𝑥 2 − 4
2
1
9𝑥 2 − 3𝑥 + 2
= 1−
+
3𝑥 + 2 3𝑥 − 2
9𝑥 2 − 4
1−
2
1
+
3𝑥 + 2
3𝑥 − 2
2
3𝑥 + 2
1
=𝑥
Integrate
separately
= 2
=
1
3𝑥 − 2
1
ln |3𝑥 + 2|
3
=
1
ln |3𝑥 − 2|
3
1
ln | 3𝑥 + 2 2 |
3
1
= 𝑥 − ln 3𝑥 + 2
3
2
1
+ ln 3𝑥 − 2 + 𝐶
3
1 (3𝑥 − 2)
=𝑥+
+𝐶
3 (3𝑥 + 2)2
You can combine the
natural logarithms (be
careful, the negative
goes on the bottom…)
6D
Integration
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
1
2𝑥 + 3
Including using partial fractions where
an expression can be factorised
However, this method will not work for
integrals of the form:
1
𝑥2 + 1
Some expressions like this can by
integrated by using the ‘standard
patterns’ technique
6E
Integration
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
2𝑥
𝑑𝑥
𝑥2 + 1
Remember…
𝑖𝑓 𝑦 = ln |𝑓 𝑥 |
Some expressions can by
integrated by using the ‘standard
patterns’ technique
𝑑𝑦 𝑓′(𝑥)
=
𝑑𝑥 𝑓(𝑥)
Find:
𝑦 = ln |𝑥 2 + 1|
2𝑥
𝑑𝑥
𝑥2 + 1
Notice that the denominator would
differentiate to become the numerator
 This is a pattern we can use to figure
out what the integral is…
𝑑𝑦
2𝑥
= 2
𝑑𝑥 𝑥 + 1
So imagine starting with ln|denominator|
In this case, we get straight
to the answer!
2𝑥
𝑑𝑥
𝑥2 + 1
= ln 𝑥 2 + 1 + 𝐶
6E
𝑖𝑓 𝑦 = ln |𝑓 𝑥 |
𝑑𝑦 𝑓′(𝑥)
=
𝑑𝑥 𝑓(𝑥)
Integration
𝑐𝑜𝑠𝑥
𝑑𝑥
3 + 2𝑠𝑖𝑛𝑥
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
Start by trying
y = ln|denominator|
𝑦 = ln |3 + 2𝑠𝑖𝑛𝑥|
Differentiate
𝑑𝑦
2𝑐𝑜𝑠𝑥
=
𝑑𝑥 3 + 2𝑠𝑖𝑛𝑥
This is double what we want so multiply the ‘guess’ by 1/2
𝑐𝑜𝑠𝑥
𝑑𝑥
3 + 2𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
𝑑𝑥
3 + 2𝑠𝑖𝑛𝑥
=
1
ln 3 + 2𝑠𝑖𝑛𝑥 + 𝐶
2
6E
Integration
You can Integrate by using
standard patterns
In this case consider the power of
sine.
 If it has been differentiated, it
must have been sin3x originally…
3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2 𝑥 𝑑𝑥
You have seen how to integrate
fractions of the form:
𝑦 = 𝑠𝑖𝑛3 𝑥
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2 𝑥 𝑑𝑥
Write as a cubed bracket
𝑦 = (𝑠𝑖𝑛𝑥)3
𝑑𝑦
= 3(𝑠𝑖𝑛𝑥)2 (𝑐𝑜𝑠𝑥)
𝑑𝑥
𝑑𝑦
= 3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2 𝑥
𝑑𝑥
Differentiate using
the chain rule
Rewrite – this has given us
exactly what we wanted!
3𝑐𝑜𝑠𝑥𝑠𝑖𝑛2 𝑥 𝑑𝑥
Don’t forget the + C!
3
= 𝑠𝑖𝑛 𝑥 + 𝐶
6E
Integration
Consider the power on the bracket
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
𝑥(𝑥 2 + 5)3 𝑑𝑥
𝑥(𝑥 2 + 5)3 𝑑𝑥
 As it is a power 3, it must have
been a power 4 before differentiation
𝑦 = (𝑥 2 + 5)4
𝑑𝑦
= 4(𝑥 2 + 5)3 (2𝑥)
𝑑𝑥
Differentiate the bracket to the
power 4 using the chain rule
Simplify
𝑑𝑦
= 8𝑥(𝑥 2 + 5)3
𝑑𝑥
This is 8 times too big so multiply the ‘guess’ by 1/8
𝑥(𝑥 2 + 5)3 𝑑𝑥
1
= (𝑥 2 + 5)4 + 𝐶
8
Don’t forget to add C!
6E
Integration
𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
(2 + 𝑐𝑜𝑡𝑥)3
You can Integrate by using
standard patterns
(𝑐𝑜𝑠𝑒𝑐 2 𝑥)(2 + 𝑐𝑜𝑡𝑥)−3 𝑑𝑥
You have seen how to integrate
fractions of the form:
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
(2 + 𝑐𝑜𝑡𝑥)3
Write using powers
Imagine how we could end up
with a -3 as a power...
𝑦 = (2 + 𝑐𝑜𝑡𝑥)−2
𝑑𝑦
= −2(2 + 𝑐𝑜𝑡𝑥)−3 (−𝑐𝑜𝑠𝑒𝑐 2 𝑥)
𝑑𝑥
Use the chain rule
Rewrite
𝑑𝑦
= 2𝑐𝑜𝑠𝑒𝑐 2 𝑥(2 + 𝑐𝑜𝑡𝑥)−3
𝑑𝑥
𝑑𝑦
2𝑐𝑜𝑠𝑒𝑐 2 𝑥
=
𝑑𝑥 (2 + 𝑐𝑜𝑡𝑥)3
This is double what we
want so multiply the
‘guess’ by 1/2
𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
(2 + 𝑐𝑜𝑡𝑥)3
=
1
(2 + 𝑐𝑜𝑡𝑥)−2 + 𝐶
2
6E
Integration
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥 𝑑𝑥
Consider using a
power 5
𝑦 = 𝑠𝑒𝑐 5 𝑥
𝑦 = (𝑠𝑒𝑐𝑥)5
Write as a bracket
to the power 5
𝑑𝑦
= 5(𝑠𝑒𝑐𝑥)4 (𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)
𝑑𝑥
𝑑𝑦
= 5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 5 𝑥
𝑑𝑥
Differentiate using
the chain rule
We have an extra
secx
5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥 𝑑𝑥
HOWEVER:
 We cannot just add this to our ‘guess’ as before, as the
differentiation will need to be performed using the
product rule from C3, rather than the Chain rule!
 We need to find another way!
6E
Integration
You can Integrate by using
standard patterns
You have seen how to integrate
fractions of the form:
Some expressions can by
integrated by using the ‘standard
patterns’ technique
Find:
5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥 𝑑𝑥
5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥 𝑑𝑥
Consider using a
power 4
𝑦 = 𝑠𝑒𝑐 4 𝑥
𝑦 = (𝑠𝑒𝑐𝑥)4
Write as a bracket
to the power 4
𝑑𝑦
= 4(𝑠𝑒𝑐𝑥)3 (𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)
𝑑𝑥
Differentiate using
the chain rule
𝑑𝑦
= 4𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥
𝑑𝑥
This is what we want, but 4/5 of the amount
 Multiply by the guess by 5/4
5𝑡𝑎𝑛𝑥𝑠𝑒𝑐 4 𝑥 𝑑𝑥
𝑦=
5
𝑠𝑒𝑐 4 𝑥 + 𝐶
4
6E
Integration
It is sometimes possible to simplify an
integral by changing the variable. This
is known as integration by substitution.
𝑥 2𝑥 + 5 𝑑𝑥
𝑢 = 2𝑥 + 5
Differentiate
Use the substitution:
𝑢 = 2𝑥 + 5
To find:
Rearrange to
get dx
To integrate this, you need to
replace the x terms with
equivalent u terms, and replace
the dx with an equivalent du
Rearrange to find x
𝑑𝑢
=2
𝑑𝑥
𝑢 − 5 = 2𝑥
1
(𝑢 − 5) = 𝑥
2
𝑑𝑢
= 𝑑𝑥
2
𝑥 2𝑥 + 5 𝑑𝑥
𝑥 2𝑥 + 5 𝑑𝑥
𝑑𝑢
1
(𝑢 − 5) 𝑢
2
2
1
𝑢−5
4
𝑢 𝑑𝑢
1 3 5 1
𝑢2 − 𝑢2 𝑑𝑢
4
4
Replace each ‘x’ term with
an equivalent ‘u’ term
Rearrange – you should
leave ‘du’ at the end
Combine terms including the
square root, changed to a
power ‘1/2’
6F
Integration
1 3 5 1
𝑢2 − 𝑢2 𝑑𝑢
4
4
It is sometimes possible to simplify an
integral by changing the variable. This
is known as integration by substitution.
=
Use the substitution:
𝑢 = 2𝑥 + 5
=
To find:
𝑥 2𝑥 + 5 𝑑𝑥
=
5
𝑢2
1
4 5
2
−
1 2 5
𝑢2 −
4 5
5
4
3
𝑢2
3
2
Differentiate
terms separately
+ 𝐶
5 2 5
𝑢2 + 𝐶
4 3
Flip the dividing
fractions
1 5
5 5
𝑢2 − 𝑢2 + 𝐶
5
6
5
5
1
5
2
= (2𝑥 + 5) − (2𝑥 + 5)2 + 𝐶
5
6
Calculate the
fraction parts
Finally,
replace with
u with its
equivalent
from the
start!
6F
Integration
𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1 + 𝑠𝑖𝑛𝑥)3 𝑑𝑥
It is sometimes possible to simplify an
integral by changing the variable. This
is known as integration by substitution.
Use the substitution:
Differentiate
𝑑𝑢
= 𝑐𝑜𝑠𝑥
𝑑𝑥
Rearrange to
get dx
𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1 + 𝑠𝑖𝑛𝑥)3 𝑑𝑥
Rearrange to
find Sinx
𝑢 − 1 = 𝑠𝑖𝑛𝑥
𝑑𝑢
= 𝑑𝑥
𝑐𝑜𝑠𝑥
𝑢 = 𝑠𝑖𝑛𝑥 + 1
To find:
𝑢 = 𝑠𝑖𝑛𝑥 + 1
𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1 + 𝑠𝑖𝑛𝑥)3 𝑑𝑥
𝑐𝑜𝑠𝑥 (𝑢 − 1) 𝑢3 𝑑𝑢
𝑐𝑜𝑠𝑥
Replace each ‘x’ term with
an equivalent ‘u’ term
Cancel the Cosx terms
𝑢 − 1 𝑢3 𝑑𝑢
Multiply out
4
3
𝑢 − 𝑢 𝑑𝑢
1
1
= 𝑢5 − 𝑢4 + 𝐶
5
4
Integrate
1
1
= (𝑠𝑖𝑛𝑥 + 1)5 − (𝑠𝑖𝑛𝑥 + 1)4 +𝐶
5
4
Replace u with x
terms again!
6F
Integration
2
𝑥(𝑥 + 1)3 𝑑𝑥
It is sometimes possible to simplify
an integral by changing the
variable. This is known as
integration by substitution.
0
𝑢 =𝑥+1
Differentiate
𝑑𝑢
=1
𝑑𝑥
Rearrange
to get dx
𝑢−1=𝑥
𝑑𝑢 = 𝑑𝑥
Use integration by substitution to find:
𝑢 =𝑥+1
You also need to
recalculate limits 𝑥 = 2,
𝑢=3
in terms of u
2
𝑥(𝑥 + 1)3 𝑑𝑥
0
Rearrange to
find x
𝑥 = 0,
2
𝑢=1
3
𝑥(𝑥 + 1) 𝑑𝑥
Sometimes you will have to decide on a
substitution yourself.
Replace x limits with u limits and
the x terms with u terms
0
3
(𝑢 − 1) 𝑢3 𝑑𝑢
1
In this case, the bracket would be hardest
to integrate so it makes sense to use the
substitution:
𝑢 =𝑥+1
An alternative method is to replace the
‘u’ terms with x terms at the end and
then just use the original ‘x’ limits –
either way is fine!
Multiply out bracket
3
𝑢4 − 𝑢3 𝑑𝑢
1
5
=
𝑢
𝑢
−
5
4
5
=
1
4
3
3
−
5
4
= 28.4
Integrate
4 3
5
−
4
1
1
−
5
4
Sub in limits
and calculate
6F
Integration
You can use integration by parts to
integrate some expressions
This is the differential of
two functions multiplied
together
 You could think of it as:
𝑑(𝑢𝑣)
𝑑𝑥
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
In C3 you met the following: (the product rule)
𝑑
𝑢𝑣
𝑑𝑥
𝑢
= 𝑢
𝑑𝑣
𝑑𝑢
+𝑣
𝑑𝑥
𝑑𝑥
Rearrange by
subtracting vdu/dx
𝑑𝑣
𝑑
𝑑𝑢
=
𝑢𝑣 − 𝑣
𝑑𝑥
𝑑𝑥
𝑑𝑥
Integrate each term
with respect to x
𝑑𝑢
𝑑𝑥
This is the formula used for
Integration by parts!
 You get given this in the
booklet
𝑢
𝑑𝑣
=
𝑑𝑥
𝑑
𝑢𝑣 −
𝑑𝑥
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
The middle term is just
a differential.
 Integrating a
differential cancels
them both out!
The other terms do not cancel as only part of them are
differentiated…
6G
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
Integration
You can use integration by parts to
integrate some expressions
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
Unlike when using the product
rule, we now have one function
to differentiate, and one to
integrate…
𝑑𝑢
𝑑𝑥
Find:
𝑣 = 𝑠𝑖𝑛𝑥
𝑑𝑢
=1
𝑑𝑥
𝑑𝑣
= 𝑐𝑜𝑠𝑥
𝑑𝑥
Differentiate
𝑥𝑐𝑜𝑠𝑥 𝑑𝑥
You can recognise that Integration by
parts is needed as we have two
functions multiplied together…
𝑢=𝑥
Integrate
Now replace the relevant parts to find the integral…
= (𝑥)(𝑠𝑖𝑛𝑥) −
(𝑠𝑖𝑛𝑥)(1)
= 𝑥𝑠𝑖𝑛𝑥 − (−𝑐𝑜𝑠𝑥)
= 𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝐶
The integral
here is simpler!
Be careful with
negatives here!
As a general rule, it is easiest to let u = anything of the form xn. The exception
is when there is a lnx term, in which case this should be used as u
6G
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
Integration
You can use integration by parts to
integrate some expressions
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
𝑢 = 𝑙𝑛𝑥
𝑑𝑢 1
=
𝑑𝑥 𝑥
Find:
Differentiate
2
𝑥 𝑙𝑛𝑥 𝑑𝑥
You can recognise that Integration by
parts is needed as we have two
functions multiplied together…
Let u be lnx!
𝑣=
𝑥3
3
𝑑𝑣
= 𝑥2
𝑑𝑥
Integrate
Now replace the relevant parts to find the integral…
= (𝑙𝑛𝑥)
𝑥3
3
𝑥3
= 𝑙𝑛𝑥 −
3
−
𝑥3
3
1
𝑥
Simplify terms
1 2
𝑥
3
𝑥3
𝑥2
= 𝑙𝑛𝑥 −
+ 𝐶
3
9
Integrate the second
part
6G
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
Integration
You can use integration by parts to
integrate some expressions
Find:
𝑥 2 𝑒 𝑥 𝑑𝑥
You can recognise that Integration by
parts is needed as we have two
functions multiplied together…
𝑑𝑣
𝑢
= 𝑢𝑣 −
𝑑𝑥
𝑑𝑢
𝑣
𝑑𝑥
𝑣 = 𝑒𝑥
𝑑𝑢
= 2𝑥
𝑑𝑥
𝑑𝑣
= 𝑒𝑥
𝑑𝑥
Differentiate
Integrate
Now replace the relevant parts to find the integral…
= (𝑥 2 )(𝑒 𝑥 ) −
= 𝑥 2𝑒 𝑥 −
(𝑒 𝑥 )(2𝑥)
2𝑥𝑒 𝑥
𝑢 = 2𝑥
𝑑𝑢
=2
𝑑𝑥
Sometimes you will have to use the
process twice!
This happens if the new integral still
has two functions multiplied together…
𝑢 = 𝑥2
Differentiate
= 𝑥 2𝑒 𝑥 −
2𝑥𝑒 𝑥 −
= 𝑥 2𝑒 𝑥 −
2𝑥𝑒 𝑥 − 2𝑒 𝑥
2 𝑥
=𝑥 𝑒
− 2𝑥𝑒
𝑥
2𝑒 𝑥
+ 2𝑥𝑒
𝑥
𝑣 = 𝑒𝑥
𝑑𝑣
= 𝑒𝑥
𝑑𝑥
Integrate
Work out the square bracket
which is the second
integration by parts
+ 𝐶
Careful with negatives!!
6G
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
Integration
You can use integration by parts to
integrate some expressions
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
When integrating lnx, you MUST
think of it as ‘lnx times 1, and
use lnx as ‘u’ and 1 as ‘dv/dx’
𝑑𝑢
𝑑𝑥
𝑢 = 𝑙𝑛𝑥
𝑑𝑢 1
=
𝑑𝑥 𝑥
Evaluate:
2
𝑙𝑛𝑥 𝑑𝑥
Differentiate
𝑣=𝑥
𝑑𝑣
=1
𝑑𝑥
Integrate
1
Leave your answer in terms of natural
logarithms…
 You will be asked to leave exact
answers a lot so make sure you
know your log laws!!
Now replace the relevant parts to find the integral…
= (𝑙𝑛𝑥)(𝑥) −
= 𝑥𝑙𝑛𝑥 −
1
= 𝑥𝑙𝑛𝑥 − 𝑥
2
1
(𝑥)
1
𝑥
Simplify terms
Integrate and use a square
bracket with limits
Sub in the limits
= 2𝑙𝑛2 − 2 − (1𝑙𝑛1 − 1)
= 2𝑙𝑛2 − 1
Calculate and
leave in terms of
ln2
6G
𝑢
𝑑𝑣
= 𝑢𝑣 −
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑥
Integration
You can use integration by parts to
integrate some expressions
You may need the following Integrals,
which you are given in the formula
booklet…
𝑡𝑎𝑛𝑥 𝑑𝑥 = ln 𝑠𝑒𝑐𝑥 + 𝐶
𝑐𝑜𝑡𝑥 𝑑𝑥 = ln 𝑠𝑖𝑛𝑥 + 𝐶
𝑠𝑒𝑐𝑥 𝑑𝑥 = ln 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 + 𝐶
𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥 = −ln 𝑐𝑜𝑠𝑒𝑐𝑥 + 𝑐𝑜𝑡𝑥 + 𝐶
6G
Integration
You can use numerical integration
𝑏
𝑎
In C2 you saw how to estimate the
area under a curve by using the
trapezium rule
𝑦 𝑑𝑥 ≈
ℎ=
1
ℎ 𝑦𝑜 + 2 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛
2
𝑏−𝑎
𝑛
n is the number of strips the
area is split into!
𝑦0 = The first value for y
𝑦𝑛 = The last value for y
This method can take time and is
only an approximation, but it allows
you to find areas of functions that
otherwise can be extremely hard
to Integrate
The central bracket will contain all the values
of y in-between. The more vales there are,
the more accurate the approximation is!
y
y0 y1
 This is because using the
trapezium rule actually avoids
Integration altogether!
a
y2
y3 y
4 y
5
h h h h h
b
x
6H
Integration
𝜋
3
You can use numerical
integration
0
h is the height of each strip
 In the table it is given by the gaps between the x values used
 The y values correspond to y0, y1 etc…
Complete the table of values and
use it to find an estimate for:
𝜋
3
1
ℎ 𝑦𝑜 + 2 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛
2
𝑠𝑒𝑐𝑥 𝑑𝑥 ≈
𝑠𝑒𝑐𝑥 𝑑𝑥
≈
1
ℎ 𝑦𝑜 + 2 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛
2
≈
1 𝜋
2 12
≈
𝜋
24
0
h
x
y
0
1
y0
h
π
12
h
h
π
6
π
4
1.035 1.155 1.414
y1
y2
y3
π
3
2
Be very careful
here – it is easy to
make an error on
your calculator!
y4
Calculate the values in the table by
substituting the x-values into the equation
above…
Calculate and
round the answer!
1 + 2 1.035 + 1.155 + 1.414 + 2
10.208
≈ 1.34
The trapezium rule isn’t any different
to in C2 really, just slightly more
tricky functions to use!
6H
Integration
You can use numerical
integration
Use the trapezium rule with 4
strips to find an approximation
for:
ℎ=
𝑏−𝑎
𝑛
ℎ=
2−0
4
ℎ = 0.5
Sub in b, a and n
(number of strips)
We can then find the
height of each strip!
2
≈
1
ℎ 𝑦𝑜 + 2 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛
2
≈
1
0.5 0 + 2 0.24 + 0.84 + 1.50 + 1.82
2
𝑥𝑠𝑖𝑛𝑥 𝑑𝑥
0
x
y
0
0
y0
0.5
1
1.5
0.24 0.84 1.50
y1
y2
y3
2
1.82
y4
Be very
careful
here!
Calculate
and round
the answer!
≈
0.25 6.98
≈ 1.745
6H
Integration
You can use Integration to find
areas and volumes
y
y
𝑦= 𝑥
This would be the
solid formed
a
b
x
You already know how to find the
area under a curve by Integration
Imagine we rotated the area shaded around the
x-axis
 What would be the shape of the solid formed?
x
In this section you will learn how to find the volume
of any solid created in this way. It also involves
Integration!
6I
Integration
You can use Integration to find
areas and volumes
y
y
dx
dx
y
y
a
b
x
a
b
x
In the trapezium rule we thought of the area under a
curve being split into trapezia.
 To simplify this explanation, we will use rectangles
now instead
For the volume of revolution, each rectangle in the area
would become a ‘disc’, a cylinder
 The height of each rectangle is y at its x-coordinate
 The width of each is dx, the change in x values
 So the area beneath the curve is the sum of ydx
(base x height)
 The radius of each cylinder would be equal to y
 The height of each cylinder is dx, the change in x
 So the volume of each cylinder would be given by
πy2dx
 The EXACT value is calculated by integrating y with
respect to x (y dx)
 The EXACT value is calculated by integrating y2 with
respect to x, then multiplying by π.
6J
𝑐𝑜𝑠2𝐴 ≡ 1 − 2𝑠𝑖𝑛2 𝐴
Integration
You can use Integration to find
areas and volumes
The volume of revolution of a solid
rotated 2π radians around the x-axis
between x = a and x = b is given by:
𝑏
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
1
𝑠𝑖𝑛2 2𝐴 ≡ (1 − 𝑐𝑜𝑠4𝐴)
2
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝜋
2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
Sub in a, b and y
(𝑠𝑖𝑛2𝑥)2 𝑑𝑥
0
𝜋
2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
Square the bracket
𝑠𝑖𝑛2 2𝑥 𝑑𝑥
0
𝜋
21
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
Find the volume of the solid formed
when the region is rotated 2π radians
about the x-axis.
2𝑠𝑖𝑛2 2𝐴 ≡ 1 − 𝑐𝑜𝑠4𝐴
𝑏
𝑎
1) The region R is bounded by the
curve y = sin2x, the x-axis and the
vertical lines x = 0 and x = π/2.
𝑐𝑜𝑠4𝐴 ≡ 1 − 2𝑠𝑖𝑛2 2𝐴
0
1
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
2
2
(1 − 𝑐𝑜𝑠4𝑥) 𝑑𝑥
𝜋
2
The 1/2 can be put
outside the integral
(1 − 𝑐𝑜𝑠4𝑥) 𝑑𝑥
0
1
1
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋 𝑥 − 𝑠𝑖𝑛4𝑥
2
4
1
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
2
𝜋2
𝑉𝑜𝑙𝑢𝑚𝑒 =
4
Using the identity Cos2A = 1 –
2sin2A, replace sin22x with
something equivalent
𝜋
2
Integrate and use a square
bracket with the limits
0
𝜋 1
1
− 𝑠𝑖𝑛2𝜋 − 0 − 𝑠𝑖𝑛0
2 4
4
Sub in the two limits
And finally we have
the volume!
6J
𝑏
Integration
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
You can use Integration to find
areas and volumes
The volume of revolution can also be
calculated when x and y are given
parametrically. In this case you must
also include dx/dt in the integral.
𝑏
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑑𝑥
𝑑𝑡
𝑑𝑡
You will also need to change limits so
they are in terms of t rather than x!
6J
𝑏
Integration
𝑏
You can use Integration to find
areas and volumes
The volume of revolution can also be
calculated when x and y are given
parametrically. In this case you must
also include dx/dt in the integral.
𝑎
𝑏
𝑎
𝑏
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑦=
𝑥 = 𝑡 + 𝑡2
1
1+𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑡
1
1+𝑡
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
The curve C has parametric equations:
𝑥 = 𝑡(1 + 𝑡)
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
2
The region R is bounded by C, the xaxis and the lines x = 0 and x = 2.
1 + 2𝑡
𝑑𝑡
(1 + 𝑡)2
Find the volume of the solid formed
when R is rotated 2π radians about
the x-axis.
1 + 2𝑡 = 𝐴 + 𝐵(1 + 𝑡)
t = -1
t=0
−1 = 𝐴
1=𝐴+𝐵
2=𝐵
𝑏
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑑𝑥
𝑑𝑡
𝑑𝑡
1 + 2𝑡 𝑑𝑡
𝐵
1 + 2𝑡
𝐴
+
≡
1+𝑡
(1 + 𝑡)2
(1 + 𝑡)2
Where t ≥ 0.
𝑎
Replace y, and
calculate dx/dt
1 + 2𝑡
𝐴 + 𝐵(1 + 𝑡)
≡
(1 + 𝑡)2
(1 + 𝑡)2
𝑑𝑥
= 1 + 2𝑡
𝑑𝑡
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
Square the bracket
and combine them
We need to use
partial fractions here
Combine with a
common denominator
Sub in values to find
A and B
We know A = -1 from before
2−1
12
1 + 2𝑡
− +
≡
1(1++𝑡 𝑡)2 (1 +
𝑡)2𝑡
(1 + 𝑡)2
1+
6J
𝑏
Integration
𝑏
You can use Integration to find
areas and volumes
The volume of revolution can also be
calculated when x and y are given
parametrically. In this case you must
also include dx/dt in the integral.
𝑎
𝑏
𝑎
𝑏
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑥 = 𝑡(1 + 𝑡)
𝑥 = 𝑡 + 𝑡2
𝑑𝑥
= 1 + 2𝑡
𝑑𝑡
𝑎
𝑏
𝑎
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
2
1 + 2𝑡 𝑑𝑡
Square the bracket
and combine them
2
1
−
1 + 𝑡 (1 + 𝑡)2
𝑥 = 𝑡(1 + 𝑡)
Where t ≥ 0.
The region R is bounded by C, the xaxis and the lines x = 0 and x = 2.
Find the volume of the solid formed
when R is rotated 2π radians about
the x-axis.
x=0
0 = 𝑡(1 + 𝑡)
𝑡 = 0 𝑜𝑟 − 1
𝑡=0
𝑑𝑥
𝑑𝑡
𝑑𝑡
Replace y, and
calculate dx/dt
1 + 2𝑡
𝑑𝑡
(1 + 𝑡)2
𝑏
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑑𝑥
𝑑𝑡
𝑑𝑡
1
1+𝑡
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
The curve C has parametric equations:
1
𝑦=
1+𝑡
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
x=2
We need to use
partial fractions here
We also need to calculate
the limits for t rather than x
 Sub in the x limits and
solve for t
2 = 𝑡(1 + 𝑡)
0 = 𝑡2 + 𝑡 − 2
0 = (𝑡 + 2)(𝑡 − 1)
𝑡 = −2 𝑜𝑟 1
𝑡=1
6J
𝑏
Integration
𝑏
You can use Integration to find
areas and volumes
The volume of revolution can also be
calculated when x and y are given
parametrically. In this case you must
also include dx/dt in the integral.
𝑎
𝑏
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑏
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
The curve C has parametric equations:
1
𝑦=
1+𝑡
𝑥 = 𝑡(1 + 𝑡)
𝑥 = 𝑡 + 𝑡2
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑏
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
1
𝑑𝑥
= 1 + 2𝑡
𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
0
𝑑𝑥
𝑑𝑡
𝑑𝑡
1
1+𝑡
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑏
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
𝑑𝑥
𝑑𝑡
𝑑𝑡
Replace y, and
calculate dx/dt
2
1 + 2𝑡 𝑑𝑡
1 + 2𝑡
𝑑𝑡
(1 + 𝑡)2
2
1
−
1 + 𝑡 (1 + 𝑡)2
2
1
−
1 + 𝑡 (1 + 𝑡)2
Square the bracket
and combine them
We need to use
partial fractions here
Use t-limits
Where t ≥ 0.
The region R is bounded by C, the xaxis and the lines x = 0 and x = 2.
Find the volume of the solid formed
when R is rotated 2π radians about
the x-axis.
6J
𝑏
Integration
You can use Integration to find
areas and volumes
The volume of revolution can also be
calculated when x and y are given
parametrically. In this case you must
also include dx/dt in the integral.
1
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
0
=𝜋
2𝑙𝑛2 +
The curve C has parametric equations:
1
𝑦=
1+𝑡
𝑥 = 𝑡(1 + 𝑡)
𝑥 = 𝑡 + 𝑡2
𝑎
𝑏
𝑦2
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
𝑎
2
1
−
1 + 𝑡 (1 + 𝑡)2
1
= 𝜋 2 ln 1 + 𝑡 +
1+𝑡
𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋
1
0
1
− (2𝑙𝑛1 + 1)
2
𝑑𝑥
𝑑𝑡
𝑑𝑡
Integrate and write as a
square bracket
Sub in limits
separately
Simplify/Calculate
= 𝜋 2𝑙𝑛2 −
1
2
𝑑𝑥
= 1 + 2𝑡
𝑑𝑡
Where t ≥ 0.
The region R is bounded by C, the xaxis and the lines x = 0 and x = 2.
Find the volume of the solid formed
when R is rotated 2π radians about
the x-axis.
6J
Integration
You can use Integration to solve
differential equations
You can solve differential equations by a
process known as the ‘separation of
variables.’
When:
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
you can write
1
𝑑𝑦 =
𝑔(𝑦)
𝑓 𝑥 𝑑𝑥
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
Multiply both sides by dx
𝑑𝑦 = 𝑓 𝑥 𝑔 𝑦 𝑑𝑥
1
𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
𝑔(𝑦)
1
𝑑𝑦 =
𝑔(𝑦)
Divide by g(y)
Take Integrals of both sides
𝑓 𝑥 𝑑𝑥
Effectively you are just putting the x terms on
one side with dx, and the y terms on the other
side, with dy
6J
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
1
𝑑𝑦 =
𝑔(𝑦)
𝑓 𝑥 𝑑𝑥
Integration
You can use Integration to solve
differential equations
You can solve differential equations by a
process known as the ‘separation of
variables.’
Find a general solution to the following
differential equation:
1 + 𝑥2
𝑑𝑦
= 𝑥𝑡𝑎𝑛𝑦
𝑑𝑥
Finding the general solution means
finding a formula for y in terms of x and
C (the unknown part).
1 + 𝑥2
𝑑𝑦
= 𝑥𝑡𝑎𝑛𝑦
𝑑𝑥
𝑑𝑦
𝑥𝑡𝑎𝑛𝑦
=
𝑑𝑥 (1 + 𝑥 2 )
Divide by (1 + x2)
Divide by tany
1 𝑑𝑦
𝑥
=
𝑡𝑎𝑛𝑦 𝑑𝑥 (1 + 𝑥 2 )
1
𝑥
𝑑𝑦 =
𝑑𝑥
𝑡𝑎𝑛𝑦
(1 + 𝑥 2 )
𝑐𝑜𝑡𝑦 𝑑𝑦 =
𝑐𝑜𝑡𝑦 𝑑𝑦 =
𝑥
𝑑𝑥
(1 + 𝑥 2 )
𝑥
𝑑𝑥
(1 + 𝑥 2 )
Multiply by dx
Replace 1/tany with coty
As the x and y terms are
separate, we can now
integrate
 Effectively working back to the
equation that generated the curve
originally (Integrating!)
6J
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
1
𝑑𝑦 =
𝑔(𝑦)
𝑓 𝑥 𝑑𝑥
Integration
𝑥
𝑑𝑥
(1 + 𝑥 2 )
You can use Integration to solve
differential equations
𝑐𝑜𝑡𝑦 𝑑𝑦 =
You can solve differential equations by a
process known as the ‘separation of
variables.’
ln 𝑠𝑖𝑛𝑦 =
1
ln 1 + 𝑥 2 + 𝐶
2
ln 𝑠𝑖𝑛𝑦 =
1
ln 1 + 𝑥 2 + 𝑙𝑛𝑘
2
Find a general solution to the following
differential equation:
1 + 𝑥2
𝑑𝑦
= 𝑥𝑡𝑎𝑛𝑦
𝑑𝑥
Finding the general solution means
finding a formula for y in terms of x and
C (the unknown part).
 Effectively working back to the
equation that generated the curve
originally (Integrating!)
Integrate each side
using whatever methods
you have (including the
formula booklet!)
As C is just a number, it
can be written as lnk,
the natural logarithm of
another number
Use the power law
1
ln 𝑠𝑖𝑛𝑦 = ln (1 + 𝑥 2 )2 + 𝑙𝑛𝑘
Use the
multiplication law
ln 𝑠𝑖𝑛𝑦 = ln 𝑘 1 + 𝑥 2
𝑠𝑖𝑛𝑦 = 𝑘 1 + 𝑥 2
As the whole of both sides
is written as natural
logarithms, we can remove
them
You can only remove the natural
logarithm one the whole of each side is
combined in this way!
6J
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
1
𝑑𝑦 =
𝑔(𝑦)
𝑓 𝑥 𝑑𝑥
Integration
You can use Integration to solve
differential equations
You can solve differential equations by a
process known as the ‘separation of
variables.’
Find the particular solution of the
differential equation:
𝑑𝑦
−3(𝑦 − 2)
=
𝑑𝑥 (2𝑥 + 1)(𝑥 + 2)
given that x = 1 when y = 4
Finding the particular solution means
you are also able to find the unknown
value C (or lnk or whatever it is called!)
You start by finding the general solution
as before…
𝑑𝑦
−3(𝑦 − 2)
=
𝑑𝑥 (2𝑥 + 1)(𝑥 + 2)
Divide by (y – 2)
Multiply by dx
1
−3
𝑑𝑦 =
𝑑𝑥
𝑦−2
(2𝑥 + 1)(𝑥 + 2)
1
𝑑𝑦 =
𝑦−2
−3
𝑑𝑥
(2𝑥 + 1)(𝑥 + 2)
1
𝑑𝑦 =
𝑦−2
1
2
−
𝑑𝑥
𝑥 + 2 2𝑥 + 1
We need to Integrate
each side
Separate the right hand
side into partial fractions
Now integrate and
include lnk
𝑙𝑛 𝑦 − 2 = 𝑙𝑛 𝑥 + 2 − 𝑙𝑛 2𝑥 + 1 + 𝑙𝑛𝑘
𝑙𝑛 𝑦 − 2 = 𝑙𝑛
𝑙𝑛 𝑦 − 2 = 𝑙𝑛
𝑦−2=
𝑥+2
+ 𝑙𝑛𝑘
2𝑥 + 1
𝑘(𝑥 + 2)
2𝑥 + 1
𝑘(𝑥 + 2)
2𝑥 + 1
Combine 2 terms
using the division law
Include the lnk using
the multiplication law
Finally remove the logarithms
(you could also move the -2
across by adding 2)
6J
𝑑𝑦
= 𝑓 𝑥 𝑔(𝑦)
𝑑𝑥
1
𝑑𝑦 =
𝑔(𝑦)
𝑓 𝑥 𝑑𝑥
Integration
You can use Integration to solve
differential equations
You can solve differential equations by a
process known as the ‘separation of
variables.’
Find the particular solution of the
differential equation:
𝑦−2=
𝑦=
𝑘(𝑥 + 2)
2𝑥 + 1
𝑘(𝑥 + 2)
+2
2𝑥 + 1
4=
𝑘(1 + 2)
+2
2(1) + 1
4=
3𝑘
+2
3
𝑑𝑦
−3(𝑦 − 2)
=
𝑑𝑥 (2𝑥 + 1)(𝑥 + 2)
3𝑘
2=
3
given that x = 1 when y = 4
6 = 3𝑘
Finding the particular solution means
you are also able to find the unknown
value C (or lnk or whatever it is called!)
2=𝑘
You start by finding the general solution
as before…
Rearrange to get y = f(x) (this
isn’t essential but can help!)
𝑘(𝑥 + 2)
𝑦=
+2
2𝑥 + 1
General Solution
Sub in y = 4 and x = 1
from the question
Simplify the
fraction parts
Subtract 2
Multiply by 3
Divide by 3
k=2
𝑦=
2(𝑥 + 2)
+2
2𝑥 + 1
Particular Solution for
y = 4 when x = 1
6J
Integration
Sometimes the differential equation
will arise out of context and the
solution might need interpretation
The rate of increase of a population of
micro-organisms at time t is given by:
𝑑𝑃
= 𝑘𝑃
𝑑𝑡
Given that at t = 0 the population was
size 8, and at t = 1 the population was
56, find the size of the population at
time t = 2.
𝑑𝑃
= 𝑘𝑃
𝑑𝑡
1
𝑑𝑃 = 𝑘 𝑑𝑡
𝑃
We need to integrate both sides
1
𝑑𝑃 =
𝑃
𝑘 𝑑𝑡
𝑙𝑛 𝑃 = 𝑘𝑡 + 𝐶
𝑙𝑛 8 = 𝐶
𝑙𝑛 𝑃 = 𝑘𝑡 + 𝑙𝑛 |8|
You are looking to find the original
formula linking P and t, so you need to
integrate.
Once you have this formula you can
then use it…
Separate the variables (remember that k
just represents an unknown number, not
a variable like x or y…)
𝑙𝑛 56 = 𝑘 + 𝑙𝑛 |8|
𝑙𝑛 7 = 𝑘
𝑙𝑛 𝑃 = 𝑘𝑡 + 𝐶
As k is just a number,
the integral is kt
Sub in t = 0 and P = 8, this gives us C
We can rewrite the equation,
replacing C with ln8
Sub in the second pair of
values
Rearrange to find k
𝑙𝑛 𝑃 = 𝑡𝑙𝑛7 + 𝑙𝑛8
We now have an equation linking P and t
6K
Integration
Sometimes the differential equation
will arise out of context and the
solution might need interpretation
The rate of increase of a population of
micro-organisms at time t is given by:
𝑑𝑃
= 𝑘𝑃
𝑑𝑡
Given that at t = 0 the population was
size 8, and at t = 1 the population was
56, find the size of the population at
time t = 2.
𝑙𝑛 𝑃 = 𝑡𝑙𝑛7 + 𝑙𝑛8
𝑙𝑛 𝑃 = 2𝑙𝑛7 + 𝑙𝑛8
𝑙𝑛 𝑃 = 𝑙𝑛49 + 𝑙𝑛8
𝑙𝑛 𝑃 = 𝑙𝑛392
𝑃 = 392
Sub in t = 2
Use the power law
Use the
multiplication law
Remove the
logarithms
You are looking to find the original
formula linking P and t, so you need to
integrate.
Once you have this formula you can
then use it…
6K
Summary
• We have covered how to Integrate
many more functions
• We have seen how Integration can be
used to calculate volumes
• We have also seen how it links in with
differential equations