C4 Chapter 6: Integration
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 25th August 2014
Overview
You haven’t probably seen integration since C1! You’ll be able to integrate a significantly
greater variety of expressions, and be able to solve differential equations (which we
encountered earlier in C4).
Integration by standard result
(There’s certain expressions you’re expected to
know straight off.)
2
sec 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
Integration by substitution
(We make a substitution to hopefully
make the expression easier to integrate)
Integration by ‘reverse chain’ rule
(We imagine what would have differentiated
to get the expression.)
1 4
sin 𝑥 cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
4
3
Integration by parts
(Allows us to integrate a product, just as the
product rule allowed us to differentiate one)
𝑥 cos 𝑥 𝑑𝑥
𝑥 2𝑥 + 5 𝑑𝑥
𝑢−5
Let 𝑢 = 2𝑥 + 5
→ 𝑥=
2
𝑑𝑢
1
= 2 → 𝑑𝑥 = 𝑢
𝑑𝑥
2
𝑢 − 51
𝑥 2𝑥 + 5 𝑑𝑥 =
𝑑𝑢 = ⋯
2 2
𝑢=𝑥
𝑑𝑢
=1
𝑑𝑥
𝑑𝑣
= cos 𝑥
𝑑𝑥
𝑣 = sin 𝑥
𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 −
= 𝑥 sin 𝑥 + cos 𝑥
sin 𝑥 𝑑𝑥
Overview
Volumes of revolution
(Find the volume of a solid formed by spinning
a curve around the 𝑥 axis)
𝑏
𝑉=𝜋
Finding an area using parametric
equations
(Find the volume of a solid formed by spinning
a curve around the 𝑥 axis)
𝑦 2 𝑑𝑥
𝑏
𝑦 𝑑𝑥 =
𝑎
𝑎
Solving Differential Equations
(Solving here means to find one variable in terms of
another without derivatives present)
𝑑𝑉
= −𝑘𝑉
𝑑𝑡
1
∫ 𝑑𝑉 = −𝑘 𝑑𝑡
𝑉
ln 𝑉 = −𝑘𝑡 + 𝐶
𝑉 = 𝑒 −𝑘𝑡+𝐶 = 𝐴𝑒 −𝑘𝑡
𝑡2
𝑡2
𝑑𝑥
𝑦
𝑑𝑡
𝑑𝑡
SKILL #1: Integrating Standard Functions
There’s certain results you should be able to integrate straight off, by just thinking about
the opposite of differentiation.
𝒚
∫ 𝒚 𝒅𝒙
𝑥𝑛
1
𝑥?𝑛+1 + 𝐶
𝑛+1
𝑒𝑥?
+𝐶
𝑒𝑥
1
𝑥
cos 𝑥
sin 𝑥
sec 2 𝑥
𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥
𝑐𝑜𝑠𝑒𝑐 2 𝑥
sec 𝑥 tan 𝑥
ln |𝑥| + 𝐶
?
sin 𝑥?+ 𝐶
− cos?𝑥 + 𝐶
tan 𝑥?+ 𝐶
−𝑐𝑜𝑠𝑒𝑐
?𝑥 + 𝐶
− cot?𝑥 + 𝐶
sec 𝑥?+ 𝐶
The |𝑥| has to do with problems
when 𝑥 is negative (as ln 𝑥 is not
defined in the real domain)
Remember my memorisation
trick of picturing sin above cos
from C3, so that ‘going down’ is
differentiating and ‘going up’ is
integrating, and we change the
sign if the wrong way round.
It’s vital you remember this one.
Have a good stare at this slide before turning your
paper over – let’s see how many you remember…
Quickfire Questions (without cheating!)
?
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
?𝐶
sin 𝑥 𝑑𝑥 = − cos 𝑥 +
?𝐶
𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = − cot 𝑥 +
− cos 𝑥 𝑑𝑥 = − sin 𝑥?+ 𝐶
Quickfire Questions (without cheating!)
sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥 𝑑𝑥 = 𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝐶
1
𝑑𝑥 = ln |𝑥| + 𝐶
𝑥
− sin 𝑥 𝑑𝑥 = cos 𝑥 + 𝐶
Quickfire Questions (without cheating!)
𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = − cot 𝑥 +? 𝐶
sin 𝑥 𝑑𝑥 = − cos 𝑥 +
?𝐶
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + ?𝐶
cos 𝑥 𝑑𝑥 = sin 𝑥 +?𝐶
Test Your Understanding
3
2 3
2 cos 𝑥 + − 𝑥 𝑑𝑥 = 2 sin 𝑥 + 3 ln |𝑥|
? − 𝑥2 + 𝐶
𝑥
3
cos 𝑥
𝑑𝑥 =
2
sin 𝑥
cosec 𝑥 cot 𝑥 𝑑𝑥? = −𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝐶
Hint: What ‘reciprocal’ trig
functions does this simplify to?
Exercise 6A
1
a
c
e
g
i
2
a
c
Integrate the following.
5 2
2
3 sec 𝑥 + + 2
𝑥 𝑥
2 sin 𝑥 − cos 𝑥 + 𝑥
2
5𝑒 𝑥 + 4 cos 𝑥 − 2
𝑥
1 1
1
+ 2+ 3
𝑥 𝑥
𝑥
2𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥 − sec 2 𝑥
𝟐
𝟑 𝐭𝐚𝐧 𝒙 + 𝟓 𝐥𝐧 𝒙 −
?𝒙 + 𝑪
− 𝟐 𝐜𝐨𝐬 𝒙 − 𝟐 𝐬𝐢𝐧?𝒙 + 𝒙𝟐 + 𝑪
𝟐
𝟓𝒆𝒙 + 𝟒 𝐬𝐢𝐧 𝒙 +? + 𝑪
𝒙
𝟏
𝟏
𝐥𝐧 𝒙 − − 𝟐?+ 𝑪
𝒙 𝟐𝒙
− 𝟐𝒄𝒐𝒔𝒆𝒄 𝒙 − 𝐭𝐚𝐧
? 𝒙+𝑪
Calculate the following.
1
1
𝟏
∫
+
𝑑𝑥
=
𝒕𝒂𝒏
𝒙
−
+𝑪 ?
cos 2 𝑥 𝑥 2
𝒙
1 + cos 𝑥 1 + 𝑥
𝟏
∫
+
𝑑𝑥
=
−
𝐜𝐨𝐭
𝒙
−
𝒄𝒐𝒔𝒆𝒄
𝒙
−
? 𝒙 + 𝐥𝐧 𝒙 + 𝑪
sin2 𝑥
𝑥2
e
𝑠𝑖𝑛𝑥 1 + sec 2 𝑥 𝑑𝑥 = − 𝐜𝐨𝐬 𝒙 + 𝐬𝐞𝐜 𝒙 +?𝑪
g
𝑐𝑜𝑠𝑒𝑐 2 𝑥 1 + tan2 𝑥 𝑑𝑥 = − 𝐜𝐨𝐭 𝒙 + 𝐭𝐚𝐧 𝒙 + ?
𝑪
i
sec 2 𝑥 1 + 𝑒 𝑥 cos 2 𝑥 𝑑𝑥 = 𝐭𝐚𝐧 𝒙 + 𝒆𝒙 + 𝑪
?
SKILL #2: Integrating by Reverse Chain Rule
Therefore:
𝑑
sin 3𝑥 + 1
𝑑𝑥
cos 3𝑥 + 1 𝑑𝑥 =
= 3 cos 3𝑥?+ 1
𝟏
𝐬𝐢 𝐧 𝟑𝒙?+ 𝟏 + 𝑪
𝟑
! For any expression where inner function is 𝑎𝑥 + 𝑏, integrate as before and ÷ 𝑎.
1
′
𝑓 𝑎𝑥 + 𝑏 𝑑𝑥 = 𝑓 𝑎𝑥 + 𝑏 + 𝐶
𝑎
Quickfire:
1 3𝑥
?
𝑒 +𝐶
3
1
1
𝑑𝑥 = ln 5𝑥 +?2 + 𝐶
5𝑥 + 2
5
2
2sec 2 3𝑥 − 2 𝑑𝑥 = tan 3𝑥?− 2 + 𝐶
3
𝑒 3𝑥 𝑑𝑥 =
1
3𝑥 +?4 4 + 𝐶
12
1
sin 1 − 5𝑥 𝑑𝑥 = cos 1 −
? 5𝑥 + 𝐶
5
1
1
−1 + 𝐶
𝑑𝑥
=
−
4𝑥
−
2
?
3 4𝑥 − 2 2
12
1
10𝑥 + 11 12 =
10𝑥 +
11 13 + 𝐶
?
130
3𝑥 + 4
3
𝑑𝑥 =
Check Your Understanding
𝟏 𝟑𝒙+𝟏
= 𝒆
+ ?𝑪
𝟑
1
𝟏
𝑑𝑥 = − 𝐥𝐧 𝟏 −?𝟐𝒙 + 𝑪
1 − 2𝑥
𝟐
𝟏
5
? 𝟔+𝑪
4 − 3𝑥 𝑑𝑥 = −
𝟒 − 𝟑𝒙
𝟏𝟖
𝟏
sec 3𝑥 tan(3𝑥) 𝑑𝑥 = 𝐬𝐞𝐜 𝟑𝒙 ?+ 𝑪
𝟑
𝑒 3𝑥+1 𝑑𝑥
Exercise 6B
Page 86
𝟏
sin 2𝑥 + 1 𝑑𝑥 = − 𝒄𝒐𝒔 𝟐𝒙 + 𝟏 + 𝑪
𝟐
1 a
?
3 a
?
c
4𝑒 𝑥+5 𝑑𝑥 = 𝟒𝒆𝒙+𝟓 + 𝑪
b
𝟏
𝑐𝑜𝑠𝑒𝑐 2 3𝑥 𝑑𝑥 = − 𝐜𝐨𝐭 𝟑𝒙 + 𝑪
𝟑
𝟏
sec 4𝑥 tan 4𝑥 𝑑𝑥 = 𝐬𝐞𝐜 𝟒𝒙 + 𝑪
𝟒
1
𝟏
3 sin 𝑥 + 1 𝑑𝑥 = −𝟔 𝐜𝐨𝐬 𝒙 + 𝟏
2
𝟐
𝟏
𝑐𝑜𝑠𝑒𝑐 2𝑥 cot 2𝑥 𝑑𝑥 = − 𝒄𝒐𝒔𝒆𝒄 𝟐𝒙 + 𝑪
𝟐
?
e
f
c
?
g
d
?
?
h
f
1
𝑒 2𝑥 − sin 2𝑥 − 1 𝑑𝑥
2
𝟏
𝟏
= 𝒆𝟐𝒙 + 𝐜𝐨𝐬 𝟐𝒙 − 𝟏 + 𝑪
𝟐
𝟒
𝟏
𝑒 𝑥 + 1 2 𝑑𝑥 = 𝒆𝟐𝒙 + 𝟐𝒆𝒙 + 𝒙 + 𝑪
𝟐
𝟏
𝟏
sec 2 2𝑥 1 + sin 2𝑥 𝑑𝑥 = 𝐭𝐚𝐧 𝟐𝒙 + 𝐬𝐞𝐜 𝟐𝒙
𝟐
𝟐
2 a
?
?
b
?
c
d
e
∫
1
2
3−2 cos 𝑥
1
sin2 2𝑥
𝑑𝑥 = −𝟔 𝐜𝐨𝐭
𝟏
𝟐
?
𝒙 + 𝟒𝒄𝒐𝒔𝒆𝒄
𝟏
𝟐
𝑒 3−𝑥 + sin 3 − 𝑥 + cos 3 − 𝑥 𝑑𝑥
?
= −𝒆𝟑−𝒙 + 𝐜𝐨𝐬 𝟑 − 𝒙 − 𝐬𝐢𝐧 𝟑 − 𝒙 + 𝑪
h
j
4 a
1
𝟏
𝑑𝑥 = 𝐥𝐧 𝟐𝒙 +
?𝟏 + 𝑪
2𝑥 + 1
𝟐
1
𝟏
𝑑𝑥
=
−
? +𝑪
2𝑥 + 1 2
𝟐 𝟐𝒙 + 𝟏
𝟏
2
2𝑥 + 1 𝑑𝑥 = 𝟐𝒙 +?
𝟏 𝟑+𝑪
𝟔
3
𝟑
𝑑𝑥 = 𝐥𝐧 𝟒𝒙 −
?𝟏 + 𝑪
4𝑥 − 1
𝟒
3
𝟑
𝑑𝑥 =
+𝑪
1 − 4𝑥 2
𝟒 𝟏 − 𝟒𝒙?
3
𝟑
𝑑𝑥
=
+𝑪
1 − 2𝑥 3
𝟒 𝟏 − 𝟐𝒙? 𝟐
5
𝟓
𝑑𝑥 = − 𝐥𝐧 𝟑 − 𝟐𝒙 + 𝑪
?
3 − 2𝑥
𝟐
3 sin 2𝑥 + 1 +
4
𝑑𝑥
2𝑥 + 1
𝟑
= − 𝐜𝐨𝐬 𝟐𝒙 + 𝟏 + 𝟐 𝐥𝐧 𝟐𝒙 + 𝟏 + 𝑪
𝟐
1
1
1
+
+
𝑑𝑥
sin2 2𝑥 1 + 2𝑥
1 + 2𝑥 2
𝟏
𝟏
𝟏
= − 𝐜𝐨𝐭 𝟐𝒙 + 𝐥𝐧 𝟏 + 𝟐𝒙 −
𝟐
𝟐
𝟐 𝟏 + 𝟐𝒙
?
𝒙
c
?
SKILL #2: Integrating using Trig Identities
Some expressions, such as sin2 𝑥 and sin 𝑥 cos 𝑥 can’t be integrated directly, but we can
use one of our trig identities to replace it with an expression we can easily integrate.
Q
Find ∫ sin2 𝑥 𝑑𝑥
Q
sin 6𝑥 ≡ 2 sin 3𝑥 cos 3𝑥
1
sin 3𝑥 cos 3𝑥 = sin 6𝑥
2?
Do you know a trig identity involving
sin2 𝑥?
cos 2𝑥 = ?
1 − 2 sin2 𝑥
1 1
sin2 𝑥 = − cos 2𝑥
2 2
1
1
sin2 𝑥 𝑑𝑥 = 𝑥 − sin 2𝑥
2
4
Q
Find ∫ cos 2 𝑥 𝑑𝑥
cos 2𝑥 =
2 cos2 𝑥
sin 3𝑥 cos 3𝑥 𝑑𝑥 = −
Q
1
cos 6𝑥 + 𝐶
12
Find ∫ sin 3𝑥 cos 2𝑥 𝑑𝑥
From formula booklet:
−1
1 1
+ cos 2𝑥
2 ?2
1
1
cos 2 𝑥 𝑑𝑥 = 𝑥 + sin 2𝑥
2
4
cos 2 𝑥 =
Find ∫ sin 3𝑥 cos 3𝑥 𝑑𝑥
𝐴+𝐵
𝐴−𝐵
cos
2
2
𝐴 − 𝐵 = 4𝑥
sin 𝐴 + sin 𝐵 = 2 sin
𝐴 + 𝐵 = 6𝑥,
𝐴 = 5𝑥, 𝐵 = 𝑥
?
1
sin 5𝑥 + sin 𝑥
2
1
1
sin 3𝑥 cos 2𝑥 𝑑𝑥 = − cos 5𝑥 − cos 𝑥 + 𝐶
10
2
sin 3𝑥 cos 2𝑥 =
Check Your Understanding
Q
Find ∫ tan2 𝑥 𝑑𝑥
Recall that 1 + tan2 𝑥 ≡ sec 2 𝑥
Hint: What identity do you know
involving tan2 𝑥?
tan2 𝑥 𝑑𝑥 =? sec 2 𝑥 − 1 𝑑𝑥
= tan 𝑥 − 𝑥 + 𝐶
Q
Find ∫ sec 𝑥 + tan 𝑥
sec 𝑥 + tan 𝑥
=
2
2
𝑑𝑥
𝑑𝑥
sec 2 𝑥 + 2 sec?𝑥 tan 𝑥 + tan2 𝑥 𝑑𝑥
= tan 𝑥 + 2 sec 𝑥 + tan 𝑥 − 𝑥 + 𝐶
= 2 tan 𝑥 − 2 sec 𝑥 − 𝑥 + 𝐶
Bro Tip: The quick way is this.
𝐴 is the sum of the two nums.
𝐵 is the difference.
Q
Find ∫ 4 cos 7𝑥 cos 3𝑥 𝑑𝑥
From formula booklet:
𝐴+𝐵
𝐴−𝐵
cos
2
2
𝐴 + 𝐵 = 14𝑥 𝐴 − 𝐵 = 6𝑥
𝐴 = 10𝑥,
𝐵 = 4𝑥
cos 𝐴 + cos 𝐵 = 2 cos
?
4 cos 7𝑥 cos 3𝑥 𝑑𝑥
=
2(cos 10𝑥 + cos 4𝑥) 𝑑𝑥
1
1
= sin 10𝑥 + cos 4𝑥
5
2
Exercise 6C
1
b
c
d
e
f
g
h
i
Page 88
?
cot 2 𝑥 𝑑𝑥 = − cot 𝑥 − 𝑥 + 𝐶
1
1
𝑥 + sin 2𝑥 + 𝐶
2
4
1
sin 2𝑥 cos 2𝑥 𝑑𝑥 = cos 4𝑥 + 𝐶
8
3
1
1 + sin 𝑥 2 𝑑𝑥 = 𝑥 − 2 cos 𝑥 − sin 2𝑥 + 𝐶
2
4
1
tan2 3𝑥 𝑑𝑥 = tan 3𝑥 − 𝑥 + 𝐶
3
?
cos2 𝑥 𝑑𝑥 =
?
?
1 + cos 𝑥
𝑑𝑥 = −2 cot 𝑥 − 𝑥 −?2𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝐶
sin 𝑥
cot 𝑥 − tan 𝑥 𝑑𝑥 = − cot 𝑥 − 4𝑥 + tan 𝑥 + 𝐶
?
c
2
?
e
?
?
?
?
?
cos 𝑥 − sec 𝑥 2 𝑑𝑥
3
1
= − 𝑥 + sin 2𝑥 + tan 𝑥 + 𝐶
2
4
cot 𝑥 − 𝑐𝑜𝑠𝑒𝑐 𝑥 𝑑𝑥 = 2 cot 𝑥 − 𝑥 + 2 cosec 𝑥 + 𝐶
1
sin 𝑥 + cos 𝑥 𝑑𝑥 = 𝑥 − cos 2𝑥 + 𝐶
2
1
1
sin2 𝑥 cos2 𝑥 𝑑𝑥 = 𝑥 −
sin 4𝑥 + 𝐶
8
32
1
𝑑𝑥 = −2 cot 2𝑥 + 𝐶
sin2 𝑥 cos2 𝑥
2
i
?
2
2
g
?
2
1 − sin 𝑥
𝑑𝑥 = tan 𝑥 − sec 𝑥 + 𝐶
cos2 𝑥
cos 2𝑥
𝑑𝑥 = 2𝑥 − tan 𝑥 + 𝐶
cos2 2𝑥
2 a
3 a
cos 2𝑥 cos 𝑥 𝑑𝑥 =
c
2 sin 3𝑥 cos 5𝑥 𝑑𝑥
1
1
sin 3𝑥 + sin 𝑥
6
2
?
1
1
= − cos 8𝑥 + cos 2𝑥
8
2
?
4 cos 3𝑥 cos 7𝑥 𝑑𝑥
e
=
g
1
1
sin 10𝑥 + sin 4𝑥
5
2
?
1
2 cos 4𝑥 sin 4𝑥 𝑑𝑥 = − cos 8𝑥 + 𝐶
8
?
SKILL #3: Using Partial Fractions
We saw earlier that we can split some expressions into partial fractions. This allows us
to differentiate some expressions with more complicated denominators.
Find
2
∫ 𝑥 2 −1
𝑑𝑥
2
2
𝐴
𝐵
=
=
+
𝑥2 − 1
𝑥+1 𝑥−1
𝑥+1 𝑥−1
2=𝐴 𝑥−1 +𝐵 𝑥+1
2 = 𝐴𝑥 − 𝐴 + 𝐵𝑥 + 𝐵
𝐴+𝐵 =0
𝐵−𝐴=2
𝐵 = 1 𝐴 = −1
?
2
1
1
∫ 2
𝑑𝑥 = −
+
𝑑𝑥
𝑥 −1
𝑥+1 𝑥−1
= − 𝐥𝐧 𝒙 + 𝟏 + 𝐥𝐧 𝒙 − 𝟏 + 𝑪
𝒙−𝟏
= 𝒍𝒏
+𝑪
𝒙+𝟏
Test Your Understanding
Find ∫
𝑥−5
𝑥+1 𝑥−2
𝑑𝑥
𝑥−5
𝐴
𝐵
≡
+
𝑥+1 𝑥−2
𝑥+1 𝑥−2
𝐴 = 2 𝑎𝑛𝑑 𝐵 = −1
𝑥−5
2
1
𝑑𝑥 = ∫
−
𝑑𝑥
𝑥+1 𝑥−2
𝑥+1 𝑥−2
= 2 ln 𝑥 + 1 − ln 𝑥 − 2 + 𝐶
𝒙+𝟏 𝟐
= 𝐥𝐧
+𝑪
𝒙−𝟐
?
Find ∫
8𝑥 2 −19𝑥+1
2𝑥+1 𝑥−2 2
8𝑥 2 − 19𝑥 + 1
2𝑥 + 1 𝑥 − 2 2
𝐴
𝐵
=
+
2𝑥 + 1
𝑥−2
𝑑𝑥
+
2
𝐶
𝑥−2
𝐴 = 2, 𝐵 =?−1, 𝐶 = 3
8𝑥 2 − 19𝑥 + 1
𝑑𝑥
2𝑥 + 1 𝑥 − 2 2
= 𝐥𝐧 𝟐𝒙 + 𝟏 𝒙 − 𝟐
𝟑
+
𝟏
+𝑪
𝒙−𝟐
Exercise 6D
Page 92
Use partial fractions to integrate the following.
3𝑥 + 5
𝑑𝑥 = 𝐥𝐧 𝒙 + 𝟏 𝟐 𝒙?+ 𝟐 + 𝑪
1 a
𝑥+1 𝑥+2
2𝑥 − 6
𝒙+𝟑 𝟑
c
?𝑪
𝑑𝑥 = 𝐥𝐧
+
𝑥+3 𝑥−1
𝒙−𝟏
4
𝟐𝒙 + 𝟏
e
𝑑𝑥 = 𝐥𝐧
+𝑪
2𝑥 + 1 1 − 2𝑥
𝟏 − 𝟐𝒙 ?
g
i
2 a
c
e
𝟏
𝟑
3 − 5𝑥
𝟐 − 𝟑𝒙
?+ 𝑪
𝑑𝑥 = 𝐥𝐧
𝟐
1 − 𝑥 2 − 3𝑥
𝟏−𝒙
5 + 3𝑥
𝒙+𝟏
𝟐
𝑑𝑥
=
𝐥𝐧
−
+𝑪
𝑥+2 𝑥+1 2
𝒙+𝟐
𝒙?
+𝟏
2 𝑥 2 + 3𝑥 − 1
𝑑𝑥 = 𝒙 + 𝐥𝐧 𝒙 + 𝟏 𝟐? 𝟐𝒙 − 𝟏 + 𝑪
𝑥 + 1 2𝑥 − 1
𝑥2
𝒙−𝟐
𝑑𝑥 = 𝒙 + 𝐥𝐧
+𝑪
𝑥2 − 4
𝒙+𝟐 ?
6 + 3𝑥 − 𝑥 2
𝟑
𝑑𝑥 = − − 𝐥𝐧 𝒙 + 𝟐?+ 𝑪
𝑥 3 + 2𝑥 2
𝒙
SKILL #4: ‘Consider and Scale’
There’s certain more complicated expressions which look like the result of having
applied the chain rule. The process is then simply:
1. Consider some expression that will differentiate to something similar to it.
2. Differentiate, and adjust for any scale difference.
𝑥 𝑥2 + 5
3
𝑑𝑥
The first 𝑥 looks like it arose
from differentiating the 𝑥 2
inside the brackets.
𝟒
Consider 𝒚 = 𝒙𝟐 + 𝟓
𝒅𝒚
𝟑
= 𝟒 𝒙𝟐 + 𝟓 × 𝟐𝒙
𝒅𝒙
𝟑
= 𝟖𝒙 𝒙𝟐 + 𝟓
?
𝟑
Thus ∫ 𝒙 𝒙𝟐 + 𝟓 𝒅𝒙
𝟏 𝟐
𝟒
=
𝒙 +𝟓 +𝑪
𝟖
2𝑥
𝑑𝑥
2
𝑥 +1
cos 𝑥 sin2 𝑥 𝑑𝑥
The cos 𝑥 probably arose from
differentiating the 𝑠𝑖𝑛.
The 2𝑥 probably arose from
differentiating the 𝑥 2 .
Consider 𝒚 = 𝐬𝐢𝐧𝟑 𝒙
𝒅𝒚
= 𝟑 𝐬𝐢𝐧𝟐 𝒙 𝐜𝐨𝐬 𝒙
𝒅𝒙
Consider 𝒚 = 𝒍𝒏|𝒙𝟐 + 𝟏|
𝒅𝒚
𝟐𝒙
= 𝟐
𝒅𝒙 𝒙 + 𝟏
Thus ∫ 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏𝟐 𝒙 𝒅𝒙
𝟏
= 𝐬𝐢𝐧𝟑 𝒙 + 𝑪
𝟑
Thus ∫
?
𝟐𝒙
𝒙𝟐 +𝟏
𝒅𝒙
?
= 𝒍𝒏 𝒙𝟐 + 𝟏 + 𝑪
SKILL #4: ‘Consider and Scale’
! Integration by Inspection: Use common sense to consider some expression that
would differentiate to the expression given. Then scale appropriately.
Common patterns:
In words: “If the bottom of a
′
fraction differentiates to give the
𝑓 𝑥
top (forgetting scaling), try ln of
𝑘
𝑑𝑥 → 𝑇𝑟𝑦 ln 𝑓 𝑥
the bottom”.
𝑓 𝑥
∫ 𝑘𝑓 ′ 𝑥 𝑓 𝑥 𝑛 → 𝑇𝑟𝑦 𝑓 𝑥 𝑛+1
𝑥2
𝑑𝑥
3
𝑥 +1
Consider 𝑦 = ln |𝑥 3 + 1|
𝑑𝑦
3𝑥 2
=
𝑑𝑥 𝑥 3 + 1
Therefore: ?
𝑥2
𝑑𝑥
𝑥3 + 1
1
= 𝑙𝑛 𝑥 3 + 1 + 𝐶
3
𝑥
2 +1
𝑥
𝑒
𝑑𝑥
2
Consider 𝑦 = 𝑒 𝑥 +1
𝑑𝑦
2
= 2𝑥 𝑒 𝑥 +1
𝑑𝑥
Therefore: ?
1 2
2
𝑥 𝑒 𝑥 +1 𝑑𝑥 = 𝑒 𝑥 +1 + 𝐶
2
Quickfire
In your head!
4𝑥 3
𝑑𝑥 = 𝐥𝐧 𝒙𝟒 − ?
𝟏 +𝑪
4
𝑥 −1
cos 𝑥
𝑑𝑥 = 𝒍𝒏 𝐬𝐢𝐧 𝒙 +? 𝟐 + 𝑪
sin 𝑥 + 2
cos 𝑥 𝑒 sin 𝑥 𝑑𝑥 = 𝒆𝐬𝐢𝐧 𝒙 + 𝑪 ?
cos 𝑥 sin 𝑥 − 5
𝑥2
𝑥3
+5
7
7
𝟏
𝑑𝑥 = 𝐬𝐢𝐧 𝒙 −?𝟓
𝟖
𝟏 𝟑
=
𝒙 +?
𝟓
𝟐𝟒
𝟖
𝟖
+𝑪
Not in your head…
𝑥
𝑥2 + 5
3 𝑑𝑥 = −
𝟏 𝟐
𝒙 +
?𝟓
𝟒
−𝟐
Bro Tip: If there’s as power around the
whole denominator, DON’T use 𝑙𝑛:
reexpress the expression as a product.
e.g. 𝑥 𝑥 2 + 5
−3
Test Your Understanding
sin 𝑥 cos 𝑥 + 1
5
𝑑𝑥
Consider 𝑦 = cos 𝑥 + 1 6
𝑑𝑦
= −6 sin 𝑥 cos 𝑥 + 1
𝑑𝑥
Therefore:
?
sin 𝑥 cos 𝑥 + 1
𝟏
= − 𝐜𝐨𝐬 𝒙 + 𝟏
𝟔
𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
2 + cot 𝑥 3
5
𝟔
𝑑𝑥
+𝑪
5
Consider 𝑦 = 2 + cot 𝑥 −2
𝑑𝑦
= −2 2 + cot 𝑥 −3 × −𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
= 2𝑐𝑜𝑠𝑒𝑐 2 𝑥 2 +?cot 𝑥 −3
Thus ∫
𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑑𝑥
2+cot 𝑥 3
sec 2 2𝑥
𝑑𝑥
tan 2𝑥 + 1
𝟏
= 𝟐 𝟐 + 𝐜𝐨𝐭 𝒙
−𝟐
+𝑪
5 tan 𝑥 sec 2 𝑥 𝑑𝑥
𝑑
1
= ln | tan 2𝑥 + 1| + 𝐶
?
2
Note that 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥, so the power
of sec 𝑥 stays the same.
?
Try 𝑦 = sec 2 𝑥
𝑑𝑦
Then 𝑑𝑥 = 2 sec 𝑥 × sec 𝑥 tan 𝑥 = 2 sec 2 𝑥 tan 𝑥
5
Thus ∫ 5 tan 𝑥 sec 2 𝑥 𝑑𝑥 = 2 sec 2 𝑥 + 𝐶
Exercise 6E
𝑥
𝟏
𝑑𝑥 = 𝐥𝐧 𝒙𝟐 + 𝟒?+ 𝑪
1
2
𝑥 +4
𝟐
𝑒 2𝑥
𝟏
𝑑𝑥 = 𝐥𝐧 𝒆𝟐𝒙 + 𝟏
? +𝑪
2𝑥
𝑒 +1
𝟐
−𝟐
𝑥
𝟏 𝟐
𝑑𝑥 = − 𝒙 + 𝟒? + 𝑪
𝑥2 + 4 3
𝟒
𝑒 2𝑥
𝟏 𝟐𝒙
−𝟐
𝑑𝑥
=
−
𝒆
+
𝟏
? +𝑪
𝑒 2𝑥 + 1 3
𝟒
cos 2𝑥
𝟏
𝑑𝑥 = 𝐥𝐧 𝟑 + 𝐬𝐢𝐧
?𝟐𝒙 + 𝑪
3 + sin 2𝑥
𝟐
𝟏 𝟐
2
𝑥 𝑒 𝑥 𝑑𝑥 = 𝒆𝒙 + 𝑪 ?
𝟐
𝟏
cos 2𝑥 1 + sin 2𝑥 4 𝑑𝑥 =
𝟏 + 𝐬𝐢𝐧
? 𝟐𝒙
𝟏𝟎
𝟏
2
2
∫ sec 𝑥 tan 𝑥 𝑑𝑥 = 𝐭𝐚𝐧𝟑 𝒙 +?𝑪
𝟑
Page 94
2
𝑥 + 1 𝑥 2 + 2𝑥 + 3 4 𝑑𝑥 =
𝟏 𝟐
=
𝒙 + 𝟐𝒙
? +𝟑
𝟏𝟎
𝟓
𝑐𝑜𝑠𝑒𝑐 2 2𝑥 cot 2𝑥 𝑑𝑥
𝟏
= − 𝐜𝐨𝐭 𝟐 𝟐𝒙?+ 𝑪
𝟒
sin5 3𝑥 cos 3𝑥 𝑑𝑥
𝟏
= 𝟏𝟖 𝐬𝐢𝐧𝟔 𝟑𝒙 ?
+𝑪
cos 𝑥 𝑒 sin 𝑥 𝑑𝑥 = 𝒆𝐬𝐢𝐧
𝟓
+𝑪
𝑥
𝑥2
+1
3
2 𝑑𝑥
sin 𝑥 cos 𝑥
𝑑𝑥
cos 2𝑥 + 3
𝟏
𝒙
+
?𝑪
𝟏 𝟐
= 𝒙 +?𝟏
𝟓
𝟓
𝟐
= − 𝟒 𝐥𝐧 𝐜𝐨𝐬 𝟐𝒙
? +𝟑 +𝑪
+𝑪
SKILL #5: Integration by Substitution
For some integrations involving a complicated expression, we can make a substitution
to turn it into an equivalent integration that is simpler. We wouldn’t be able to use
‘reverse chain rule’ on the following:
Q
Use the substitution 𝑢 = 2𝑥 + 5 to find ∫ 𝑥 2𝑥 + 5 𝑑𝑥
The aim is to completely remove any reference to 𝑥, and replace it with 𝑢. We’ll
have to work out 𝑥 and 𝑑𝑥 so that we can replace them.
STEP 1: Using
substitution, work out
𝑥 and 𝑑𝑥 (or variant)
STEP 2: Substitute
these into expression.
STEP 3: Integrate
simplified expression.
STEP 4: Write answer
in terms of 𝑥.
𝑑𝑢
=2 →
𝑑𝑥
𝑢−5
𝑥=
2
1
𝑑𝑥 = 𝑑𝑢
2
𝑥 2𝑥 + 5 𝑑𝑥 =
1
=
4
3
𝑢2
−
𝑢−5
1
1
𝑢 𝑑𝑢 =
2
2
4
?
1
5𝑢2
=
2𝑥 + 5
10
−
5 2𝑥 + 5
6
𝑢 𝑢 − 5 𝑑𝑢
Bro Tip: If you have a constant
factor, factor it out of the integral.
1 2 5 10 3
=
𝑢2 −
𝑢2 + 𝐶
4 5
3
5
2
?
Bro Tip: Be careful about
ensuring you reciprocate
when rearranging.
?
3
2
+?
𝐶
How can we tell what substitution to use?
In Edexcel you will usually be given the substitution!
However in some other exam boards, and in STEP, you often aren’t.
There’s no hard and fast rule, but it’s often helpful to replace to replace
expressions inside roots, powers or the denominator of a fraction.
Sensible substitution:
cos 𝑥 1 + sin 𝑥 𝑑𝑥
2
𝒖 = 𝟏 + 𝐬𝐢𝐧 𝒙?
6𝑥 𝑒 𝑥 𝑑𝑥
𝒖 = 𝒙𝟐
𝑥𝑒 𝑥
𝑑𝑥
1+𝑥
𝒖=𝟏+𝒙
?
1−𝑥
𝑒 1+𝑥
𝟏−𝒙
𝒖=
𝟏+𝒙
?
𝑑𝑥
But this can be integrated
?
by inspection.
Another Example
Q
Use the substitution 𝑢 = sin 𝑥 + 1 to find ∫ cos 𝑥 sin 𝑥 1 + sin 𝑥
STEP 1: Using
substitution, work out
𝑥 and 𝑑𝑥 (or variant)
STEP 2: Substitute
these into expression.
STEP 3: Integrate
simplified expression.
STEP 4: Write answer
in terms of 𝑥.
3
𝑑𝑥
𝑢 = sin 𝑥 + 1
𝑑𝑢
= cos 𝑥 → 𝑑𝑢 = cos 𝑥 𝑑𝑥
𝑑𝑥
sin 𝑥 = 𝑢 − 1
? find 𝒙 or 𝒅𝒙. We could,
Notice this time we didn’t
but then 𝒙 = 𝐚𝐫𝐜𝐬𝐢𝐧(𝒖 − 𝟏), and it would be slightly
awkward simplifying the expression (although is still
very much a valid method!)
𝑢 − 1 𝑢3 𝑑𝑢
=
?
=
𝑢4 − 𝑢3 𝑑𝑢
=
1 5 1 4
𝑢 −
?4 𝑢 + 𝐶
5
1
= sin 𝑥 + 1
5
5
1
−
? 4 sin 𝑥 + 1
4
+𝐶
Using substitutions involving implicit differentiation
When a root is involved, it makes thing much tidier if we use 𝑢2 = ⋯
Q
Use the substitution 𝑢2 = 2𝑥 + 5 to find ∫ 𝑥 2𝑥 + 5 𝑑𝑥
STEP 1: Using
substitution, work out
𝑥 and 𝑑𝑥 (or variant)
STEP 2: Substitute
these into expression.
STEP 3: Integrate
simplified expression.
STEP 4: Write answer
in terms of 𝑥.
𝑑𝑢
=2 →
𝑑𝑥
𝑢2 − 5
𝑥=
2
2𝑢
𝑑𝑥 = 𝑢 𝑑𝑢
𝑥 2𝑥 + 5 𝑑𝑥 =
1 4 5 2
=
𝑢 − 𝑢 𝑑𝑢
2
2
1 5 5 3
=
𝑢 − 𝑢 +𝐶
10
6
2𝑥 + 5
=
10
5
2
?
𝑢2 − 5
𝑢 × 𝑢𝑑𝑢
2
5 2𝑥 + 5
−
6
?
?
3
2
+?𝐶
This was marginally less tedious than when we used 𝑢 = 2𝑥 + 5, as we
didn’t have fractional powers to deal with.
Test Your Understanding
Edexcel C4 Jan 2012 Q6c
Hint: You might want to use
your double angle formula first.
𝑑𝑢
1
= − sin 𝑥
→ 𝑑𝑥 = −
𝑑𝑢
𝑑𝑥
sin 𝑥
cos 𝑥 = 𝑢 − 1 (As before 𝑥 = arccos 𝑢 − 1 is going to be messy)
2 sin 2𝑥
4 sin 𝑥 cos 𝑥
4 sin 𝑥 𝑢 − 1 1
𝑑𝑥 =
𝑑𝑥?= −
𝑑𝑢
1 + cos 𝑥
1 + cos 𝑥
𝑢
sin 𝑥
𝑢−1
1
= −4
𝑑𝑢 = −4 1 − 𝑑𝑢
𝑢
𝑢
= −4 𝑢 − ln 𝑢 + 𝑐 = −4 1 + cos 𝑥 − ln cos 𝑥 + 1 + 𝑐 = ⋯
Definite Integration
Now consider:
𝜋
2
Calculate ∫0 cos 𝑥 1 + sin 𝑥 𝑑𝑥
Q
Use substitution: 𝒖 = 𝟏 +?𝐬𝐢𝐧 𝒙
𝑑𝑢
= cos 𝑥
𝑑𝑥
sin 𝑥 = 𝑢 − 1
→
𝑑𝑢 = cos 𝑥 𝑑𝑥
?
Now because we’ve changed from 𝒙 to 𝒖, we have to work out what values of 𝒖
would have given those limits for 𝒙:
𝜋
When 𝑥 = 2 , 𝑢 = 2
When 𝑥 = 0, 𝑢 = 1
𝜋
2
?
2
cos 𝑥 1 + sin 𝑥 𝑑𝑥 =
0
1
=
1
𝑢2
𝑑𝑢
2 3 2
𝑢2
3
1
?
2
= 2 2−1
3
Test Your Understanding
Edexcel C4 June 2011 Q4
𝒅𝒖
= 𝟐𝒙 → 𝒅𝒖 = 𝟐𝒙 𝒅𝒙
𝒅𝒙
𝒙𝟐 = 𝒖 − 𝟐
𝟐
When 𝒙 = 𝟐, 𝒖 = 𝟐 + 𝟐 = 𝟒
𝟐
When 𝒙 = 𝟎, 𝒖 =
?𝟎 + 𝟐 = 𝟐
𝟐
𝒙𝟐 𝐥𝐧 𝒙𝟐 + 𝟐 𝒙 𝒅𝒙
𝟎
𝟒
=
𝟐
𝟏
(𝒖 − 𝟐) 𝐥𝐧 𝒖 𝒅𝒖
𝟐
Exercise 6F
Use an appropriate substitution.
Use the given substitution to integrate.
1 a
𝟐
𝟐
= 𝟏+𝒙 − 𝟏+𝒙
𝟓
𝟑
1 + sin 𝑥
𝑑𝑥; 𝑢 = sin 𝑥
cos 𝑥
= − 𝐥𝐧 𝟏 − 𝐬𝐢𝐧 𝒙 + 𝑪
𝟓
𝟐
c
5
𝑥 1 + 𝑥 𝑑𝑥 ; 𝑢 = 1 + 𝑥
?
𝟑
𝟐
3 a
+𝑪
c
𝟏
𝐜𝐨𝐬 𝟑 𝒙 − 𝐜𝐨𝐬 𝒙 + 𝑪
𝟑
e
?
e
𝑥 2 + 𝑥 𝑑𝑥 ; 𝑢 = 2 + 𝑥
𝟓
𝟐
𝟒
− 𝟐+𝒙
𝟑
?
𝟑
𝟐
+𝑪
sec 2 𝑥 tan 𝑥 1 + tan 𝑥 𝑑𝑥 ; 𝑢2 = 1 + tan 𝑥
𝟐
= 𝟏 + 𝐭𝐚𝐧 𝒙
𝟓
𝟓
𝟐
𝟐
− 𝟏 + 𝐭𝐚𝐧 𝒙
𝟑
?
sec 4 𝑥 𝑑𝑥 ; 𝑢 = tan 𝑥
𝟏
= 𝐭𝐚𝐧 𝒙 + 𝐭𝐚𝐧𝟑 𝒙 + 𝑪
𝟑
?
𝟑
𝟐
+𝑪
1
𝑑𝑥
1+ 𝑥−1
𝟐
= 𝟐 + 𝟐 𝐥𝐧
?𝟑
1
0
2
𝟐
= 𝟐+𝒙
𝟓
c
2
sin3 𝑥 𝑑𝑥 ; 𝑢 = cos 𝑥
=
2 a
0
5
?
e
𝑥 𝑥 + 4 𝑑𝑥 =
𝑥 2+𝑥
3
= 𝟗. 𝟕
?
𝑑𝑥
𝟓𝟎𝟔
𝟏𝟑
?
SKILL #6: Integration by Parts
𝑥 cos 𝑥 𝑑𝑥 =?
Just as the Product Rule was used to differentiate the product of two
expressions, we can often use ‘Integration by Parts’ to integrate a product.
! To integrate by parts:
𝑑𝑣
𝑢
𝑑𝑥 = 𝑢𝑣 −
𝑑𝑥
𝑑𝑢
𝑣
𝑑𝑥
𝑑𝑥
The Product Rule:
𝑑
𝑑𝑢
𝑑𝑣
𝑢𝑣 = 𝑣
+𝑢
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑢
𝑑𝑣
On the right-hand-side, both 𝑣 𝑑𝑥 and 𝑢 𝑑𝑥 are the product of two expressions. So if we
Proof
?
𝑑𝑣
made either the subject, we could use 𝑢 𝑑𝑥 say to represent 𝑥 cos 𝑥 in the example.
(not needed for exam)
Rearranging:
𝑑𝑣
𝑑
𝑑𝑢
𝑢
=
𝑢𝑣 − 𝑣
𝑑𝑥 𝑑𝑥
𝑑𝑥
Integrating both sides with respect to 𝑥, we get the desired formula.
SKILL #6: Integration by Parts
𝑥 cos 𝑥 𝑑𝑥 =?
𝑢=𝑥
𝑑𝑣
?𝑥
= cos
𝑑𝑥
𝑑𝑣
𝑢
𝑑𝑥 = 𝑢𝑣 −
𝑑𝑥
𝑑𝑢
𝑣
𝑑𝑥
𝑑𝑥
STEP 1: Decide which thing
𝑑𝑣
will be 𝑢 (and which ).
𝑑𝑥
𝑑𝑣
You’re about to differentiate 𝑢 and integrate , so the
𝑑𝑥
idea is to pick them so differentiating 𝑢 makes it
𝑑𝑣
‘simpler’, and can be integrated easily.
𝑑𝑥
𝑢 will always be the 𝑥 𝑛 term UNLESS one term is ln 𝑥.
𝑑𝑢
=1
𝑑𝑥
STEP 2: Find
𝑣 = sin?𝑥
𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 −
?
1 sin 𝑥 𝑑𝑥
= 𝑥 sin 𝑥 + cos 𝑥 + 𝐶
𝑑𝑢
𝑑𝑥
and 𝑣.
STEP 3: Use the formula.
I just remember it as “𝑢𝑣 minus
the integral of the two new things
timesed together”
Another Example
Q Find ∫ 𝑥 2 ln 𝑥 𝑑𝑥
𝑢 = ln 𝑥
𝑑𝑢 1
=
𝑑𝑥 𝑥
𝑑𝑣
= 𝑥2
𝑑𝑥
𝑣=
1 3
𝑥
3
1 3
𝑥 ln 𝑥 −
3
𝟏
𝟏
= 𝒙𝟑 𝐥𝐧 𝒙 − 𝒙𝟑 + 𝑪
𝟑
𝟗
𝑥 2 ln 𝑥 𝑑𝑥 =
This time, we
choose ln 𝑥 to be
the 𝑢 because it
differentiates
nicely.
?
1 2
𝑥 𝑑𝑥
3
STEP 1: Decide which thing
𝑑𝑣
will be 𝑢 (and which ).
𝑑𝑥
STEP 2: Find
𝑑𝑢
𝑑𝑥
and 𝑣.
STEP 3: Use the formula.
IBP twice! 
Q Find ∫ 𝑥 2 𝑒 𝑥 𝑑𝑥
𝑢 = 𝑥2
𝑑𝑢
= 2𝑥
𝑑𝑥
𝑑𝑣
= 𝑒𝑥
𝑑𝑥
𝑣 = 𝑒𝑥
𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 −
2𝑥𝑒 𝑥 𝑑𝑥
We have to apply IBP again!
𝑢 = 2𝑥
𝑑𝑢
=2
𝑑𝑥
𝑑𝑣
= 𝑒𝑥
𝑑𝑥
?𝑥
𝑣=𝑒
2𝑥𝑒 𝑥 𝑑𝑥 = 2𝑥𝑒 𝑥 −
2𝑒 𝑥 𝑑𝑥
= 2𝑥𝑒 𝑥 − 2𝑒 𝑥
Therefore
𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2𝑥𝑒 𝑥 − 2𝑥𝑒 𝑥 + 𝐶
= 𝑥 2 𝑒 𝑥 − 2𝑥𝑒 𝑥 + 2𝑒 𝑥 + 𝐶
Bro Tip: I tend to write
out my working for any
second integral
completely separately,
and then put the result
back into the original
integral later.
Test Your Understanding
Q Find ∫ 𝑥 2 sin 𝑥 𝑑𝑥
= 2𝑥 sin 𝑥 − 𝑥 2 cos 𝑥 + 2 cos 𝑥 + 𝐶
?
Integrating ln 𝑥 and definite integration
Q
Find ∫ ln 𝑥 𝑑𝑥, leaving your answer in terms of natural logarithms.
𝑑𝑣
=1
𝑑𝑥
𝑢 = ln 𝑥
𝑑𝑢 1
=
𝑑𝑥 𝑥
𝑣=𝑥
?
ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 −
1 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 + 𝐶
Q
2
Find ∫1 ln 𝑥 𝑑𝑥, leaving your answer in terms of natural logarithms.
2
ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥
1
2
1
= 𝟐 𝐥𝐧 𝟐 − 𝟏
If we were doing it from scratch:
𝑢 = ln 𝑥
𝑑𝑢 1
=
𝑑𝑥 𝑥
𝑑𝑣
=1
𝑑𝑥
2
ln 𝑥 𝑑𝑥 = 𝒙 𝒍𝒏 𝒙
1
?
𝑣=𝑥
𝟐
𝟏
2
−
2
1
1 𝑑𝑥
1
= 2 ln 2 − 1 ln 1 − 𝑥
= 2 ln 2 − (2 − 1) = 2 ln 2 − 1
In general:
𝑏
𝑑𝑣
𝑢
𝑑𝑥 = 𝑢𝑣
𝑑𝑥
𝑎
𝑏
𝑎
𝑏
−
𝑣
𝑎
𝑑𝑢
𝑑𝑥
𝑑𝑥
Test Your Understanding
Q
𝜋
2
Find ∫0 𝑥 sin 𝑥 𝑑𝑥
𝑢=𝑥
𝑑𝑢
=1
𝑑𝑥
𝜋
2
𝜋
4
𝑑𝑣
= sin 𝑥
𝑑𝑥
𝑣 = − cos 𝑥
𝑥 ln 𝑥 𝑑𝑥 = −𝑥 cos 𝑥
𝜋
2
𝜋
4
−
𝜋
2
𝜋
4
− cos 𝑥 𝑑𝑥
𝜋
𝜋 𝜋? 𝜋
= − cos + cos
− − sin 𝑥
2
2 4
4
𝜋
𝜋
𝜋
=0+
− − sin + sin
2
4
4 2
𝜋
1
=
− −1 +
4 2
2
𝜋
1
=
−
+1
4 2
2
𝜋
2
𝜋
4
Exercise 6G
1
a
3 a
𝑥 sin 𝑥 𝑑𝑥
?
= −𝒙 𝐜𝐨𝐬 𝒙 + 𝐬𝐢𝐧 𝒙 + 𝑪
𝑥 sec 2 𝑥 𝑑𝑥
c
e
?
c
?
𝟔
?
𝑥 2 2 sec 2 𝑥 tan 𝑥 𝑑𝑥 = 𝒙𝟐 𝐬𝐞𝐜 𝟐 𝒙 − 𝟐𝒙 𝐭𝐚𝐧 𝒙 + 𝟐 𝐥𝐧 𝐬𝐞𝐜 𝒙 + 𝑪
𝟏
𝟏
e
= 𝒙𝟑 𝐥𝐧 𝒙 − 𝒙𝟑 + 𝑪
𝟑
𝟗
ln 𝑥
𝐥𝐧 𝒙
𝟏
𝑑𝑥
=
−
−
+𝑪
𝑥3
𝟐𝒙𝟐 𝟒𝒙𝟐
𝑑𝑥 = 𝒙𝟐 𝟑 + 𝟐𝒙
𝟕
𝟏
𝟑 + 𝟐𝒙
𝟏𝟏𝟐
e
c
𝑥 2 ln 𝑥 𝑑𝑥
5
𝟏
− 𝒙 𝟑 + 𝟐𝒙
𝟕
c
?
2
a
?
12𝑥 2 3 + 2𝑥
= 𝒙 𝐭𝐚𝐧 𝒙 − 𝐥𝐧 𝐬𝐞𝐜 𝒙 + 𝑪
𝑥
𝑑𝑥
4
sin2 𝑥
a
= −𝒙 𝐜𝐨𝐭 𝒙 + 𝐥𝐧 𝐬𝐢𝐧 𝒙 + 𝑪
?
𝑥 2 𝑒 −𝑥 𝑑𝑥 = −𝒆−𝒙 𝒙𝟐 − 𝟐𝒙 𝒆−𝒙 − 𝟐𝒆−𝒙 + 𝑪
+
𝟖
+𝑪
?
ln 2
0
𝜋
2
?
𝑥 𝑒 2𝑥 𝑑𝑥 = 𝟐 𝐥𝐧 𝟐 −
𝑥 cos 𝑥 𝑑𝑥 =
0
1
4𝑥 1 + 𝑥
0
g
𝜋
3
3
𝟑
𝟒
𝝅
−𝟏
𝟐
Bonus Question:
arcsin 𝑥 𝑑𝑥
?
?
𝑑𝑥 = 𝟗. 𝟖
sin 𝑥 ln(sec 𝑥) 𝑑𝑥 =
0
=
?
𝟏 − 𝒙𝟐 + 𝒙 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 + 𝑪
𝑑𝑣
(hint: 𝑑𝑥
= 1)
𝟏
𝟏 − 𝐥𝐧 𝟐
𝟐
?
𝑥 2 + 1 ln 𝑥 𝑑𝑥
e
𝟏 𝟑
𝟏
𝒙 𝐥𝐧 𝒙 − 𝒙𝟑
𝟑
𝟗
+𝒙 𝐥𝐧 𝒙 − 𝒙 + 𝑪
=
?
You will need the following standard results (given in your formula
booklet). We’ll prove them later. !
tan 𝑥 𝑑𝑥 = ln sec 𝑥 + 𝐶
cot 𝑥 𝑑𝑥 = ln s𝑖𝑛 𝑥 + 𝐶
sec 𝑥 𝑑𝑥 = ln sec 𝑥 + tan 𝑥 + 𝐶
cosec 𝑥 𝑑𝑥 = ln cosec 𝑥 + cot 𝑥 + 𝐶
SKILL #7: Integrating top-heavy algebraic fractions
𝑥2
𝑑𝑥 = ?
𝑥+1
How would we deal with this? (the clue’s in the title)
𝑥 −1
𝑥+1
𝑥 2 + 0𝑥 + 0
Some manipulation
𝑥 2 + to
𝑥 simplify
?
−𝑥 + 0
−𝑥 − 1
1
𝑥2
𝑑𝑥
𝑥+1
1
= Now
𝑥 − 1integrate
+
𝑑𝑥
𝑥
+
1
?
1 2
= 𝑥 − 𝑥 + ln 𝑥 + 1
2
+𝐶
Test Your Understanding
𝑥
𝑑𝑥
𝑥−1
= 𝒙 + 𝐥𝐧 𝒙 −?𝟏 + 𝑪
𝑥3 + 2
𝑑𝑥
𝑥+1
𝟏 𝟑 𝟏 𝟐
= 𝒙 − 𝒙 +𝒙
𝟑
𝟐?
+ 𝐥𝐧 𝒙 + 𝟏 + 𝑪
Trapezium Rule Revisited
𝑏
1
Trapezium Rule: ∫𝑎 𝑦 𝑑𝑥 ≈ 2 ℎ 𝑦0 + 2 𝑦1 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛
You already know the Trapezium Rule from C2. You may be asked similar questions in a
C4 exam, except now the expressions involve integrations rules you’ve just learnt.
𝜋
3
Given 𝐼 = ∫0 sec 𝑥 𝑑𝑥
a) Find the exact value of 𝐼.
Q b) Use the trapezium rule with two strips to estimate 𝐼.
c) Use the trapezium rule with four strips to find a second estimate of 𝐼.
d) Find the percentage error in using each estimate.
a
b
𝐼 = ln sec 𝑥 + tan 𝑥
?
= ln 2 + 3
𝒙
𝒚
𝜋
6
1 1.155
?
0
𝜋
3
0
c
𝜋
3
2
1𝜋
𝐼≈
1 + 2 1.155 + 2 = 1.39
26
𝐼≈
𝒙
0
𝒚
1
𝜋
12
1.035
𝜋
6
?1.155
𝜋
4
1.414
𝜋
3
2
1 𝜋
1 + 2 1.035 + 1.155 + 1.414 + 2 = 1.34
2 12
d
Error is 5.6% and 1.5%.
?
Test Your Understanding
a)
Q b)
c)
d)
3
Find the exact value of 𝐼 = ∫0 𝑥 2 − 𝑥 𝑑𝑥.
Use the trapezium rule with 4 strips to estimate 𝐼.
Use the trapezium rule with 6 strips to find a second estimate of 𝐼.
Compare the percentage error in using each estimate.
a
16 2
15
?
b
1.34
?
c
1.42
?
d
11.4% and 61.%
?
Exercise 6I
1 Find the (i) area of region and (ii) solid of revolution bound by lines 𝑥 = 𝑎 and 𝑥 = 𝑏 for:
a
2
𝑓 𝑥 =
; 𝑎 = 0, 𝑏 = 1
1+𝑥
c
𝑓 𝑥 = ln 𝑥 ; 𝑎 = 1, 𝑏 = 2
(Hint: use integration by parts with
e
3
𝑨 = 𝟐 𝐥𝐧?𝟐
𝑽 = 𝟐𝝅?
𝑨 = 𝟐 𝒍𝒏?
𝟐−𝟏
𝑑𝑣
𝑑𝑥
𝑓 𝑥 = 𝑥 4 − 𝑥 2 ; 𝑎 = 0, 𝑏 = 2
𝑽 = 𝟐 𝐥𝐧 𝟐
?− 𝟒 𝐥𝐧 𝟐 𝜋
= 1 for ∫ ln2 𝑥)
𝑨=
𝟖
?
𝟑
𝑽=
𝟔𝟒
?𝝅
𝟏𝟓
The region 𝑅 is bound by the curve 𝐶, the 𝑥-axis and the lines 𝑥 = −8 and 𝑥 = +8. The
parametric equations for 𝐶 are 𝑥 = 𝑡 3 and 𝑦 = 𝑡 2 . Find:
𝟏𝟗𝟐
a) The area of 𝑅. 𝟓 ?
b) The volume of the solid of revolution formed about the 𝑥-axis.
4
𝟐
𝟕𝟔𝟖
𝝅
𝟕
?
The curve 𝐶 has parametric equations 𝑥 = sin 𝑡 and 𝑦 = sin 2𝑡, 0 ≤ 𝑡 <
a) Find the area of the region bounded by 𝐶 and the 𝑥-axis.
b) Find the volume of the solid of revolution.
𝟖
𝝅
𝟏𝟓
?
𝟐
𝟑
?
𝜋
2
SKILL #8: Volumes of Revolution
𝑏
∫𝑎 𝑦
𝑑𝑥 gives the area bounded between 𝑦 = 𝑓(𝑥), 𝑥 = 𝑎,
𝑥 = 𝑏 and the 𝑥-axis. Why?
If we split up the area into thin rectangular strips, each
with width 𝑑𝑥 and each with height the 𝑦 = 𝑓 𝑥 for
that particular value of 𝑥. Each has area 𝑓 𝑥 × 𝑑𝑥.
𝑦
If we had ‘discrete’ strips, the total area would be:
𝑏
Σ𝑥=𝑎
(𝑓 𝑥 𝑑𝑥)
But because the strips are infinitely small and we
have to think continuously, we use ∫ instead of Σ.
Integration therefore can be thought of as a
continuous version of summation.
𝑦1
𝑎
𝑦2
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑏
𝑥
SKILL #8: Volumes of Revolution
Now suppose we spun the line 𝑦 = 𝑓(𝑥)
about the 𝑥 axis to form a solid (known
as a volume of revolution):
𝑦
Click to Start
Bromanimation
How could we use the same principle as
before to express the volume? What
should we use instead of strips?
𝑦
𝑎
𝑏
𝑑𝑥
𝑥
Reveal >>
We’re summing a bunch of infinitely thin
cylinders, each of width 𝑑𝑥 and radius 𝑦.
Each has a volume of:
𝜋𝑦 2?𝑑𝑥
Thus the volume is: !
𝒃
𝑽=𝝅
?
𝒂
𝒚𝟐 𝒅𝒙
i.e. Square the function and slap a 𝜋 on front.
Examples
The region 𝑅 is bounded by the curve with equation 𝑦 = 𝑥, the 𝑥-axis and
the lines 𝑥 = 0 and 𝑥 = 2.
Q a) Find the area of 𝑅.
b) Find the volume of the solid formed when the region 𝑅 is rotated
through 2𝜋 radians about the 𝑥-axis.
2
a
b
0
1
𝑥 𝑑𝑥 = 𝑥 2
2
2
2
0
= 2?
1 3
2
𝑥 𝑑𝑥 = 𝜋 𝑥
3
𝜋
0
2
8𝜋
?= 3
0
Test Your Understanding: Do the same when 𝑦 = sin 2𝑥, bound between
𝜋
Q
𝑥 = 0 and 𝑥 = .
2
𝜋
2
a
0
b
1
sin 2𝑥 𝑑𝑥 = − cos
? 2𝑥
2
2
𝜋
0
𝜋
2
0
=1
1 1
sin2 2𝑥 𝑑𝑥 = 𝜋 ?− cos 4𝑥
2 2
𝜋
2
0
𝜋2
=
4
Area/Volume with Parametric Equations
I once again have parametric equations:
𝑥 = 𝑡2
𝑦=𝑡+1
I want the integration to be in terms of 𝑑𝑡 rather than 𝑑𝑥.
∫ 𝑦 𝑑𝑥 =
Area:
𝑑𝑥
∫ 𝑦 𝑑𝑡
𝑑𝑡?
𝑑𝑥
Volume: 𝜋 ∫ 𝑦 2 𝑑𝑥 = 𝜋 ∫ 𝑦 2 ? 𝑑𝑡
𝑑𝑡
Bro Memory Tip: No need to remember
the whole new formulae. All you need to
𝑑𝑥
remember is that 𝑑𝑡 = 𝑑𝑥 , which is
𝑑𝑡
obvious since the 𝑑𝑡’s cancel.
Bro Tip: Remember that you have to
change the bounds!
Example: Find the volume of revolution and the area bound by the curve and 𝑥 = 0 and 𝑥 = 2.
𝑑𝑥
= 2𝑡
𝑑𝑡
𝑊ℎ𝑒𝑛 𝑥 = 2, 𝑡 = 2
?
𝒅𝒙
2
𝐴=
2
2𝑡 2 + 2𝑡 𝑑𝑡
𝑡 + 1 2𝑡 𝑑𝑡 =
0
2
= 𝑡3 + 𝑡2
3
2
𝑉=𝜋
0
2
0
𝑡+1
0
?
𝑊ℎ𝑒𝑛 𝑥 = 0, 𝑡 = 0
=
2
4√2
+2
3
?
2
2𝑡 3 + 4𝑡 2 + 2𝑡 𝑑𝑡 = 𝜋 8 + 6 2
2𝑡 𝑑𝑡 = 𝜋
0
STEP 1: Find
𝒅𝒕
STEP 2: Change limits
STEP 3: Integrate
Test Your Understanding
June 2011 Q7
𝒅𝒙
= 𝐬𝐞𝐜 𝟐 𝜽
𝒅𝜽
𝝅
When 𝒙 = 𝟑, 𝜽 =
𝟑
When 𝒙 = 𝟎, 𝜽 = 𝟎
𝝅
𝝅
𝟑
𝟎
=𝝅
=𝝅
=𝝅
=𝝅
𝒚𝟐
𝝅
𝟑
𝒅𝒙
𝒅𝜽
𝒅𝜽
𝒔𝒊𝒏𝟐 𝜽 𝒔𝒆𝒄𝟐 𝜽 𝒅𝜽
𝟎
𝝅
𝟑 𝒔𝒊𝒏𝟐 𝜽
𝟎
𝝅
𝟑
𝟎
𝝅
𝟑
𝟎
𝒄𝒐𝒔𝟐 𝜽
𝟐
𝒅𝜽
?
𝒕𝒂𝒏 𝜽 𝒅𝜽
𝒔𝒆𝒄𝟐 𝜽 − 𝟏 𝒅𝜽
𝝅
𝟑
𝟎
𝝅 𝐭𝐚𝐧 𝜽 − 𝜽
𝝅 𝝅
= 𝝅 𝐭𝐚𝐧 −
𝟑 𝟑
𝝅
=𝝅 𝟑−
𝟑
𝟏
= 𝝅 𝟑 − 𝝅𝟐
𝟑
SKILL #9: Differential Equations
(We’re on the home straight!)
Earlier in C4 we encountered differential equations. These are equations
𝒅𝒚
involving a mix of variables and derivatives, e.g. 𝒚, 𝒙 and 𝒅𝒙.
𝑑𝑦
‘Solving’ these equations means to get 𝑦 in terms of 𝑥 (with no 𝑑𝑥 ).
Q
Find the general solution to
𝑑𝑦
=𝑦 𝑥+1
𝑑𝑥
?
1 𝑑𝑦
= 𝑥+1
𝑦 𝑑𝑥
1
𝑑𝑦 =? 𝑥 + 1 𝑑𝑥
𝑦
1 2
ln 𝑦 = 𝑥 + 𝑥 + 𝐶
2
1 2 ?
𝑦 = 𝑒 2𝑥 +𝑥 + 𝐶
𝑑𝑦
𝑑𝑥
= 𝑥𝑦 + 𝑦
𝑑𝑦
STEP 1: Get 𝑦 to the side of 𝑑𝑥 by dividing
and 𝑥 to the other side.
(you may need to factorise)
STEP 2: Multiply both sides by 𝑑𝑥, then
integrate both sides.
STEP 3: Make 𝑦 the subject, if the question
asks.
Another Example
Q
Find the general solution to 1 + 𝑥 2
1 𝑑𝑦
𝑥
=
tan 𝑦 𝑑𝑥 ?1 + 𝑥 2
𝑑𝑦
𝑑𝑥
= 𝑥 tan 𝑦
𝑑𝑦
STEP 1: Get 𝑦 to the side of 𝑑𝑥 by dividing
and 𝑥 to the other side.
(you may need to factorise)
cot 𝑦 𝑑𝑦 =
𝑥
𝑑𝑥
1?+ 𝑥 2
1
ln 1 + 𝑥 2 + 𝐶
2
1
ln sin 𝑦 = ln 1 + 𝑥 2 + ln 𝑘
2
? 2
ln sin 𝑦 = ln 𝑘 1 + 𝑥
STEP 2: Multiply both sides by 𝑑𝑥, then
integrate both sides.
ln sin 𝑦 =
sin 𝑦 = 𝑘 1 + 𝑥 2
𝑦 = arcsin 𝑘 1?+ 𝑥 2
STEP 2b: If possible, try to combine your
constant of integration with other terms
(e.g. by letting 𝐶 = ln 𝑘 where 𝑘 is another constant)
STEP 3: Make 𝑦 the subject, if the question
asks.
Test Your Understanding
The rate of increase of a rabbit population (with population 𝑦, where time
Q is 𝑥) is proportional to the current population.
Form a differential equation, and find its general solution.
𝑑𝑦
= 𝑘𝑦
𝑑𝑥
1 𝑑𝑦
=𝑘
𝑦 𝑑𝑥
1
𝑑𝑦 = 𝑘 𝑑𝑥
?
𝑦
ln 𝑦 = 𝑘𝑥 + 𝐶
𝑦 = 𝑒 𝑘𝑥+𝐶
𝒚 = 𝑨𝒆𝒌𝒙 (𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝑒 𝐶 )
𝑥
(Notice by the way that 𝑒 𝑘𝑥 = 𝑒 𝑘 , and since 𝑒 𝑘 is a constant,
we could always write 𝑦 = 𝐴 ⋅ 𝐵 𝑥 . i.e. The general solution is
‘any generic exponential function’, not just restricted to those
with 𝑒 as the base. However it is customary to write 𝐴𝑒 𝑘𝑥 )
Differential Equations with Boundary Conditions
Q
Find the general solution to
𝑑𝑦
𝑑𝑥
=−
3 𝑦−2
2𝑥+1 𝑥+2
Given that 𝑥 = 1 when 𝑦 = 4. Leave your answer in the form 𝑦 = 𝑓 𝑥
1
−3
𝑑𝑦 =
𝑑𝑥
𝑦−2
2𝑥 + 1 𝑥 + 2
Use partial fractions to split up RHS.
−3
𝐴
𝐵
≡
+
2𝑥 + 1 𝑥 + 2
2𝑥 + 1 𝑥 + 2
𝐴 = −2, 𝐵 = 1
1
1
2
𝑑𝑦 =
−
𝑑𝑥
𝑦−2
𝑥 + 2 2𝑥
+
1
?
ln 𝑦 − 2 = ln 𝑥 + 2 − ln 2𝑥 + 1 + ln 𝑘
𝑘 𝑥+2
ln 𝑦 − 2 = ln
2𝑥 + 1
𝑘 𝑥+2
𝑦−2=
2𝑥 + 1
𝑘 1+2
When 𝑥 = 1, 𝑦 = 4: 4 − 2 = 2+1 → 𝑘 = 2
𝟑
𝒚=𝟑+
𝟐𝒙 + 𝟏
Bro Tip: While it
doesn’t matter when
you determine your
constant using the
boundary conditions,
usually it’s easiest to
determine it as soon
as possible.
Key Points
• Get 𝑦 on to LHS by dividing (possibly factorising first).
• If after integrating you have 𝑙𝑛 on the RHS, make your
constant of integration ln 𝑘.
• Be sure to combine all your 𝑙𝑛’s together just as you did in C2.
E.g.:
𝑥+1 2
?
2 ln 𝑥 + 1 − ln 𝑥
→ ln
𝑥
• Sub in boundary conditions to work out your constant –
better to do sooner rather than later.
• Exam questions  partial fractions.
Exercise 6J
1
a
c
e
2
a
c
e
4
a
c
e
Find the general solution in the form 𝑦 = 𝑓(𝑥).
𝑑𝑦
𝟐
= 1 + 𝑦 1 − 2𝑥
𝒚 = 𝑨𝒆𝒙−𝒙 − 𝟏
𝑑𝑥
𝑑𝑦
cos2 𝑥
= 𝑦 2 sin2 𝑥
𝒚 = −𝟏/(𝐭𝐚𝐧 𝒙 − 𝒙 + 𝑪)
𝑑𝑥
𝟏
𝑑𝑦
𝑥2
= 𝑦 + 𝑥𝑦
𝒚 = 𝑨𝒙 𝒆−𝒙
𝑑𝑥
?
?
?
Find the general solution (no need to put in form 𝑦 = 𝑓 𝑥 )
𝑑𝑦
= tan 𝑦 tan 𝑥
𝐬𝐢𝐧 𝒚 = 𝒌 𝐬𝐞𝐜 𝒙
𝑑𝑥
𝑑𝑦
𝟏+𝒚
1 + 𝑥2
= 𝑥 1 − 𝑦2
= 𝒌 𝟏 + 𝒙𝟐
𝑑𝑥
𝟏−𝒚
𝑑𝑦
𝑒 𝑥+𝑦
= 𝑥 2 + 𝑒𝑦
𝐥𝐧 𝟐 + 𝒆𝒚 = −𝒙𝒆−𝒙 − 𝒆−𝒙 + 𝑪
𝑑𝑥
?
?
?
Find particular solutions using boundary conditions.
𝑑𝑦
𝜋
𝟏
𝟏
= 𝑠𝑖𝑛𝑥 cos 2 𝑥 ; 𝑦 = 0, 𝑥 =
− 𝐜𝐨𝐬 𝟑 𝒙
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝜋
3
= 2 cos2 𝑦 cos2 𝑥 ; 𝑦 = , 𝑥 = 0
4
𝑑𝑦
1
2 1+𝑥
= 1 − 𝑦 2 ; 𝑥 = 5, 𝑦 =
𝑑𝑥
2
?
𝟏
𝐭𝐚𝐧 𝒚 = 𝐬𝐢𝐧 𝟐𝒙
?+ 𝒙 + 𝟏
𝟐
𝟏+𝒚 𝟏+𝒙
=
𝟏−𝒚
𝟐 ?
𝟐𝟒
𝟑
Exercise 6K
1
The size of a certain population at time 𝑡 is given by 𝑃. The rate of increase of 𝑃 I given
𝑑𝑃
by 𝑑𝑡 = 2𝑃. Given that a time 𝑡 = 0, the population was 3, find the population at time
𝑡 = 2. 𝟑𝒆𝟒
?
3
The mass 𝑀 at time 𝑡 of the leaves of a certain plant varies according to:
𝑑𝑀
= 𝑀 − 𝑀2
𝑑𝑡
a) Given that at time 𝑡 = 0, 𝑀 = 0.5, find an expression for 𝑀 in terms of 𝑡.
𝑴 = 𝒆𝒕 /(𝟏
? + 𝒆𝒕)
b) Find a value for 𝑀 when 𝑡 = ln 2. 𝟐/𝟑
?
c) Explain what happens to the value of 𝑀 as a 𝑡 increases 𝑴 approaches 1.
?
5
𝑑𝑥
𝟒𝟔
7
1
The thickness of ice 𝑥 mm on a pond is increasing and 𝑑𝑡 = 20𝑥 2, where 𝑡 is measured
in hours. Find how long it takes the thickness of ice to increase from 1mm to 2mm.
𝟐
𝟑
?
𝑑𝑟
The rate of increase of the radius 𝑟 km of an oil slick is given by 𝑑𝑡 = 𝑘/𝑟 2 , where 𝑘 is a
positive constant. When the slick was first observed the radius was 3km. Two days later
it was 5km. Find, to the nearest day when the radius will be 6.
𝟒?
Summary of Functions
𝒇(𝒙)
𝒔𝒊𝒏 𝒙
How to deal with it
Standard result
𝒄𝒐𝒔 𝒙
Standard result
𝒕𝒂𝒏 𝒙
In formula booklet, but use
sin 𝑥
∫ cos 𝑥 𝑑𝑥 which is of the form
∫
𝒔𝒊𝒏𝟐 𝒙
𝑘𝑓′ 𝑥
𝑓 𝑥
?
?
𝒕𝒂𝒏𝟐 𝒙
𝒄𝒐𝒔𝒆𝒄 𝒙
𝒔𝒆𝒄 𝒙
𝒄𝒐𝒕 𝒙
No
ln sec 𝑥
Yes
1
1
𝑥 − sin 2𝑥
2
4
No
?
1
1
𝑥 + sin 2𝑥
2
4
No
?
tan 𝑥 − 𝑥
No
?
−ln 𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥
Yes
?
ln sec 𝑥 + tan 𝑥
Yes
?
ln 𝑠𝑖𝑛 𝑥
Yes
?
𝑑𝑥
For both sin2 𝑥 and cos2 𝑥 use
identities for cos 2𝑥
cos 2𝑥 = 1 − 2 sin2 𝑥
1 1
sin2 𝑥 = − cos 2𝑥
2 2
cos 2𝑥 = 2 cos 2 𝑥 − 1
1 1
cos 2 𝑥 = + cos 2𝑥
2 2
1 + tan2 𝑥 ≡ sec 2 𝑥
tan2 𝑥 ≡ sec 2 𝑥 − 1
Would use substitution 𝑢 =
𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥, but too hard for
exam.
Would use substitution 𝑢 = sec 𝑥 +
tan 𝑥, but too hard for exam.
cos 𝑥
∫ sin 𝑥 𝑑𝑥 which is of the form
𝑓′ 𝑥
Formula booklet?
No
sin 𝑥
?
𝒄𝒐𝒔𝟐 𝒙
∫ 𝒇 𝒙 𝒅𝒙 (+constant)
− cos 𝑥
Summary of Functions
𝒇(𝒙)
𝒄𝒐𝒔𝒆𝒄𝟐 𝒙
𝒔𝒆𝒄𝟐 𝒙
𝒄𝒐𝒕𝟐 𝒙
∫ 𝒇 𝒙 𝒅𝒙 (+constant) Formula booklet?
How to deal with it
By observation.
?
By observation.
?
1 + cot 2 𝑥 ≡ 𝑐𝑜𝑠𝑒𝑐 2 𝑥
?
𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔 𝟐𝒙 For any product of sin and cos with
same coefficient of 𝑥, use double
angle.
1
sin 2𝑥 cos 2𝑥 ≡ sin 4𝑥
2
𝟏
𝒙
𝑑𝑣
𝐥𝐧 𝒙
Use IBP, where 𝑢 = ln 𝑥 , = ln 𝑥
𝒙
𝒙+𝟏
𝟏
𝒙 𝒙+𝟏
Use algebraic division.
Use partial fractions.
𝑥
𝑥+1
tan 𝑥
− cot 𝑥 − 𝑥
No!
Yes
(but memorise)
No
1
− cos 4𝑥
8
No
?
ln 𝑥
No
?
𝑥 ln 𝑥 − 𝑥
No
?
𝑑𝑥
− 𝐜𝐨𝐭 𝒙
≡1−
1
𝑥+1
?
?
𝑥 − ln 𝑥 + 1
ln 𝑥 − ln 𝑥 + 1
Summary of Functions
𝒇(𝒙)
𝟒𝒙
𝒙𝟐 + 𝟏
𝒙
𝒙𝟐 + 𝟏 𝟐
𝒆𝟐𝒙+𝟏
𝟏
𝟏 − 𝟑𝒙
𝒙 𝟐𝒙 + 𝟏
𝐬𝐢𝐧𝟓 𝒙 𝒄𝒐𝒔 𝒙
∫ 𝒇 𝒙 𝒅𝒙 (+constant)
How to deal with it
Reverse chain rule. ?
Of form
𝑘𝑓′ 𝑥
∫ 𝑓𝑥
Power around denominator so NOT of
form ∫
𝑘𝑓′
𝑥
𝑓 𝑥
. Rewrite as product.
𝑥 𝑥2 ?
+ 1 −2
Reverse chain rule (i.e. “Consider 𝑦 =
𝑥 2 + 1 −1 " and differentiate)
For any function where ‘inner function’
is linear expression, divide by coefficient
?
of 𝑥
Use sensible substitution. 𝑢 = 2𝑥 + 1
or even better, 𝑢2 =?2𝑥 + 1.
Reverse chain rule.
?
2 ln 𝑥 2 + 1
1 2
− 𝑥 +1
2
−1
1 2𝑥+1
𝑒
2
1
− ln 1 − 3𝑥
3
3
1
2𝑥 + 1 2 3𝑥 − 1
15
1 6
sin 𝑥
6