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Compiled by: D
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van der Berg

Understanding Integration
Introduction
What is integration?
•

Reverse differentiation (anti-differentiation)

Why use it?
•

Used to find volumes, area and central points
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∫ (2𝑥𝑥 (3+1) + 1𝑥𝑥 (2+1) + 7𝑥𝑥 (1+1) )𝑑𝑑𝑑𝑑
∫ �(2 ÷ 4)𝑥𝑥 4 + (1 ÷ 3)𝑥𝑥 3 + (7 ÷ 2)𝑥𝑥 2 �𝑑𝑑𝑑𝑑
1

IMPORTANT NOTE:

1

7

= ∴ 𝑓𝑓 ′ (𝑥𝑥) = 𝑥𝑥 4 + 𝑥𝑥 3 + 𝑥𝑥 2
2

3

2

You can check your answer by finding its derivative!

Compiled by: D
...
van der Berg

Rules of integration
Includes 4 major rules (ISTP)
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3
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 depends on the given functions

Inspection
Substitution
Trigonometric substitution
Parts

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By Substitution:
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When a function and its derivative appear in the integral:

Let 𝑢𝑢 = 𝑓𝑓(𝑥𝑥)

∫ 𝑓𝑓(𝑥𝑥)
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2
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J
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By Trigonometric Substitution:
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Compound-angle formulae
Integrating a product of trigonometric functions:

∫ sin 4𝑥𝑥 cos 3𝑥𝑥 𝑑𝑑𝑑𝑑
and

(1) sin(4𝑥𝑥 + 3𝑥𝑥) = sin 4𝑥𝑥 cos 3𝑥𝑥 + cos 4𝑥𝑥 sin 3𝑥𝑥
(2) sin(4𝑥𝑥 − 3𝑥𝑥) = sin 4𝑥𝑥 cos 3𝑥𝑥 − cos 4𝑥𝑥 sin 3𝑥𝑥
Add these two expansions (1) and (2):

∴ sin 7𝑥𝑥 + sin 𝑥𝑥 = 2 sin 4𝑥𝑥 cos 3𝑥𝑥
1

(our substitution)

∴ sin 4𝑥𝑥 cos 3𝑥𝑥 = 2 (sin 7𝑥𝑥 + sin 𝑥𝑥)
So,

∫ sin 4𝑥𝑥 cos 3𝑥𝑥 𝑑𝑑𝑑𝑑
1

= 2 ∫ (sin 7𝑥𝑥 + sin 𝑥𝑥) 𝑑𝑑𝑑𝑑

=

− cos 7𝑥𝑥
14

−

cos 𝑥𝑥
2

+ 𝑐𝑐

Shortcut rules to study:
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(a) sin 𝐴𝐴 cos 𝐵𝐵 = 2 [sin(𝐴𝐴 + 𝐵𝐵) + sin(𝐴𝐴 − 𝐵𝐵) ]
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(b) sin 𝐴𝐴 sin 𝐵𝐵 = 2 [cos(𝐴𝐴 − 𝐵𝐵) − cos(𝐴𝐴 + 𝐵𝐵) ]
1

(c) cos 𝐴𝐴 cos 𝐵𝐵 = 2 [cos(𝐴𝐴 − 𝐵𝐵) + cos(𝐴𝐴 + 𝐵𝐵) ]
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Double-angle formulae
Useful for squared trigonometric functions:

∫ sin2 𝑥𝑥 𝑑𝑑𝑑𝑑

or

∫ cos 2 𝑥𝑥 𝑑𝑑𝑑𝑑

We use the following rules:
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(a) sin2 𝑥𝑥 = 2 (1 − cos 2𝑥𝑥)
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(b) cos2 𝑥𝑥 = 2 (1 + cos 2𝑥𝑥)

Derivation from
and from

cos 2𝑥𝑥 = 2 cos2 𝑥𝑥 − 1
sin2 𝑥𝑥 + cos2 𝑥𝑥 = 1

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...
van der Berg

4
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That is, the one is NOT the
derivative of the other:
∫ 𝑓𝑓(𝑥𝑥)𝑔𝑔(𝑥𝑥)𝑑𝑑𝑑𝑑
Developed from the reverse product rule for differentiation
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Carefully consider this:
Option (1):

if we make our choices as below:
then 𝑑𝑑𝑑𝑑 = cos 𝑥𝑥𝑥𝑥𝑥𝑥

𝑢𝑢 = sin 𝑥𝑥,

And 𝑑𝑑𝑑𝑑 = 𝑥𝑥𝑥𝑥𝑥𝑥,

then 𝑣𝑣 =

Substitute our choices into
∴ ∫ 𝑥𝑥 sin 𝑥𝑥 𝑑𝑑𝑑𝑑 =
Option (2):

𝑥𝑥 2
2

𝑥𝑥 2

𝑥𝑥 2
2

(we integrate dv)

∫ 𝑢𝑢𝑢𝑢𝑢𝑢 = 𝑢𝑢𝑢𝑢 − ∫ 𝑣𝑣𝑣𝑣𝑣𝑣:

sin 𝑥𝑥 − ∫ � 2 cos 𝑥𝑥� 𝑑𝑑𝑑𝑑

 this is complicated!! Try Option (2)

if we make our choices as below:

𝑢𝑢 = 𝑥𝑥, then 𝑑𝑑𝑑𝑑 = 1𝑑𝑑𝑑𝑑
And 𝑑𝑑𝑑𝑑 = sin 𝑥𝑥 𝑑𝑑𝑑𝑑,

then 𝑣𝑣 = − cos 𝑥𝑥

∴ ∫ 𝑥𝑥 sin 𝑥𝑥 𝑑𝑑𝑑𝑑 = −𝑥𝑥 cos 𝑥𝑥 − ∫ (− cos 𝑥𝑥)𝑑𝑑𝑑𝑑

=

−𝑥𝑥 cos 𝑥𝑥 − sin 𝑥𝑥 + 𝑐𝑐

Compiled by: D
...
van der Berg

Another useful rule – Reverse Chain Rule:
Used for expressions with difficult exponent-based brackets:
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∫ (𝑥𝑥 2 − 1)2 𝑑𝑑𝑑𝑑

Example

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Divide entire expression by new exponent

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sin 5𝜃𝜃 𝑑𝑑𝑑𝑑

(d) ∫ �𝑦𝑦�𝑦𝑦 + 3� 𝑑𝑑𝑑𝑑
(e) ∫ 𝑥𝑥 2 sin𝑥𝑥 𝑑𝑑𝑑𝑑
(f) ∫ �

𝑥𝑥+1

� 𝑑𝑑𝑑𝑑

√2𝑥𝑥−1

(g) ∫ 𝑥𝑥 2 (𝑥𝑥 3 − 1)5 𝑑𝑑𝑑𝑑

Check your answers by finding its derivative



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