Definite integrals

If 𝑓(𝑥) is a continuous function defined on a closed interval[𝑎, 𝑏], and if 𝐹(𝑥) is the ant-
𝑑
derivative of 𝑓(𝑥) i.e. [𝐹(𝑥)] = 𝑓(𝑥), then the definite integral of 𝑓(𝑥) over[𝑎, 𝑏] is
𝑑𝑥
denoted by
𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥

and is equal to
[ℱ(𝑏) − ℱ(𝑎)] i.e.

𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥 = [ℱ(𝑥)]𝑏𝑎 = [(ℱ𝑏) − ℱ(𝑎)]

Where "𝑎" is the lower limit and "𝑏" is the upper limit.
Note:
Taking ℱ(𝑥) + 𝐶 instead of ℱ(𝑥) as the anti-derivative of (𝑥) , we have;

𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = [ℱ(𝑥) + 𝑐]𝑏𝑎

= [ℱ(𝑏) + 𝑐] − [ℱ(𝑎) + 𝑐]
= ℱ(𝑏) − ℱ(𝑎)
𝑏
As the constant of integration disappears, so ∫𝑎 𝑓(𝑥)𝑑𝑥 has a definite value

[ℱ(𝑏) − ℱ(𝑎)].
Examples
Evaluate the following integrals:
−1 1 5 1 1
a) ∫−4 𝑥
𝑑𝑥 b) ∫2 (𝑥 3 + 𝑥)𝑑𝑥 c) ∫0 2𝑥−3
𝑑𝑥

Solution

−1 1 −1
a) ∫−4 𝑑𝑥 = − ∫−4 𝑥 −1 𝑑𝑥
𝑥

= ln 𝑥|−4
−1

= (ln|−1| − ln |(−4)|)

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 = 0 − ln |(−4)|
= − ln 4

5
b) ∫2 (𝑥 3 + 𝑥)𝑑𝑥

5 5
= ∫2 𝑥 3 𝑑𝑥 + ∫2 𝑥 𝑑𝑥
5 5
𝑥4 𝑥2
= | + |
4 2 2 2

54 24 52 22
= (4 − )+(4 − )
4 2
651
= 4
1 1
c) ∫0 𝑑𝑥
2𝑥−3

1
= ∫0 (2𝑥 − 3)−1 𝑑𝑥
1
= ∙ ln(2𝑥 − 3)|10
2
1
= [𝑙𝑛|−1| − 𝑙𝑛|−3|]
2
1
= [ln 1 − ln 3]
2
1
= − ln 3
2

4 3
d) ∫0 (𝑥 + 𝑥 2 ) 𝑑𝑥

2 𝑥 2 −3𝑥+2
e) ∫1 𝑑𝑥
𝑥4

1 1
f) ∫0 𝑑𝑥
√1+𝑥+√𝑥

𝜋
g) ∫02 cos 2𝑥 𝑑𝑥

𝜋
h) ∫02 sin2 𝑥 𝑑𝑥

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 𝜋
𝑠𝑖𝑛𝑥
i) ∫02 𝑑𝑥 (hint: use half angle identities)
√1+cos 𝑥

𝜋 𝜋 2
𝑥 𝑠𝑖𝑛𝑥
j) ∫02 sin2 (1+𝑐𝑜𝑠𝑥)2 𝑑𝑥 hint : ⟹ ∫02 (1+𝑐𝑜𝑠𝑥) 𝑑𝑥

𝜋
k) ∫02 sin4 𝑥 𝑑𝑥

Integration by substitution

Indefinite integrals of the form ∫ 𝑓(𝑔(𝑥) ∙ 𝑔′ (𝑥)𝑑𝑥 can sometimes be evaluated by

making the 𝑢 − substitution;
𝑢 = 𝑔(𝑥) 𝑑𝑢 = 𝑔′ (𝑥)𝑑𝑥 → Substitution (1),
Which converts the integral to the form

∫ 𝑓(𝑢).
To apply this method for the effect that the substitution has on the 𝑥 −limits of
integration. We can do this in two ways:
Method 1
First evaluate the indefinite integral

∫ 𝑓(𝑔(𝑥)) ∙ 𝑔′(𝑥)𝑑𝑥

By substitution, and then use the relationship

𝑏 𝑏
∫𝑎 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = [∫ 𝑓(𝑔(𝑥) ∙ 𝑔′ (𝑥)𝑑𝑥]𝑎

To evaluate the definite integral. This procedure does not require any modification of the
𝑥 − limits of integration.
Method 2
Make substitution (1) directly in the definite integral, and then use the relationship 𝑢 =
𝑔(𝑥) to replace 𝑥 − limits, 𝑥 = 𝑎 and 𝑥 = 𝑏 by corresponding 𝑢 −limits, 𝑢 = 𝑔(𝑎) and
𝑢 = 𝑔(𝑏). This produces a new definite integral
𝑔(𝑏)
∫𝑔(𝑎) 𝑓(𝑢)𝑑𝑢

That is expressed entirely in terms of𝑢. This method is sometimes refeered to as the
change of limits method.

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Examples
Use the two methods above to evaluate
2
∫0 𝑥(𝑥 2 + 1)3 𝑑𝑥

Method 1:
Let
𝑢 = 𝑥 2 + 1 so that
𝑑𝑢 = 2𝑥𝑑𝑥
Then replacing we have;
𝑑𝑢
∫ 𝑥(𝑥 2 + 1)3 𝑑𝑥 = ∫ 𝑥 ∙ 𝑢3 ∙
2𝑥
1
= ∫ 2 ∙ 𝑢3 𝑑𝑢
1
= 2 ∫ 𝑢3 ∙ 𝑑𝑢

1 𝑢4
=2∙ +𝑐
4
4
𝑢4 (𝑥 2 +1)
= +𝑐 = +𝑐
8 8

Thus;
2
2 2 𝑋 2 +1
∫0 𝑥(𝑥 2 + 1)3 𝑑𝑥 = [∫ 𝑥(𝑥 2 + 1)3 𝑑𝑥]0 = 8
|
0
625 1
= −8
8

= 78
Method 2
If we make the substitution
𝑢 = 𝑥2 + 1
In (1), then
𝑢 = 1 if 𝑥 = 0
𝑢 = 5 if 𝑥 = 2
Thus;
5
2 1 5 𝑢4
∫0 𝑥(𝑥 2 + 1)3 𝑑𝑥 = 2 ∫1 𝑢3 𝑑𝑢 = 8
|
𝑢=1

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 625 1
= −8
8

= 78 (same as the result for method 1)

Example
Evaluate:
3 𝜋
𝑑𝑥
a) ∫04 1−𝑥 b) ∫08 sin5 2𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥

Solutions

a) Let 𝑢 = 1 − 𝑥 so that 𝑑𝑢 = −𝑑𝑥

⟹ 𝑢 = 1 if 𝑥 = 0
3
𝑢 = 1/4 if 𝑥 = 4
Thus:
3 1 1
𝑑𝑥 𝑑𝑢 1
∫04 1−𝑥 = ∫14 − 𝑢
= − ∫14 𝑢 𝑑𝑢
1
= −ln|𝑢| |14
1
= − [ln (4) − ln(1)]
1
= − ln (4)
= ln 4

b) Let 𝑢 = sin 2𝑥
So that 𝑑𝑢 = 2 cos 2𝑥 𝑑𝑥
1
⟹ 2 𝑑𝑢 = cos 2𝑥𝑑𝑥
⟹ 𝑢 = sin(0) = 0 if 𝑥 = 0
𝜋 1 𝜋
𝑢 = sin ( ) = if 𝑥 =
4 √2 8

Therefore;
𝜋 1
𝑑𝑢
∫0 sin 2𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥 = ∫0 𝑢5 ∙ cos 2𝑥 ∙ 2 cos 2𝑥
8 5 √2

1
1
6 √2
1 1 𝑢 1
= 2 ∫0 𝑢5 𝑑𝑢 = 2 ∙
√2
| = 96
6 0

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Determine:

a) ∫ 6𝑥(5𝑥 2 + 1)5 𝑑𝑥

Let 𝑢 = 5𝑥 2 + 1
So that 𝑑𝑢 = 10𝑥𝑑𝑥
Thus;
𝑑𝑢
∫ 6𝑥 (5𝑥 2 + 1)5 𝑑𝑥 = ∫ 6𝑥 ∙ 𝑢5 ∙ 10𝑥
3
= 5 ∫ 𝑢5 𝑑𝑢

3 𝑢6
= ∙ +𝑐
5 6
1
= 10 (5𝑥 2 + 1)6 + 𝑐

b) Evaluate ∫0 24 sin5 𝜃 cos 𝜃𝑑𝜃

√6

Let 𝑢 = sin 𝜃 then

𝑑𝑢 = 𝑐𝑜𝑠𝜃𝑑𝜃
Thus;
𝜋 𝜋
𝑑𝑢
∫0 24 sin5 𝜃𝑐𝑜𝑠𝜃𝑑𝜃 = ∫06 24 ∙ 𝑢5 𝑐𝑜𝑠𝜃 ∙ 𝑐𝑜𝑠𝜃
6

𝜋
= 24 ∫0 𝑢5 𝑑𝑢
6

𝜋
𝑢6 6 𝜋 1
= 24 ∙ [ 6 ] (sin (6 ) = 2)
0
1
= 16
∫ 𝑐𝑜𝑠√𝑥
c) 𝑑𝑥
√𝑥

∫ 𝑐𝑜𝑠𝑥
d) 𝑑𝑥
sin4 𝑥

e) 𝑥 2 cot ∫ 𝑥 3 𝑑𝑥

4𝜋 2 1
f) ∫𝜋2 ∙ 𝑠𝑖𝑛√𝑥 𝑑𝑥
√𝑥

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 𝜋
g) ∫𝜋2 𝑠𝑖𝑛𝜃√1 − cos 2 𝜃 𝑑𝜃 (𝑢 = 2𝑐𝑜𝑠𝜃)
3

Integration by trig substitution

Example
Integrate ∫ √1 − 𝑥 2 𝑑𝑥

Solution

Let 𝑥 = 𝑠𝑖𝑛𝜃 , 𝑑𝑥 = 𝑐𝑜𝑠𝜃𝑑𝜃

⟹ 𝑥 2 = sin2 𝜃
Therefore,
1 − 𝑥 2 = 1 − sin2 𝜃 = cos2 𝜃
Thus;
∫ √1 − 𝑥 2 𝑑𝑥 = ∫ √cos2 𝜃 𝑑𝜃 ∙ 𝑐𝑜𝑠𝜃 𝑑𝜃
= ∫ cos2 𝜃 𝑑𝜃
1
= ∫ 2 (𝑐𝑜𝑠2𝜃 + 1)𝑑𝜃
1 1
= 4 𝑠𝑖𝑛2𝜃 + 2 𝜃 + 𝑐 ⋯ (1)

But 𝑥 = 𝑠𝑖𝑛𝜃
⟹ 𝜃 = sin−1 (𝑥) and 𝑠𝑖𝑛2𝜃 = 2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 = 2√1 − sin2 𝜃 ∙ 𝑠𝑖𝑛𝜃
Replacing in (1) we have
1 1
∙ 2 √1 − sin2 𝜃 ∙ 𝑠𝑖𝑛𝜃 + 2 sin−1 𝑥 + 𝑐
4

1 1
= 2 √1 − 𝑥 2 ∙ 𝑥 + 2 sin−1 𝑥 + 𝑐

𝑥 sin−1 𝑥
= 2 √1 − 𝑥 2 + +𝑐
2

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 Summary table of integrals that require use of trigonometric substitution

𝑓(𝑥) ∫ 𝑓(𝑥)𝑑𝑥 method

1 1
1. cos 2 𝑥 (𝑥 + 2 𝑠𝑖𝑛2𝑥) + 𝑐 use cos 2𝑥 = 2 cos2 𝑥 − 1
2

1 1
2. sin2 𝑥 (𝑥 − 2 𝑠𝑖𝑛2𝑥) + 𝑐 use cos 2𝑥 = 1 − 2 sin2 𝑥
2

3. tan2 𝑥 tan 𝑥 − 𝑥 + 𝑐 use 1 + tan2 𝑥 = sec 2 𝑥

4. cot 2 𝑥 − cot 𝑥 − 𝑥 + 𝑐 use cot 2 𝑥 + 1 = 𝑐𝑜𝑠𝑒𝑐 2 𝑥

5. cos 𝑚 𝑥𝑠𝑖𝑛𝑛 𝑥 when m or n is odd but not both use cos 2 𝑥 + sin2 𝑥 = 1

When m and n are both even use either 𝑐𝑜𝑠2𝑥 = 2 cos2 𝑥 − 1

or
cos 2𝑥 = 1 − sin2 𝑥

1
6. sin 𝐴𝑐𝑜𝑠𝐵 use 2 [sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)]

1
7. 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 use 2 [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)]

1
8. 𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 use 2 [cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)]

1
9. 𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐵 use − 2 [cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵)]

1 𝑥
10. √𝑎2 sin−1 (𝑎) + 𝑐 use 𝑥 = 𝑎 𝑠𝑖𝑛𝜃 substitution
−𝑥2

𝑥 𝑎2 𝑥
11. √𝑎2 − 𝑥 2 √𝑎2 − 𝑥 2 + sin−1 (𝑎) + 𝑐 use 𝑥 = 𝑎 𝑠𝑖𝑛𝜃 substitution
2 2

1 1 𝑥
12. 𝑎2 +𝑥 2 tan−1 (𝑎) + 𝑐 use 𝑥 = tan 𝜃 substitution
𝑎

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