INTEGRATION
Tutorial Manual
MUT
Maths 1
Basic Integration
Learning Outcomes
After completing this section, you should be able to
Appreciate that integration is the reverse process of differentiation
Recognise the need for a constant of integration
Evaluate the indefinite integrals of standard forms
Evaluate the indefinite integrals of polynomials
Evaluate the indefinite functions of composite functions
Integrate all algebraic functions, transcendental and trigonometric
functions
Differentiation Revisited
When 𝑥 3 is differentiated with respect to x, the derivative is 3𝑥 2 .
Conversely, if the derivative of an unknown function is 3𝑥 2 then it is clear that
the unknown function could be 𝑥 3 .
This process of finding a function from its derivative, which reverses the
operation of differentiating, is called integrating.
The Constant of Integration
As seen above, 3𝑥 2 is the derivative of 𝑥 3 , but it is also the derivative of 𝑥 3 + 5,
𝑥 3 − 7, and in fact, the derivative of any 𝑥 3 + any constant.
Therefore the result of integrating 3𝑥 2 , which is called the integral of 3𝑥 2 , is not
a unique function but it is of the form
𝑥3 + 𝐶
where 𝐶 is any constant. Note: any letter can be used to represent a constant.
1
This is written
∫ 3𝑥 2 𝑑𝑥 = 𝑥 3 + 𝐶
where ∫ ⋯ 𝑑𝑥 means the integral of ⋯ w.r.t. x.
Integrating any function reverses the process of differentiating so, for any
function f(x), we have
𝑑
∫ 𝑓(𝑥) 𝑑𝑥 = 𝑓(𝑥) + 𝐶
𝑑𝑥
Since as we have noted above,
∫ 3𝑥 2 𝑑𝑥 = 𝑥 3 + 𝐶
it follows that
2
𝑥3
∫ 𝑥 𝑑𝑥 = +𝐶
3
1
C represents any constant so there is no need to re-write it as 𝐶.
3
In general, the derivative of 𝑥 𝑛+1 is (𝑥 + 1)𝑥 𝑛 therefore
1
∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 + 𝐶
(𝑛 + 1)
where 𝑛 ≠ −1. The case for 𝑛 = −1 is deal with later in this content area.
Integrating a Sum or Difference of Functions
We have already seen that to differentiate a sum or difference of functions, we
differentiate the function term by term. Likewise for integration, which is the
reverse for differentiation, integration of a sum or difference of functions is also
done term by term.
2
Example
1
Find the integral of 𝑥 3 + √𝑥 + .
√𝑥
Solution
1 1 1
∫ (𝑥 3 + √𝑥 + ) 𝑑𝑥 = ∫ (𝑥 3 + 𝑥 2 + 𝑥 −2 ) 𝑑𝑥
√𝑥
3 1
𝑥4 𝑥2 𝑥2
= + + +𝐶
4 (3) (1)
2 2
3
𝑥 4 2𝑥 2 1
= + + 2𝑥 2 + 𝐶
4 3
Integrate the following with respect to x
a) 𝑥 5
4
b) √𝑥
1
c) 5
𝑥2
d) 𝑥 −3
Integrating (𝒂𝒙 + 𝒃)𝒏
Let us start by considering the function 𝑓(𝑥) = (2 + 3𝑥)6 . To differentiate f(x),
we refer to Formula no. 1 (page 39 study guide) to get
𝑑
(2 + 3𝑥)6 = 𝑛[𝑓(𝑥)]𝑛−1 𝑓 ′ (𝑥) = 6(2 + 3𝑥)5 (3) = (6)(3)(2 + 3𝑥)5
𝑑𝑥
Hence
∫(6)(3)(2 + 3𝑥)5 𝑑𝑥 = (2 + 3𝑥)6 + 𝐶
or
1
∫(2 + 3𝑥)5 𝑑𝑥 = (2 + 3𝑥)6 + 𝐶
(3)(6)
3
Considering 𝑓(𝑥) = (𝑎𝑥 + 𝑏)𝑛+1 in a similar way gives the general result
1
∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = (𝑎𝑥 + 𝑏)𝑛+1 + 𝐶
(𝑎)(𝑛 + 1)
An alternative way to evaluate ∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 will be to use substitution as
follows
Let
𝑢 = 𝑎𝑥 + 𝑏
Therefore
𝑑𝑢
=𝑎
𝑑𝑥
𝑑𝑢
If we treat as a fraction, we can cross-multiply to get
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
𝑎
Now, substituting for 𝑑𝑥 and 𝑢 in the integral ∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 gives
𝑑𝑢 1
∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = ∫ 𝑢𝑛 ∙ ( ) = ∫ 𝑢𝑛 𝑑𝑢
𝑎 𝑎
You can see that we have changed the variables of integration from x to u and
just like for differentiation, the integration of the variables with respect to u
follows the same rules as the integration with respect to x that we are used to.
Therefore
1
𝑛 𝑛
1 𝑢𝑛+1 1
∫(𝑎𝑥 + 𝑏) 𝑑𝑥 = ∫ 𝑢 𝑑𝑢 = +𝐶 = 𝑢𝑛+1 + 𝐶
𝑎 𝑎 (𝑛 + 1) (𝑎)(𝑛 + 1)
However, since 𝑢 = 𝑎𝑥 + 𝑏, we do the reverse substitution for u to finally get
1
∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = (𝑎𝑥 + 𝑏)𝑛+1 + 𝐶
(𝑎)(𝑛 + 1)
4
I find the substitution method quite straight forward and really handy and will be
using it everywhere necessary in the rest of this section.
NB: The importance of writing 𝑑𝑥 in the integral expression should apparent
from the substitution. The 𝑑𝑥 term multiplies each term under the integral sign
hence the function to be integrated is put into brackets. You may be penalised if
you leave out the 𝒅𝒙 term from an integral expression.
Example
1
Integrate (4𝑥 − 1)3 with respect to x.
Solution
Let 𝑢 = 4𝑥 − 1, therefore
𝑑𝑢
=4
𝑑𝑥
𝑑𝑢
→ = 𝑑𝑥
4
Therefore
1 1 𝑑𝑢 1 1
∫(4𝑥 − 1)3 𝑑𝑥 = ∫ 𝑢3 ∙ ( ) = ∫ 𝑢3 𝑑𝑢
4 4
4
1 𝑢3 3 4
= ( )+𝐶 = 𝑢3 + 𝐶
4 4 16
3
By reverse substitution therefore,
1 3 4
∫(4𝑥 − 1)3 𝑑𝑥 = (4𝑥 − 1)3 + 𝐶
16
5
Determine the following integrals by using appropriate substitution
a) ∫(2 − 5𝑥)2 𝑑𝑥
b) ∫(4 − 3𝑥)−5 𝑑𝑥
c) ∫ √2 − 3𝑥 𝑑𝑥
1
d) ∫ 𝑑𝑥
1−𝑥
e) ∫ √3𝑥 + 7 𝑑𝑥
1
f) ∫ (𝑥 3 − (1−𝑥)3 ) 𝑑𝑥
g)
Standard Integrals
Whenever a function f(x) is recognised as the derivative of a function f(x) then
𝑑
𝑓(𝑥) = 𝑓 ′ (𝑥)𝑑𝑥 ⇒ ∫ 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑥) + 𝐶
𝑑𝑥
Thus an integral whose derivative is known can be established as a standard
integral.
Integrating Exponential Functions
𝑑
From your lectures, you already know that 𝑒𝑥 = 𝑒𝑥
𝑑𝑥
hence
∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
It also follows that
∫ 𝑐𝑒 𝑥 𝑑𝑥 = 𝑐 ∫ 𝑒 𝑥 𝑑𝑥 = 𝑐𝑒 𝑥 + 𝐶
6
We can use the standard formula for ∫ 𝑒 𝑥 𝑑𝑥 and substitution to generate the
integral of ∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 as follows.
Example
Evaluate ∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥
Solution
𝑑𝑢
Let 𝑢 = 𝑎𝑥 + 𝑏 ⇒ =𝑎
𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
𝑎
Hence
𝑑𝑢 1
∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 = ∫ 𝑒 𝑢 ( ) = ∫ 𝑒 𝑢 𝑑𝑢
𝑎 𝑎
1 𝑢
= 𝑒 +𝐶
𝑎
By reverse substitution therefore, we have
1 𝑎𝑥+𝑏
∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 = 𝑒 +𝐶
𝑎
Example
Evaluate ∫ 2𝑒 1−5𝑥 𝑑𝑥
Solution
𝑑𝑢
Let 𝑢 = 1 − 5𝑥 ⇒ = −5
𝑑𝑥
𝑑𝑢
∴ − = 𝑑𝑥
5
7
Hence the integral becomes
𝑑𝑢
∫ 2𝑒 1−5𝑥 𝑑𝑥 = 2 ∫ 𝑒 1−5𝑥 𝑑𝑥 = 2 ∫ 𝑒 𝑢 (− )
5
2 2
= − ∫ 𝑒 𝑢 𝑑𝑢 = − 𝑒 𝑢 + 𝐶
5 5
2
∴ ∫ 2𝑒 1−5𝑥 𝑑𝑥 = − 𝑒 1−5𝑥 + 𝐶
5
Integrate each function with respect to x
a) 𝑒 4𝑥
b) 4𝑒 −𝑥
c) 𝑒 3𝑥−2
d) 6𝑒 −2𝑥
e) 5𝑒 𝑥−3
f) 𝑒 (2+𝑥⁄2)
You already know from differentiation Formula no. 3 that
𝑑 𝑥
𝑎 = 𝑎 𝑓(𝑥) 𝑓 ′ (𝑥) ln 𝑎 = 𝑎 𝑥 ∙ 1 ∙ ln 𝑎 = 𝑎 𝑥 ln 𝑎
𝑑𝑥
Therefore
∫(𝑎 𝑥 ) ln 𝑎 𝑑𝑥 = ln 𝑎 ∫ 𝑎 𝑥 𝑑𝑥 = 𝑎 𝑥 + 𝐶
1 𝑥
⇒ ∫ 𝑎 𝑥 𝑑𝑥 = 𝑎 +𝐶
ln 𝑎
We can use the above standard integral to integrate functions of the form 𝑎𝑏𝑥+𝑐
8
Example
Integrate 2(3𝑥−2) with respect to x.
Solution
There are two equivalent approaches in this type of integral and I will do them
both. You are free to select the one you are confident with.
Approach (1)
𝑑𝑢
Let 𝑢 = 3𝑥 − 2 ⇒ =3
𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
3
Therefore
𝑑𝑢 1 1 1 𝑢
∫ 2(3𝑥−2) 𝑑𝑥 = ∫ 2𝑢 ( ) = ∫ 2𝑢 𝑑𝑢 = ( 2 )+𝐶
3 3 3 ln 2
1 1 (3𝑥−2)
∴ ∫ 2(3𝑥−2) 𝑑𝑥 = 2(3𝑥−2) + 𝐶 = 2 +𝐶
3 ln 2 ln 8
Approach (2)
23𝑥 1
∫ 2(3𝑥−2) 𝑑𝑥 = ∫ 2
𝑑𝑥 = ∫ 23𝑥 𝑑𝑥
2 4
𝑑𝑢
Let 𝑢 = 3𝑥 ⇒ =3
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
3
Hence
1 1 𝑑𝑢 1
∫ 23𝑥 𝑑𝑥 = ∫ 2𝑢 ( ) = ∫ 2𝑢 𝑑𝑢
4 4 3 12
1 1 𝑢 1
= ( 2 )+𝐶 = 23𝑥 + 𝐶
12 ln 2 12 ln 2
9
It is not necessary to go beyond this stage since this is a complete answer but I
will do so only to show that the two solutions are behind.
1
∴ ∫ 2(3𝑥−2) 𝑑𝑥 = 23𝑥 + 𝐶
4 × 3 ln 2
1 3𝑥
1 3𝑥−2
= 2 + 𝐶 = 2 +𝐶
22 × ln 8 ln 8
You are free to choose either of the two alternatives, both solutions are
acceptable.
Integrate each function with respect to x
a) 2𝑥
b) 4(2+𝑥)
c) 𝑎1−2𝑥
d) 2𝑥 + 𝑥 2
𝟏
How to Integrate
𝒙
You will have realised that it not possible to use the standard integral for 𝑥 𝑛 to
1
integrate . Check
𝑥
1 𝑥0
∫ 𝑑𝑥 = +𝐶
𝑥 0
which is undefined because of the division by zero. However, you do remember
from Formula no. 4 that
𝑑 1
ln 𝑥 =
𝑑𝑥 𝑥
Since we know that ln x is defined when x > 0, it therefore implies that
10
𝑑 1 1
ln 𝑥 = ⇔ ∫ 𝑑𝑥 = ln |𝑥| + 𝐶
𝑑𝑥 𝑥 𝑥
I will explain the use of |x| when we come to definite integration. You need to
carefully study this result as it is of great importance when you get to Mathematics
2 in integration by partial fractions and homogenous ordinary differential
equations.
As before, we can easily use the above result to obtain the standard integral for
1
𝑎𝑥+𝑏
Example
5
Evaluate ∫ 𝑑𝑥
6−7𝑥
Solution
𝑑𝑢
Let 𝑢 = 6 − 7𝑥 ⇒ = −7
𝑑𝑥
𝑑𝑢
∴ − = 𝑑𝑥
7
Hence
5 5 𝑑𝑢 5 1 5
∫ 𝑑𝑥 = ∫ (− ) = − ∫ 𝑑𝑢 = − ln|𝑢| + 𝐶
6 − 7𝑥 𝑢 7 7 𝑢 7
5 5
∴ ∫ 𝑑𝑥 = − ln|6 − 7𝑥| + 𝐶
6 − 7𝑥 7
5
or 𝐶 = − ln 𝐴
7
5 5 5 5 𝐴
= − ln|6 − 7𝑥| − ln 𝐴 = − ln 𝐴|6 − 7𝑥| = ln
7 7 7 7 |6 − 7𝑥|
Integrate w.r.t. x giving each answer in a form which
(a) uses C (b) uses ln 𝐴 and is simplified
11
1
1)
2𝑥
1
2)
3𝑥+1
6
3)
2+3𝑥
3
4)
4−2𝑥
4
5)
1−𝑥
Integrating Trigonometric Functions
The integration of trig functions follows a similar suit. I will illustrate a couple of
examples since the workings are the same as those for the other functions outlined
earlier.
Example
Integrate the following functions w.r.t. x
𝜋
a) sin ( + 𝑥)
4
𝜋
b) 3 cos (4𝑥 − )
2
𝜋
c) sec 2 ( + 2𝑥)
3
Solution
a)
𝑑
cos 𝑥 = − sin 𝑥 ⇔ ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶
𝑑𝑥
𝜋 𝑑𝑢
Let 𝑢 = + 𝑥 ⇒ =1
4 𝑑𝑥
∴ 𝑑𝑢 = 𝑑𝑥
12
Hence
𝜋
∫ sin ( + 𝑥) 𝑑𝑥 = ∫ sin 𝑢 𝑑𝑢 = − cos 𝑢 + 𝐶
4
𝜋 𝜋
∴ ∫ sin ( + 𝑥) 𝑑𝑥 = − cos ( + 𝑥) + 𝐶
4 4
b)
𝑑
sin 𝑥 = cos 𝑥 ⇔ ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
𝑑𝑥
𝜋 𝑑𝑢
Let 𝑢 = 4𝑥 − ⇒ =4
2 𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
4
Hence
𝜋 𝑑𝑢 3 3
∫ 3 cos (4𝑥 − ) 𝑑𝑥 = ∫ 3 cos 𝑢 ( ) = ∫ cos 𝑢 𝑑𝑢 = − sin 𝑢 + 𝐶
2 4 4 4
𝜋 3 𝜋
∫ 3 cos (4𝑥 − ) 𝑑𝑥 = − sin (4𝑥 − ) + 𝐶
2 4 2
c)
𝑑
tan 𝑥 = sec 2 𝑥 ⇔ ∫ sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
𝑑𝑥
𝜋 𝑑𝑢
Let 𝑢 = + 2𝑥 ⇒ =2
3 𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
2
Hence
𝜋 𝑑𝑢 1 1
∫ sec 2 ( + 2𝑥) 𝑑𝑥 = ∫ sec 2 𝑢 ( ) = ∫ sec 2 𝑢 𝑑𝑢 = tan 𝑢 + 𝐶
3 2 2 2
𝜋 1 𝜋
∴ ∫ sec 2 ( + 2𝑥) 𝑑𝑥 = tan ( + 2𝑥) + 𝐶
3 2 3
13
Integration Functions of the Form 𝒇′(𝒙)[𝒇(𝒙)]𝒏
The substitution method comes in handy when we want to integrate functions of
the form ∫ 𝑓 ′ (𝑥) [𝑓(𝑥)]𝑛 𝑑𝑥. Let us briefly look at a few examples since all the
methods we have been using readily apply in such instances. What counts
however is the realisation aspect that indeed the function is in this form.
Example
Evaluate the following integrals
a) 𝑥√1 − 𝑥 2 𝑑𝑥
b) ∫ cos 2𝑥 + 3 (sin 2𝑥 + 3)2 𝑑𝑥
c) ∫ 𝑒 𝑥 √1 + 𝑒 𝑥 𝑑𝑥
d) ∫ sin 𝜃 √(1 − cos 𝜃)𝑑𝜃
e) ∫ sec 2 𝑥tan3 𝑥 𝑑𝑥
f) ∫(𝑥 + 1) √𝑥 2 + 2𝑥 + 3𝑑𝑥
ln 𝑥
g) ∫ 𝑑𝑥
𝑥
h)
Solution
𝑑𝑢
a) Let 𝑢 = 1 − 𝑥 2 ⇒ = −2𝑥
𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
−2𝑥
Hence
3
𝑑𝑢 1 1 𝑢2
∫ 𝑥 √1 − 𝑥 2 𝑑𝑥 = ∫ 𝑥 √𝑢 (− ) = − ∫ √𝑢 𝑑𝑢 = − +𝐶
2𝑥 2 2 (3)
2
1 3
∴ ∫ 𝑥 √1 − 𝑥 2 𝑑𝑥 = − (1 − 𝑥 2 )2 + 𝐶
3
14
𝑑𝑢
b) Let 𝑢 = sin 2𝑥 + 3 ⇒ = 2 cos(2𝑥 + 3)
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
2 cos(2𝑥 + 3)
Hence
𝑑𝑢
∫ cos(2𝑥 + 3) (sin 2𝑥 + 3)2 𝑑𝑥 = ∫ cos(2𝑥 + 3) 𝑢2 ( )
2 cos 2𝑥 + 3
1
= ∫ 𝑢2 𝑑𝑢
2
1 𝑢3 1
= ( ) + 𝐶 = (sin 2𝑥 + 3)3 + 𝐶
2 3 6
1
∴ ∫ cos(2𝑥 + 3) (sin 2𝑥 + 3)2 𝑑𝑥 = (sin 2𝑥 + 3)3 + 𝐶
6
𝑑𝑢
c) Let 𝑢 = 1 + 𝑒 𝑥 ⇒ = 𝑒𝑥
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
𝑒𝑥
Hence
3
𝑑𝑢 𝑢2 2 3
∫ 𝑒 𝑥 √1 + 𝑒 𝑥 𝑑𝑥 = ∫ 𝑒 𝑥 √𝑢 ( 𝑥 ) = ∫ √𝑢 𝑑𝑢 = + 𝐶 = 𝑢2 + 𝐶
𝑒 3 3
(2)
2 3
∴ ∫ 𝑒 𝑥 √1 − 𝑒 𝑥 𝑑𝑥 = (1 + 𝑒 𝑥 )2 + 𝐶
3
NB: Remember C, the constant of integration
𝑑𝑢
d) Let 𝑢 = 1 − cos 𝜃 ⇒ = sin 𝜃
𝑑𝜃
𝑑𝑢
∴ = 𝑑𝜃
sin 𝜃
Hence
15
3
𝑑𝑢 𝑢2
∫ sin 𝜃√1 − cos 𝜃 𝑑𝜃 = ∫ sin 𝜃 √𝑢 ( ) = ∫ √𝑢 𝑑𝑢 = +𝐶
sin 𝜃 3
(2)
2 3 2 3
= 𝑢2 + 𝐶 = (1 − cos 𝜃)2 + 𝐶
3 3
NB: Always remember the constant of integration
e) Remember tan3 𝑥 = (tan 𝑥)3
𝑑𝑢
Let 𝑢 = tan 𝑥 ⇒ = sec 2 𝑥
𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
sec 2 𝑥
Hence
2 3
𝑑𝑢 2 3 3
𝑢4
∫ sec 𝑥 tan 𝑥 𝑑𝑥 = ∫ sec 𝑥 (𝑢 ) ( 2 ) = ∫ 𝑢 𝑑𝑥 = +𝐶
sec 𝑥 4
(tan 𝑥)4
= +𝐶
4
𝑑𝑢
f) Let 𝑢 = 𝑥 2 + 2𝑥 + 3 ⇒ = 2𝑥 + 2 = 2(𝑥 + 1)
𝑑𝑥
𝑑𝑢
∴ = 𝑑𝑥
2(𝑥 + 1)
𝑑𝑢 1
∫(𝑥 + 1)√𝑥 2 + 2𝑥 + 3 𝑑𝑥 = ∫(𝑥 + 1)√𝑢 ( ) = ∫ √𝑢 𝑑𝑥
2(𝑥 + 1) 2
3
1 𝑢2
= +𝐶
2 (3)
2
3
(𝑥 2 + 2𝑥 + 3)2
= +𝐶
3
g) Initially, this looks like a fraction but once we recognise that it is the
1 𝑑
product of and ln 𝑥, it is clear that (ln 𝑥) and we can make the
𝑥 𝑑𝑥
substitution 𝑢 = ln 𝑥.
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𝑑𝑢 1
Therefore, let 𝑢 = ln 𝑥 ⇒ =
𝑑𝑥 𝑥
𝑥𝑑𝑢 = 𝑑𝑥
Hence
ln 𝑥 𝑢 𝑢2 (ln 𝑥)2
∫ 𝑑𝑥 = ∫ (𝑥𝑑𝑢) = ∫ 𝑢 𝑑𝑢 = +𝐶 = +𝐶
𝑥 𝑥 2 2
NB: Always remember the constant of integration
Turn to page 31 of the study guide. On Question 2, do all the questions
from 2.1 to 2.10.
In the next section, we will be looking at the evaluation of definite integrals where
limits of integration are applied.
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