This document outlines a lecture on anti-derivatives given by Dr. Kamel ATTAR. The lecture covers various topics related to anti-derivatives including definitions, indefinite integrals, integration by substitution, integration by parts, and examples of evaluating definite and indefinite integrals of various functions. Dr. ATTAR provides examples and exercises for students to practice evaluating integrals using the techniques covered in the lecture.
1. Faculty of Sciences (Section V)
Lebanese University
Chapter 1: Anti-derivatives
Dr. Kamel ATTAR
attar.kamel@gmail.com
Lecture #1 F Tuesday 16/MAR/2021 F
2. 2Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
1 Anti-derivatives
Definition
Integration by Change of Variable
Integration by Parts
2 Integration of a Rational Functions
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Factor
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Factor
3 Trigonometric Integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
4 Abelian integrals
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
3. 3Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
H Anti-derivatives H
Definition
Definition
A function F is called anti-derivative (or primitive) of f on an interval I, iff:
I F is defined and differentiable on I, and
I F0
(x) = f(x) ∀x ∈ I .
Remark: We use capital letters such as F to represent an anti-derivative of a
function f.
Theorem
If F is an anti-derivative of f on I, then the set of anti-derivatives of f on I is
consisted of functions
F(x) + C ,
where C is an arbitrary constant.
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
4. 4Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find an antiderivatives for each of the following functions:
1. f(x) = 2x.
2. g(x) = cos x.
3. h(x) = 2x + cos x
Solution
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
5. 5Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find an antiderivatives for each of the following functions:
1. f(x) = 2x.
2. g(x) = cos x.
3. h(x) = 2x + cos x
Solution
We need to think backward here: What function do we know has a derivative
equal to the given function?
1. F(x) = x2
.
2. G(x) = sin x.
3. H(x) = x2
+ sin x
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
6. 6Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find an antiderivative of f(x) = 3x2
that satisfies F(1) = −1.
Solution
Since the derivative of x3
is 3x2
, the general antiderivative
F(x) = x3
+ C
gives all the antiderivative of f(x). The condition F(1) = −1 determine a
specific value for C. Substituting x = 1 into F(x) + x3
+ C gives
F(1) = (1)3
+ C = 1 + C =⇒ C = −2
So F(x) = x3
− 2 .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
7. 7Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Indefinite Integral
Definition
The collection of all anti-derivatives of f is called the indefinite integral of f
with respect to x, and is denoted by
Z
f(x) dx .
The symbol
Z
is an integral sign. The function f is the integrate of the integral,
and x is the variable of integration.
Z
f(x) dx = F(x) + C , F0
(x) = f(x) and (C is an arbitrary constant) .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
8. 8Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Commonly Occurring Indefinite Integrals
Z
0 dx = C
Z
a dx = ax + C
Z
xn
dx =
xn+1
n + 1
+ C , n 6= −1
Z
1
xn
dx =
−1
(n − 1)xn−1
+ C .
Z
1
x
dx = ln |x| + C
Z
ex
dx = ex
+ C
Z
sin(x) dx = − cos(x) + C
Z
cos(x) dx = − sin(x) + C
Z
1
a2 + x2
dx =
1
a
arctan
x
a
+ C
Z
1
a2 − x2
dx =
1
a
arg tanh
x
a
Z
1
√
a2 − x2
dx = arcsin
x
a
+ C
Z
1
√
a2 − x2
dx = − arccos
x
a
+ C
Z
1
√
x2 + a2
dx = arg sinh
x
a
+ C
Z
1
√
x2 − a2
dx = arg cosh
x
a
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
9. 9Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Operations on Primitives
Theorem
Let f1 and f2 two functions defined on I and λ1, λ2 ∈ R then:
Z
λ1f1(x) + λ2f2(x)
dx = λ1
Z
f1(x)dx + λ2
Z
f2(x)dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
10. 10Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Evaluate
Z
(x2
− 2x + 5) dx .
Solution
Z
(x2
− 2x + 5) dx =
Z
x2
dx −
Z
2x dx +
Z
5 dx
=
Z
x2
dx − 2
Z
x dx + 5
Z
1 dx
=
x3
3
+ C1
− 2
x2
2
+ C2
+ 5 (x + C3)
=
x3
3
+ C1 − x2
− 2C2 + 5x + 5C3
=
x3
3
− x2
+ 5x + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
11. 11Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Integration by Change of Variable
Theorem
If ϕ : [α, β] → R is a continuously differentiable function and if f is continuous
mapping on an interval contained ϕ
[α, β]
, then :
Z
f
ϕ(x)
ϕ0
(x) dx =
Z
f(t) dt
We can say that we perform the change of variable t = ϕ(x).
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
12. 12Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Change of variables (u Substitution)
Z
u0
u
= ln |u| + C
Z
u0
un
=
−1
(n − 1)un−1
+ C
Z
u0
un
=
un+1
n + 1
+ C
Z
u0
eu
= eu
+ C
Z
u0
cos(u) = − sin(u) + C
Z
u0
sin(u) = cos(u) + C
Z
u0
a2 + u2
=
1
a
arctan
u
a
+ C
Z
u0
a2 − u2
=
1
a
arg tanh
u
a
+ C
Z
u0
√
a2 − u2
= arcsin
u
a
+ C
Z
u0
√
a2 − u2
= − arccos
u
a
+ C
Z
u0
√
u2 + a2
= arg sinh
u
a
+ C
Z
u0
√
u2 − a2
= arg cosh
u
a
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
13. 13Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Calculate I =
Z
tan x dx =
Z
sin x
cos x
dx and J =
Z p
2x + 1 dx.
Solution
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
14. 14Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Calculate I =
Z
tan x dx =
Z
sin x
cos x
dx and J =
Z p
2x + 1 dx.
Solution
Let u = cos x. Then u0
= − sin x dx, and so
I =
Z
−
u0
u
= − ln |u| + C = − ln | cos x| + C .
Let u = 2x + 1. Then u0
= 2 dx, and so
Z p
2x + 1 dx =
Z p
2x + 1 dx =
1
2
Z
u
1
2 du =
1
3
u3/2
+ C
=
1
3
(2x + 1)3/2
+ C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
15. 15Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Calculate I =
Z
xdx
p
1 − x2
with x 6= ±1 and J =
Z
(x3
+ x)5
(3x2
+ 1) dx.
Solution
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
16. 16Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Calculate I =
Z
xdx
p
1 − x2
with x 6= ±1 and J =
Z
(x3
+ x)5
(3x2
+ 1) dx.
Solution
Let u = 1 − x2
. Then u0
= −2x, and so
I = −
Z
−2xdx
2
p
1 − x2
= −
Z
u0
2
√
u
= −
√
u = −
p
1 − x2 + C .
We set u = x3
+ x. Then u0
= (3x2
+ 1)dx , and so by substitution we have
J =
Z
(x3
+ x)5
(3x2
+ 1) dx =
Z
u5
u0
=
u6
6
+ C =
(x3
+ x)6
6
+ C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
17. 17Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Evaluate
Z
2z dz
3
√
z2 + 1
Solutions
1. Substitute u = z2
+ 1 then du = 2zdz and so
Z
2z dz
3
√
z2 + 1
=
Z
du
u1/3
=
Z
u−1/3
du =
u2/3
2/3
+ C
=
3
2
u2/3
+ C =
3
2
(z2
+ 1)2/3
+ C
2. Substitute u =
3
p
z2 + 1 instead then u3
= z2
+ 1 and 3udu = 2zdz
Z
2z dz
3
√
z2 + 1
=
Z
3u2
du
u
= 3
Z
u du = 3
u2
2
+ C =
3
2
(z2
+ 1)2/3
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
18. 18Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Exercise -A-
1.
Z
2(2x + 4)5
dx 2.
Z
(3x + 2)(3x2
+ 4x)4
dx 3.
Z
x2
sin(x3
) dx
4.
Z
4x3
(x4 + 1)2
dx 5.
Z
2x dx
3
√
x2 + 1
6.
Z
x
p
2x + 1 dx
7.
Z
7
p
7x − 1 dx 8.
Z
2x(x2
+ 5)−4
dx 9.
Z
9x2
√
1 − x3
dx
10.
Z
1
x2
r
2 −
1
x
dx 11.
Z
1 − cos
x
2
2
sin
x
2
dx 12.
Z
x sin(2x2
) dx
13.
Z
(1 +
√
x)1/3
√
x
dx 14.
Z
√
x sin2
(x3/2
− 1) dx 15.
Z
1
x2
cos2
1
x
dx
16.
Z
(x + 5)(x − 5)
1
3 dx 17.
Z
1
√
5x + 8
dx 18.
Z
x
4
p
1 − x2 dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
19. 19Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
19.
Z
1
√
x(1 +
√
x)2
dx 20.
Z
sin5 x
3
cos
x
3
dx 21.
Z
sin(2x + 1)
cos2(2t + 1)
dx
22.
Z
1
x2
cos
1
x
− 1
dx 23.
Z
1
√
x
cos(
√
x + 3) dx 24.
Z
x3
p
x2 + 1 dx
25.
Z
1
x2
sin
1
x
cos
1
x
dx 26.
Z
1
x3
r
x2 − 1
x2
dx 27.
Z
x
p
4 − x dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
20. 20Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Integration by Parts
Sometimes we meet an integration that is the product of 2 functions. We may
be able to integrate such products by using Integration by Parts
Theorem
Let u and v two differentiable functions on interval I. If vu0
admits on this
interval, then uv0
admits also a primitive and:
Z
u(x)v0
(x) dx = u(x)v(x) −
Z
v(x)u0
(x) dx
u v0
×
u0
←−
−
R v
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
21. 21Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Calculate
Z
x sin 2xdx
Solution
We need to choose u. We could let u = x or u = sin 2x, but usually only one of them
will work. In general, we choose the one that allows to be of a simpler form than u. So
for this example,
u = x v0
= sin 2x
u0
= 1 ←− v = −
1
2
cos 2x
−
Z
Substituting these 4
expressions into the integration by parts formula, we get
Z
x sin 2x dx = x
−
1
2
cos 2x
−
Z
−
1
2
cos 2x
1dx
= −
1
2
x cos 2x +
1
2
Z
cos 2x dx = −
1
2
x cos 2x +
1
4
sin 2x .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
22. 22Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find
Z
x ln x dx.
Solution
It would be natural to choose u = x so that when we differentiate it we getx0
= 1.
However v0
this choice would mean choosing v0
= lnx and we would need to be able to
integrate this. This integral is not a known standard form. So, in this Example we will
choose
u = ln x v0
= x
u0
=
1
x
←− v =
x2
2
−
Z
Then, applying the formula
Z
x ln x dx =
x2
2
ln x −
Z
x2
2
×
1
x
dx =
x2
2
ln x −
x2
4
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
23. 23Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Z
x cos xdx = x sin x + cos x + C u = x v0
= cos x
u0
= 1 v = sin x
Z
xex
dx = −xex
− ex
+ C u = x v0
= ex
u0
= 1 v = ex
Z
ln xdx = x ln x − x + C u = ln x v0
= 1
u0
=
1
x
v = x
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
24. 24Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find
Z
x2
e3x
dx.
Solution
We have to make a choice and let one of the functions in the product equal u and one
equal to v0
. As a general rule we let u be the function which will become simpler when
we differentiate it. In this case it makes sense to let
u = x2
and v0
= e3x
then
u0
= 2x and v =
1
3
e3x
Then, using the formula for integration by parts,
Z
x2
e3x
dx =
1
3
e3x
· x2
−
Z
1
3
e3x
· 2x dx =
1
3
x2
e3x
−
2
3
Z
xe3x
dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
25. 25Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
The resulting integral is still a product. It is a product of the functions
2
3
x and
e3x
. We can use the formula again. This time we choose
u =
2
3
x and v0
= e3x
then
u0
=
2
3
and v =
1
3
e3x
So
Z
x2
e3x
dx =
1
3
x2
e3x
−
2
3
Z
xe3x
dx
=
1
3
x2
e3x
−
2
3
x ·
1
3
e3x
−
Z
1
3
e3x
·
2
3
dx
=
1
3
x2
e3x
−
2
9
xe3x
+
2
27
e3x
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
26. 26Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Example
Find I =
Z
ex
sin x dx.
Solution
Whichever terms we choose for u and v0
it may not appear that integration by
parts is going to produce a simpler integral. Nevertheless, let us make a choice:
u = ex
and v0
= sin x
then
u0
= ex
and v = − cos x
Then,
I =
Z
ex
sin xdx = −ex
cos x +
Z
ex
cos x dx .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
27. 27Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
We now integrate by parts again choosing
u = ex
and v0
= cos x
then
u0
= ex
and v = sin x
Then,
I =
Z
ex
sin xdx = −ex
cos x +
Z
ex
cos x dx
= −ex
cos x +
ex
sin x −
Z
sin x · ex
dx
= −ex
cos x + ex
sin x −
Z
ex
sin x dx .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
28. 28Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
WNotice that the integral we have ended up with is exactly the same as the one
we started with. So
I = −ex
cos x + ex
sin x − I ⇐⇒ 2I = −ex
cos x + ex
sin x
⇐⇒ I =
1
2
(ex
sin x − ex
cos x) .
So Z
ex
sin xdx =
1
2
(ex
sin x − ex
cos x) + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
29. 29Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
Exercise -B-
1.
Z
x sin x dx 2.
Z
x cos 4x dx 3.
Z
xe−x
dx 4.
Z
x2
cos x dx
5.
Z
2x2
ex
dx 6.
Z
x2
ln x dx 7.
Z
arctan x dx 8.
Z
arcsin x dx
9.
Z
ex
cos x dx 10.
Z
sin3
x dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
30. 30Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Definition
Integration by Change of Variable
Integration by Parts
ANSWERS
1. −x cos x + sin x + C 2.
1
4
x sin 4x +
1
16
cos 4x + C
3. −xex
− e−x
+ C 4. x2
sin x + 2x cos x − 2 sin x + C
5. 2x2
ex
− 4xex
+ 4ex
+ C 6.
1
3
x3
ln x −
1
9
x3
+ C
7. x arctan x −
1
2
ln |1 + x2
| + C 8. x arcsin x +
p
1 − x2 + C
9.
1
2
ex
(cos x + sin x) + C 10. −
1
3
(cos x sin2
x + 2 cos x) + C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
31. 31Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
H Integration of a Rational Functions H
Z
5x − 3
(x + 1)(x − 3)
=??
The method for rewriting rational functions as a sum of simpler fractions is
called the method of partial fractions.
5x − 3
(x + 1)(x − 3)
=
A
x + 1
+
B
x − 3
To find A and B, we first clear the equation of fractions and regroup in powers
of x, obtaining
5x − 3 = A(x − 3) + B(x + 1) = (A + B)x − 3A + B
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
32. 32Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
This will be an identity in x if and only if the coefficients of like powers of x on
the two sides are equal:
A + B = 5 , −3A + B = −3
Solving these equations simultaneously gives A = 2 and B = 3.
Thus we simply sum the integrals of the fractions on the right side:
Z
5x − 3
(x + 1)(x − 3)
=
Z
2
x + 1
+
3
x − 3
dx =
Z
2
x + 1
dx +
Z
3
x − 3
dx
= 2 ln |x + 1| + 3 ln |x − 3| + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
33. 33Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
H Expressing a Fractional Function In Partial Fractions H
RULE ¬: Before a fractional function can be expressed directly in partial
fractions, the numerator must be of at least one degree less than the
denominator.
RULE : Denominator Containing Linear Factors:
f(x)
(x − r1)(x − r2) · · · (x − rn)
=
A1
(x − r1)
+
A2
(x − r2)
+ · · · +
An
(x − rn)
Example
Calculate
Z
x2
+ 4x + 1
(x − 1)(x + 1)(x + 3)
dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
34. 34Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Solution
The partial fraction decomposition has the form
x2
+ 4x + 1
(x − 1)(x + 1)(x + 3)
=
A
(x − 1)
+
B
(x + 1)
+
C
(x + 3)
.
To find the values of the undetermined coefficients A, B, and C, we use two
methods:
First method: We clear fractions and get
x2
+ 4x + 1 = A(x + 1)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x + 1)
= A(x2
+ 4x + 3) + B(x2
+ 2x − 3) + C(x2
− 1)
= (A + B + C)x2
+ (4A + 2B)x + (3A − 3B − C) .
The polynomials on both sides of the above equation are identical, so we
equate coefficients of like powers of x, obtaining
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
35. 35Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Coefficient of x2
: A + B + C = 1
Coefficient of x1
: 4A + 2B = 4
Coefficient of x0
: 3A − 3B − C = 1
The solution is A = 3/4, B = 1/2, and C = −1/4.
Second method: We clear fractions and get
x2
+ 4x + 1 = A(x + 1)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x + 1)
Now we take values for x :
• for x = 1 we get 6 = 8A then A =
3
4
.
• for x = −1 we get −2 = −4B then B =
1
2
.
• for x = −3 we get −2 = 8C then C = −
1
4
.
Hence we have
Z
x2 + 4x + 1
(x − 1)(x + 1)(x + 3)
dx =
Z
3/4
x − 1
+
1/2
x + 1
−
1/4
x + 3
dx
=
3
4
ln |x − 1| +
1
2
ln |x + 1| −
1
4
ln |x + 3| + C.
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
36. 36Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
RULE ®: Denominator Containing reducible Quadratic Factor:
Corresponding to any quadratic factor ax2
+ bx + c = a(x − x1)(x − x2) in the
denominator, there will be a partial fraction of the form
A
x − x1
+
B
x − x2
.
Example
Calculate
Z
5x − 1
x2 − x − 2
dx
Solution
First we factorise x2
− x − 2 = (x − 2)(x + 1). Then, we can write
5x − 1
x2 − x − 2
=
5x − 1
(x − 2)(x + 1)
=
2
x + 1
+
3
x − 2
Therefore
Z
5x − 1
x2 − x − 2
dx =
Z
2
x + 1
dx +
Z
3
x − 2
dx = 2 ln |x + 1| + 3 ln |x − 2| + C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
37. 37Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Example
Find
Z
x2 + 1
x2 − 5x + 6
dx
solution
In this case, the integrand is NOT a proper rational function. Hence we divide (x2
+ 1) by
(x2
− 5x + 6) and get,
x2 + 1
x2 − 5x + 6
= 1 +
(5x − 5)
(x2 − 5x + 6)
= 1 +
5x − 5
(x − 2)(x − 3)
Now, let’s look at the second half of the above equation and let
5x − 5
(x − 2)(x − 3)
=
A
x − 2
+
B
x − 3
On solving it, we get 5x − 5 = A(x − 3) + B(x − 2) =
⇒ A = −5 and B = 10 .
Hence, we have
Z
x2 + 1
x2 − 5x + 6
dx = 5 − 5 ln |x − 2| + 10 ln |x − 3| + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
38. 38Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
RULE ¯: Denominator Containing Repeated Linear Factors:
If a linear factor is repeated n times in the denominator, there will be n
corresponding partial fractions with degree 1 to n.
f(x)
(x − r)n
=
A1
(x − r)
+
A2
(x − r)2
+ · · · +
An
(x − r)n
.
Knowing that
Z
dx
(ax + b)n
=
1
a
ln |ax + b| si n = 1
−
1/a
ax + b
si n = 2
−
1/a
(n − 1)(ax + b)n−1
si n ≥ 2
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
39. 39Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Example
Calculate
Z
6x + 7
(x + 2)2
dx
Solution
The partial fraction decomposition has the form
6x + 7
(x + 2)2
=
A
x + 2
+
B
(x + 2)2
We clear fractions and get: 6x + 7 = A(x + 2) + B
• for x = −2 we get B = −5.
• for x = 0 we get 7 = 2A + B then A = 6.
Hence we have
Z
6x + 7
(x + 2)2
dx =
Z
6
x + 2
−
5
(x + 2)2
dx = 6 ln |x + 2| +
5
x + 2
+ C.
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
40. 40Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Example
Find
Z
3x − 2
(x + 1)2(x + 3)
dx
Solution
We have
3x − 2
(x + 1)2(x + 3)
=
A
x + 1
+
B
(x + 1)2
+
C
x + 3
.
On solving it, we get
3x − 2 = A(x2
+ 4x + 3) + B(x + 3) + C(x2
+ 2x + 1) .
We take x = −1 then x = −3 and then x = 0 to get A = 11/4, B − 5/2 and C = −11/4.
Hence, we have
Z
3x − 2
(x + 1)2(x + 3)
dx =
11
4
Z
1
x + 1
dx −
5
2
Z
1
(x + 1)2
dx −
11
4
Z
1
x + 3
dx
=
11
4
ln |x + 1| +
5/2
x + 1
−
11
4
ln |x + 3| + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
41. 41Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
RULE °: Denominator Containing Irreducible Quadratic Factor:
Corresponding to any quadratic factor ax2
+ bx + c with ∆ = b2
− 4c 0 in the denominator,
there will be a partial fraction of the form
Ax + B
ax2 + bx + c
.
Example
Find
Z
x2 + x + 1
(x + 2)(x2 + 1)
dx
Solution
We have
x2 + x + 1
(x + 2)(x2 + 1)
=
A
x + 2
+
Bx + C
x2 + 1
. On solving this equation, we get
x2
+ x + 1 = A(x2
+ 1) + (Bx + C)(x + 2)
Take x = −2 to get A =
3
5
, x = 0 to get C =
1
5
and x = 1 to get B =
2
5
. Therefore
Z
x2 + x + 1
(x + 2)(x2 + 1)
dx =
3
5
Z
1
x + 2
dx +
1
5
Z
2x + 1
x2 + 1
dx
=
3
5
Z
1
x + 2
dx +
1
5
Z
2x
x2 + 1
dx +
1
5
Z
1
x2 + 1
dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
43. 43Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Therefore
1
(x − 1)3(x2 + 2x + 5)
=
1/8
(x − 1)3
−
1/16
(x − 1)2
+
1/64
(x − 1)
+
− 1
16
x + 1
4
x2 + 2x + 5
.
We can now calculate the integral
Z
1
(x − 1)3(x2 + 2x + 5)
dx = −
1/16
(x − 1)2
+
1/16
x − 1
+
1
64
ln |x − 1|
+
Z
− 1
16
x + 1
4
x2 + 2x + 5
dx
We have
Z
− 1
16
x + 1
4
x2 + 2x + 5
dx = −
1
32
Z
2x − 8
x2 + 2x + 5
dx
= −
1
32
Z
2x + 2
x2 + 2x + 5
dx +
10
32
Z
1
x2 + 2x + 5
dx
= −
1
32
ln |x2
+ 2x + 5| +
10
32
Z
1
(x + 1)2 + 22
dx
= −
1
32
ln |x2
+ 2x + 5| +
5
32
arctan
x + 1
2
dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
44. 44Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Denominator Containing Linear Factors
Denominator Containing reducible Quadratic Facto
Denominator Containing Repeated Linear Factors
Denominator Containing Irreducible Quadratic Fac
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
45. 45Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
H Trigonometric Integrals H
Products of Powers of Sines and Cosines
Z
sinm
x cosn
x dx .
¶ If m is odd, we write m as 2k + 1 and use the identity sin2
x = 1 − cos2
x to obtain
sinm
x = sin2k+1
x = (sin2
x)k
sin x = (1 − cos2
x)k
sin x.
Then we combine the single sin x with dx in the integral and we take u = cos x.
Example
Calculate
Z
sin3
x cos2
x dx
Solution
Z
sin3
x cos2
x dx =
Z
sin2
x cos2
x sin x dx =
Z
(1 − cos2
x) cos2
x(sin x dx) t = cos x
= −
Z
(1 − t2
)t2
dt =
t5
−
t3
+ C =
cos5 x
−
cos3
+ C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
46. 46Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
· If m is even and n is odd, we write n as 2k + 1 and use the identity cos2
x = 1 − sin2
x to
obtain
cosn
x = cos2k+1
x = (cos2
x)k
cos x = (1 − sin2
x)k
cos x.
Then we combine the single cos x with dx in the integral and we do the change of variable
u = sin x.
Example
Find
Z
cos5
x dx .
Solution
We have m = 0 even and n = 5 odd then
Z
cos5
x dx =
Z
cos4
x cos x dx =
Z
(1 − sin2
x)2
(cos x dx) t = sin x
=
Z
(1 − t2
)2
dt =
Z
(1 + t4
− 2t2
) dt
=
t5
5
−
2t3
3
+ u + C =
sin5
x
5
−
2 sin3
x
3
+ sin x + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
47. 47Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
¸ If both m and n are even, we substitute
sin2
x =
1 − cos 2x
2
and cos2
x =
1 + cos 2x
2
to reduce the integrand to one in lower powers of cos 2x.
Example
Find
Z
sin2
x cos2
x dx .
Solution
We have m = 2 and n = 2 are even, then
Z
sin2
x cos2
x dx =
Z
1 − cos 2x
2
1 + cos 2x
2
dx =
1
4
Z
(1 − cos2
2x) dx
=
1
4
Z
1
2
−
1
2
cos 4x
dx =
1
8
x −
1
32
sin 4x + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
48. 48Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
Products of Sines and Cosines
Z
sin mx sin nx dx ,
Z
sin mx cos nx dx and
Z
cos mx cos nx dx .
We use the simpler identities:
¶ sin mx sin nx =
1
2
[cos(m − n)x − cos(m + n)x]
· sin mx cos nx =
1
2
[sin(m − n)x + sin(m + n)x]
¸ cos mx cos nx =
1
2
[cos(m − n)x + cos(m + n)x]
Example
Z
cos(5x) sin(3x) dx =
Z
sin(8x) − sin(2x)
2
dx = cos(2x)/4 − cos(8x)/16 + C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
49. 49Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
Exercise -C-
1.
Z
sin x cos x dx 2.
Z
cos3
x sin x dx 3.
Z
sin4
x cos x dx
4.
Z
cos3
x dx 5.
Z
sin3
x dx 6.
Z
sin2
2x cos3
2x dx
7.
Z
sin3
x cos3
x dx 8.
Z
cos3
2x sin5
2x dx 9.
Z
16 sin2
x cos2
x dx
10.
Z
sin4
x cos2
xdx 11.
Z
cos3
2x sin 2x dx 12.
Z
cos 3x cos 4x dx
13.
Z
sin 3x cos 2x dx 14.
Z
sin 2x cos 3x dx 15.
Z
sin 3x sin 3x dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
50. 50Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
Bioche Rule
Let f(x) =
P(sin x, cos x)
Q(sin x, cos x)
. To calculate
Z
f(x) dx we take ω(x) = f(x) dx. Then, we study the
invariance of ω(x) by the following change of variables
x 7−
→ −x , x 7−
→ π − x or x 7−
→ π + x .
a) If ω(−x) = ω(x), or if ω(x) is invariant by the operation
x 7−
→ −x
dx 7−
→ −dx
We take t = cos(x), dt = − sin(x)dx .
b) If ω(π − x) = ω(x), or if ω(x) is invariant by the operation
x 7−
→ π − x
dx 7−
→ −dx
We take t = sin(x), dt = cos(x)dx .
c) If ω(π + x) = ω(t), or if ω(x) is invariant by the operation
x 7−
→ π + x
dx 7−
→ dx
We take
t = tan(x) , dt =
1
cos2(x)
dx =
1 + tan2
(x)
dx .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
51. 51Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
In any case, the change of variable t = tan
x
2
, bring back to the calculation of a
primitive of a rational function in terms of t. We can use the following
cos(x) =
1 − t2
1 + t2
, sin(x) =
2t
1 + t2
, tan(x) =
2t
1 − t2
, x = 2 arctan(t) et dx =
2dt
1 + t2
.
Example
Calculate the following integral
a)
Z
sin x
1 + cos2 x
dx b)
Z
cos x
sin2
x − cos2 x
dx
c)
Z
1
cos2 x
1 + tan x
dx d)
Z
1
1 + cos x
dx
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
52. 52Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
Solution
a) Let ω(x) =
sin x
1 + cos2 x
dx. We have,
ω(−x) =
sin(−x)
1 + cos2(−x)
d(−x) =
− sin x
1 + cos2 x
(−dx) =
sin x
1 + cos2 x
dx = ω(x).
According to Bioche r tle we take t = cos x, and so
Z
sin x
1 + cos2 x
dx = −
Z
1
1 + t2
dt = − arctan t + C = − arctan(cos x) + C .
b) Let ω(x) =
cos x
sin2
x − cos2 x
dx. We have, ω(π − x) =
− cos x
sin2
x − cos2 x
(−dx) = ω(x).
Then we take t = sin x.
Z
cos x
sin2
x − cos2 x
dt =
Z
dt
t2 − (1 − t2)
=
Z
dt
2t2 − 1
=
1
2
Z
dt
t − 1
√
2
t + 1
√
2
=
1
2
Z − 1
√
2
t + 1
√
2
dt +
1
2
Z − 1
√
2
t − 1
√
2
dt
=
1
2
−
1
√
2
ln
69. 53Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Products of Powers of Sines and Cosines
Products of Sines and Cosines
Bioche Rule
Solution
c) Let ω(x) =
1
cos2 x(1 + tan x)
dx. We have,
ω(π + x) =
1
cos2(π + x)
1 + tan(π + x)
d(π + x) =
1
cos2 x
1 + tan x
dx = ω(x) .
Then we take t = tan x.
Z
1
cos2 x
1 + tan x
dx =
Z
dt
1 + t
= ln |t| + C = ln | tan x| + C .
d) Let ω(x) =
1
1 + cos x
dx . None of the three Bioche rules we can apply. Then we take
t = tan(x/2), and get
Z
1
1 + cos x
dx =
Z
1 + t2
2
2dt
1 + t2
= t + C = tan(x/2) + C.
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
70. 54Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
H Abelian integrals H
The aim is to calculate a primitive of the following forms:
a)
Z
F x, n
s
ax + b
cx + d
!
dx or b)
Z
F
x,
p
ax2 + bx + c
dx
a) We take t = n
s
ax + b
cx + d
then, the integral becomes the one of a rational function in terms t.
t = n
s
ax + b
cx + d
=
⇒ tn
=
ax + b
cx + d
=
⇒ x = −
dtn − b
ctn − a
=
⇒ dx = −
n(ad − bc)tn−1
(atn − c)2
dt
Thus,
Z
F x, n
s
ax + b
cx + d
!
dx =
Z
F
dtn − b
ctn − a
, t
n(ad − bc)tn−1
(atn − c)2
dt .
In the case where n = 1, i.e.
Z
F x,
s
ax + b
cx + d
!
dx we take t =
s
ax + b
cx + d
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
71. 55Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Example
Calculate the following integrals:
a)
Z
1
x +
√
x − 1
dx b)
Z
1
x
√
x − 1
dx
Solution
a) Thaking t =
p
x − 1, then x = t2
+ 1 and dx = 2tdt. Therefore
Z
dx
x +
√
x − 1
=
Z
2tdt
t2 + 1 + t
=
Z
2t + 1
t2 + 1 + t
dt −
Z
1
t + 1
2
2
+ 3
4
dt
= ln |t2
+ t + 1| −
2
√
3
arctan
2t + 1
√
3
+ C .
b) Taking t =
p
x − 1, then x = t2
+ 1 and dx = 2tdt. Therefore
Z
dx
x
√
x − 1
=
Z
2tdt
t(t2 + 1)
= 2 arctan t + C = 2 arctan
p
x − 1
+ C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
72. 56Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
b)
Z
F
x,
p
ax2 + bx + c
dx
• If a 0, we take
p
ax2 + bx + c = x
√
a + t.
• If a 0, then it is necessary that ∆ ≥ 0. Thus
ax2
+ bx + c = a(x − α)(x − β) =
⇒
p
ax2 + bx + c =
q
a(x − α)(x − β)
= |x − α|
s
a
x − β
x − α
.
On se ramène au calcul de la section précédente et on peut utiliser le changement
de variable défini par
p
ax2 + bx + c = t(x − α) =⇒ t =
r
a
x − β
x − α
.
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
73. 57Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Example
Evaluate
Z
dx
x
p
x2 + 6x + 10
.
Solution
We have a = 1 0. Then, we take
p
x2 + 6x + 10 = x
√
1 + t = x + t, and so
x =
t2 − 10
2(3 − t)
, x + t =
−t2 + 6t − 10
2(3 − t)
, and dx =
−2t2 + 12t − 20
4(3 − t)2
.
D’où
Z
dx
x
p
x2 + 6x + 10
=
Z
−2t2 + 12t − 20
(t2 − 10)(−t2 + 6t − 10)
dt =
Z
2dt
(t −
√
10)(t +
√
10)
=
1
√
10
Z
dt
(t −
√
10)
−
1
√
10
Z
dt
(t +
√
10)
=
1
√
10
ln |t −
√
10| − ln |t +
√
10|
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
74. 58Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Particular case
If we want to calculate
Z p
ax2 + bx + c dx. To calculate this type of integral we will use the
following method:
ax2
+ bx + c =
k2
− u2
(x) : We take u(x) = k cos θ or u(x) = k sin θ
k2
+ u2
(x) : We take u(x) = k sinh θ or u(x) = k tan θ
u2
(x) − k2
: We take u(x) = k cosh θ or u(x) =
k
cos θ
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
75. 59Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Example
Evaluate
Z
x2
dx
p
9 − x2
.
Solution
We take x = 3 sin t, dx = 3 cos t . Alors 9 − x2
= 9(1 − sin2
t) = 9 cos2
t.
Il s’ensuit
Z
x2
dx
p
9 − x2
=
Z
9 sin2
t · 3 cos tdt
3 cos t
= 9
Z
sin2
t dt = 9
Z
1 − cos 2t
2
dt
=
9
2
t −
sin 2t
2
+ C =
9
2
(t − sin t cos t) + C
=
9
2
arcsin
x
3
−
x
3
p
9 − x2
3
!
+ C
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives
76. 60Ú61
Anti-derivatives
Integration of a Rational Functions
Trigonometric Integrals
Abelian integrals
Example
Calculate I =
Z p
x2 + 2x + 5 dx and J =
Z p
−x2 + 4x − 3 dx .
Solution
I: We have x2
+ 2x + 5 = 22
+ (x + 1)2
. Taking x + 1 = 2 sinh t, we obtain dx = 2 cosh t dt
Z p
x2 + 2x + 5 dx =
Z q
22(1 + sinh2
t) dx = 4
Z
cosh2
t dt
= 4
Z
1 + cosh(2t)
2
dt = 2t + sinh(2t) + C .
J: We have −x2
+ 4x − 3 = 1 − (x − 2)2
. Taking x − 2 = sin t, we obtain dx = cos t dt
Z p
−x2 + 4x − 3 dx =
Z q
1 − sin2
t dx =
Z
cos2
t dt
=
Z
1 + cos(2t)
2
dt =
t
2
+
sin(2t)
4
+ C .
Dr. Kamel ATTAR | Chapter 1: Anti-derivatives