Simple integral

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

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

3Ú61
Anti-derivatives
Abelian integrals
Definition
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.

4Ú61
Anti-derivatives
Abelian integrals
Definition
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

5Ú61
Anti-derivatives
Abelian integrals
Definition
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

6Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

7Ú61
Anti-derivatives
Abelian integrals
Definition
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) .

8Ú61
Anti-derivatives
Abelian integrals
Definition
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


9Ú61
Anti-derivatives
Abelian integrals
Definition
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

10Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

11Ú61
Anti-derivatives
Abelian integrals
Definition
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).

12Ú61
Anti-derivatives
Abelian integrals
Definition
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

13Ú61
Anti-derivatives
Abelian integrals
Definition
Example
Calculate I =
Z
tan x dx =
Z
sin x
cos x
dx and J =
Z p
2x + 1 dx.
Solution

14Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

15Ú61
Anti-derivatives
Abelian integrals
Definition
Example
Calculate I =
Z
xdx
p
1 − x2
with x 6= ±1 and J =
Z
(x3
+ x)5
(3x2
+ 1) dx.
Solution

16Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

17Ú61
Anti-derivatives
Abelian integrals
Definition
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

18Ú61
Anti-derivatives
Abelian integrals
Definition
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

19Ú61
Anti-derivatives
Abelian integrals
Definition
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

20Ú61
Anti-derivatives
Abelian integrals
Definition
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

21Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

22Ú61
Anti-derivatives
Abelian integrals
Definition
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

23Ú61
Anti-derivatives
Abelian integrals
Definition
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

24Ú61
Anti-derivatives
Abelian integrals
Definition
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

25Ú61
Anti-derivatives
Abelian integrals
Definition
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

26Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

27Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

28Ú61
Anti-derivatives
Abelian integrals
Definition
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 .

29Ú61
Anti-derivatives
Abelian integrals
Definition
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

30Ú61
Anti-derivatives
Abelian integrals
Definition
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

31Ú61
Anti-derivatives
Abelian integrals
Denominator Containing reducible Quadratic Facto
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

32Ú61
Anti-derivatives
Abelian integrals
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 .

33Ú61
Anti-derivatives
Abelian integrals
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

34Ú61
Anti-derivatives
Abelian integrals
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

35Ú61
Anti-derivatives
Abelian integrals
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.

36Ú61
Anti-derivatives
Abelian integrals
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

37Ú61
Anti-derivatives
Abelian integrals
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 .

38Ú61
Anti-derivatives
Abelian integrals
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

39Ú61
Anti-derivatives
Abelian integrals
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.

40Ú61
Anti-derivatives
Abelian integrals
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 .

41Ú61
Anti-derivatives
Abelian integrals
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

42Ú61
Anti-derivatives
Abelian integrals
Example
Calculate
Z
1
(x − 1)3(x2 + 2x + 5)
dx
Solution
Let g(x) =
1
(x − 1)3(x2 + 2x + 5)
. Then we have
g(x) =
A0
(x − 1)3
+
A1
(x − 1)2
+
A2
(x − 1)
+
Bx + C
x2 + 2x + 5
A0 = lim
x→1
(x − 1)3
g(x) = lim
x→1
1
x2 + 2x + 5
=
1
8
A1 = lim
x→1
h
(x − 1)3
g(x)
i0
= lim
x→1
−
2x + 2
(x2 + 2x + 5)2
= −
1
16
A2 = lim
x→1
1
2
h
(x − 1)3
g(x)
i00
= lim
x→1
3x2 + 6x − 1
(x2 + 2x + 5)3
=
1
64
A2 + B = lim
x→+∞
xg(x) = 0
Finally −A0 + A1 − A2 +
1
5
C = lim
x→0
g(x) = −
1
5
=
⇒ C =
1
4

43Ú61
Anti-derivatives
Abelian integrals
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

44Ú61
Anti-derivatives
Abelian integrals

45Ú61
Anti-derivatives
Abelian integrals
Bioche Rule
H Trigonometric Integrals H
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 .

46Ú61
Anti-derivatives
Abelian integrals
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 .

47Ú61
Anti-derivatives
Abelian integrals
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 .

48Ú61
Anti-derivatives
Abelian integrals
Bioche Rule
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 .

49Ú61
Anti-derivatives
Abelian integrals
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

50Ú61
Anti-derivatives
Abelian integrals
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 .

51Ú61
Anti-derivatives
Abelian integrals
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

52Ú61
Anti-derivatives
Abelian integrals
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

Simple integral

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