Differential Equations
Chapter 1: Introduction
Lecture #0
Table of contents
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
2 / 58
Derivatives
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
3 / 58
Derivatives
Table of Derivatives
df
dx
f (x)
f ′ (x) =
c
0
ax + b
a
xn
nx n−1
sin(x)
cos(x)
cos(x)
−sin(x)
tan(x)
sec 2 (x) = 1 + tan2 (x)
cot(x)
−csc 2 (x)
sec(x)
sec(x)tan(x)
csc(x)
−csc(x)cot(x)
4 / 58
Derivatives
Recall
tan(x) =
sec =
sin(x)
cos(x)
1
cos(x)
cot(x) =
cos(x)
sin(x)
csc(x) =
1
sin(x)
Recall
√
1
x
1
√
m
= x −1
xm
xn
x =x2
n
xn = x m
= x m−n
√
n
xn = x 2
1
xn
= x −n
Example
Find the derivative of the following functions:
√
1) x 6
2) 1345π
3) x
4)
5)
√
x5
6)
√
3
x7
7)
√1
x
8)
1
x
1
x5
5 / 58
Derivatives
Example
Solution:
1) f (x) = x 6
f ′ (x) = 6x 6−1 = 6x 5 .
⇒
2) f (x) = 1345π = constant
3) f (x) =
4) f (x) =
5) f (x) =
6) f (x) =
√
1
1 12 −1
x
2
1 − 21
x
2
1 √1
2 x
x =x2
⇒
f ′ (x) =
= x −1
⇒
f ′ (x) = −1x −1−1 = −x −2 = − x12 .
1
x
√
f ′ (x) = 0.
⇒
5
x5 = x 2
√
3
7
x7 = x 3
7) f (x) =
√1
x
8) f (x) =
1
x5
1
= x− 2
= x −5
=
=
√
=
1
√
.
2 x
⇒
f ′ (x) =
5 25 −1
x
2
=
5 32
x
2
=
5
2
⇒
f ′ (x) =
7 73 −1
x
3
=
7 34
x
3
=
√
7 3 4
x .
3
⇒
f ′ (x) = − 21 x − 2 −1 = − 12 x − 2 = −
⇒
1
x 3.
3
√1
2 x3
f ′ (x) = −5x −6 = − x56 .
6 / 58
Rules of Derivatives
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
7 / 58
Rules of Derivatives
Rules of Derivatives
1. u + v
3. ku
5.
′
u ′
v
7. u n
′
′
= u′ + v ′
= ku ′
2. u − v
4. u.v
′
′
= u ′ v + uv ′
u ′ v −uv ′
v2
6.
1 ′
v
= nu ′ u n−1
8.
√ ′
u =
=
= u′ − v ′
′
= − vv 2
′
u
√
2 u
′
9. sin(u) = u ′ cos(u)
′
10. cos(u) = −u ′ sin(u)
′
11. sec(u) = u ′ sec(u)tan(u)
′
12. csc(u) = −u ′ csc(u)cot(u)
′
13. tan(u) = u ′ sec 2 (u)
′
14. cot(u) = −u ′ csc 2 (u)
8 / 58
Rules of Derivatives
Example
Findp
the derivative of the following functions:
p
1) x 3 − 2 + 105
2) 3 sin(x) + 1
100
3) tan(x) + cot(x) + 1
5)
√
x+1
x 2 −2
4)
1
sec(x)+csc(x)
6) x 2 sin(x)
7) sin(x 5 − x 2 + 1)
Solution:
p
x 3 − 2 + 105
1) f (x) =
f ′ (x) =
⇒
p
f ′ (x) =
p
′
x 3 − 2 + 105 , so
′ ′
√
x 3 − 2 + 105 = ( u)′ + 0 (u = x 3 − 2)
√
u′
( u)′ = √
2 u
and
u ′ = 3x 2 + 0 ⇒ f ′ (x) =
3x 2
p
.
2 x3 − 2
9 / 58
Rules of Derivatives
Example
Solution:
p
3
2) f (x) =
sin(x) + 1 =
√
3
1
u = u3
and u = sin(x) + 1. Then:
u = sin(x) + 1
⇒
u ′ = cos(x)
1 ′ 1 −1
u u3
3
− 2
1
3
f ′ (x) = cos(x) sin(x) + 1
.
3
100
3) f (x) = tan(x) + cot(x) + 1
= u 100 and u = tan(x) + cot(x) + 1.
1
f (x) = u 3
⇒
u = tan(x) + cot(x) + 1
f ′ (x) =
⇒
u ′ = sec 2 (x) − csc 2 (x)
f (x) = u 100 ⇒ f ′ (x) = 100u ′ u 99
99
f ′ (x) = 100 sec 2 (x) − csc 2 (x) tan(x) + cot(x) + 1
.
10 / 58
Rules of Derivatives
Example
Solution:
1
4) f (x) = sec(x)+csc(x)
=
1
u
and u = sec(x) + csc(x).
u = sec(x) + csc(x)
f (x) =
1
u
⇒
f ′ (x) = −
⇒
u ′ = sec(x)tan(x) − csc(x)cot(x)
u′
u2
sec(x)tan(x)
−
csc(x)cot(x)
f ′ (x) = −
(sec(x) + csc(x))2
√
5) f (x) =
x+1 u
x 2 −2 v
u′ =
′
f (x) =
√
and u =
1
√ +0
2 x
1
√
(x 2
2 x
x + 1 v = x 2 − 2. Then:
and
v ′ = 2xf (x) =
√
− 2) − 2x( x + 1)
(x 2 − 2)2
u
v
⇒
f ′ (x) =
u ′ v − uv ′
v2
.
11 / 58
Rules of Derivatives
Example
Solution:
6) f (x) = x 2 sin(x) = u.v
and u = x 2
u = x2
⇒
f (x) = u.v
⇒
v = sin(x).
u ′ = 2x and v = sin(x)
′
′
f (x) = u v + uv
⇒
v ′ = cos(x)
′
f ′ (x) = 2xsin(x) + x 2 cos(x).
7) f (x) = sin(x 5 − x 2 + 1) = sin(u)
and u = x 5 − x 2 + 1. Then:
u = x5 − x2 + 1
f (x) = sin(u)
⇒
⇒ u ′ = 5x 4 − 2x
f ′ (x) = u ′ cos(u)
f ′ (x) = 5x 4 − 2xcos(x 5 − x 2 + 1).
12 / 58
Indefinite Integrals
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
13 / 58
Indefinite Integrals
Definition
We say that F is the integral (or antiderivative) of f ⇔ F ′ (x) = f (x), and we write:
Z
f (x) dx = F (x) + C
14 / 58
Indefinite Integrals
Table of Integration
f
Z
0
C
x n (n ̸= −1)
x n+1
n+1
sin(ax)
− a1 cos(ax)
cos(ax)
1
sin(ax)
a
+C
sec 2 (ax)
1
tan(ax)
a
+C
csc 2 (ax)
− a1 cot(ax) + C
sec(ax)tan(ax)
1
sec(ax)
a
csc(ax)cot(ax)
− a1 csc(ax) + C
f dx
+C
+C
+C
15 / 58
Indefinite Integrals
Example
1)
R
3)
R
x 2 dx
2)
R√
4)
R
x dx
Find the derivative of the following functions:
Solution:
R
1) x 2 dx =
x 2+1
2+1
+C =
x3
3
R√
3)
R
sin(2x) dx = − 12 cos(2x) + C
4)
R
1
x2
dx =
R
R
1
x 2 dx =
x −2 dx =
1
x2
dx
+ C.
2)
x dx =
sin(2x) dx
1 +1
x2
1 +1
2
3
+ C = 2 x32 + C.
x −2+1
−2+1
+C =
x −1
−1
+ C = − x1 .
16 / 58
Rules of Anti-derivatives
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
17 / 58
Rules of Anti-derivatives
Rules of Anti-derivatives
Z
Z
Z
1.
f + g dx =
f dx + g dx
Z
Z
f − g dx =
2.
Z
3.
Z
f dx −
g dx
Z
kf dx = k
f dx
Remark
Two important Remarks !
Z
Z
Z
f .g dx ̸=
f dx. g dx
Z
and
R
f dx
f
dx ̸= R
g
g dx
18 / 58
Rules of Anti-derivatives
Example
Z
Evaluate: I =
sin(3x) − 2x +
√
3
x − π dx.
Solution:
Z
I=
Z
sin(3x) dx − 2
Z
x dx +
1
x 3 dx −
Z
π dx
1
x2
x 3 +1
1
+ 1
− πx + C
= − cos(3x) − 2
3
2
+1
3
1
1
x 3 +1
− πx + C
= − cos(3x) − x 2 + 1
3
+1
3
19 / 58
Definite Integral
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
20 / 58
Definite Integral
Definition
The form of a definite integrals is :
Z
J=
b
f (x) dx.
a
To evaluate J:
Z
Step 1: Solve I =
f (x) dx = F (x) + C
ib
Step 2: Solve J = F (x) = F (b) − F (a).
a
21 / 58
Definite Integral
Example
Z
Evaluate: J =
π
6
π
18
cos(3x) dx.
Solution:
Z
I=
cos(3x) dx =
1
J = sin(3x)
3
1
sin(3x) + C.
3
#π
6
π
18
1
π π
=
sin(3 ) − sin(3 )
3
6
18
1
π
π =
sin( ) − sin( )
3
2
6
1
1
1
1−
= .
=
3
2
6
22 / 58
Inverse Trigonometric Functions
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
23 / 58
Inverse Trigonometric Functions
Derivatives
1
2
3
′
arcsin(u) = √
u′
a2 −u 2
′
′
arctan(u) = a2u+u 2
′
′
arcsec(u) = √u 2
|u|
u −a2
Integrals
Z
du
u
p
1.
= arcsin
+C
2
2
a
a −u
Z
2.
1
u
du
= arctan
+C
a2 + u 2
a
a
du
Z
3.
u
p
u 2 − a2
=
1
u
arcsec
+ C with a > 0
a
a
24 / 58
Logarithm and Exponential
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
25 / 58
Logarithm and Exponential
Recall
′
u′
ln(u) =
u
eu
′
= u ′ eu
Recall
1
Exponential functions of base a (for a > 0):
ax = ex ln(a)
2
Logarithm function of base b (for b ̸= 1):
logb (x) =
ln(x)
ln(b)
26 / 58
Logarithm and Exponential
Derivatives
1
We have
au
2
′
= u ′ au ln(a)
We have
logb (u))′ =
u′
uln(b)
Integrals
1
2
We have
We have
Z
au du =
Z
logb (u)du =
1
au + C
ln(a)
1
ln(b)
Z
ln(u)du
27 / 58
Practice More
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
28 / 58
Practice More
Practice More
Find the derivative of the following functions:
1
f (x) = tan(3x).
2
g(x) =
3
h(x) = sin(x − 1)2cos(x−1) .
e3x
.
3 + e3x
4
L(x) = cos(2x)e−x .
5
K (x) = (x 2 + 1) sin(2x).
6
M(x) = √
4
L(x) = cos(2x)e−x .
5
K (x) = (x 2 + 1) sin(2x).
6
M(x) = √
x
x +1
Find the integral of the following functions:
1
f (x) = tan(3x).
2
g(x) =
3
h(x) = sin(x − 1)2cos(x−1) .
e3x
.
3 + e3x
x
x +1
29 / 58
Integration by Parts
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
30 / 58
Integration by Parts
Rule
Z
uv ′ dx = uv −
Z
u ′ vdx
Its proof follows from the derivative rule (uv )′ = u ′ v + v ′ u
Remark
To choose u and v ′ follow the LIATE rule:
For u, choose from the beginning of the list, and for v ′ chose from the end of the list.
1
L - logarithm functions
2
I - Inverse trigonometric functions
3
A - Algebraic functions (simple polynomial terms)
4
T - trigonometric functions
5
E - Exponential functions
31 / 58
Integration by Parts
Example (1)
Z
I=
ln(x)dx
1
x
v′ = 1 → v = x
Z
Z
Z
1
So I = uv − vu ′ = x ln(x) − x dx = x ln(x) − dx + C = x ln(x) − x + C.
x
u = ln(x)
→
u′ =
→
u′ = 1
Example (2)
Z
I=
x cos(x)dx
u=x
′
Z
So I = uv −
v = cos(x) → v = sin(x)
Z
vu ′ = x sin(x) − sin(x)dx = x sin(x) + cos(x) + C.
32 / 58
Integration by Parts Twice (or More)
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
33 / 58
Tabular Integration or American Way
Example
Z
I=
cos(x)ex dx
u = cos(x)
′
Z
So I = uv −
→
u ′ = −sin(x)
v =e
→ v = ex
Z
vu ′ = cos(x)ex + sin(x)ex dx again integration by parts:
So I = cos(x)ex +
Z
x
u = sin(x)
→
u ′ = cos(x)
v ′ = ex
→
v = ex
sin(x)ex dx = cos(x)ex + sin(x)ex −
Z
cos(x)ex dx
= cos(x)ex + sin(x)ex − I
⇒2I = cos(x)ex + sin(x)ex + C
1
cos(x) + sin(x) ex + C
⇒I =
2
34 / 58
Tabular Integration or American Way
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
34 / 58
Tabular Integration or American Way
Rule
Let P be a polynomial of finite degree, we use the tabular integration to evaluate integrals of
the form:
Z
Z
Z
P(x)eax dx,
P(x) sin(ax)dx,
P(x) cos(ax)dx (a ∈ R).
In fact, tabular integration is a by parts integration done several times successively.
35 / 58
Tabular Integration or American Way
Example
Z
I=
x 3 sin(x)dx
sin(x)
x3
+
3x 2
−cos(x)
−
Derive
6x
−sin(x)
Integrate
+
6
cos(x)
−
0
sin(x)
So I = −x 3 cosx + 3x 2 sinx + 6xcosx − 6sinx + C.
36 / 58
Trigo Integration: Products of Powers of Sin and Cos
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
37 / 58
Trigo Integration: Products of Powers of Sin and Cos
Rule
Z
Integrals of the form
sinn (x) cosm (x)dx where m, n ∈ N:
We have four cases:
Case 1: If n is odd:
n = 2k + 1,
Use the substitution: u = cos(x), then du = − sin(x)dx
Write sinn (x) = (sin2 (x))k sin(x) = (1 − cos2 (x))k sin(x)
Case 2: If m is odd:
m = 2k + 1,
Use the substitution: u = sin(x), then du = cos(x)dx
Write cosm (x) = (cos2 (x))k cos(x) = (1 − sin2 (x))k cos(x)
Case 3: If m, n are even:
1 + cos(2x)
,
2
Case 4: If m, n are odd: Use case 1 or case 2.
Use the rules cos2 (x) =
sin2 (x) =
1 − cos(2x)
2
38 / 58
Trigo Integration: Products of Powers of Sin and Cos
Example
=
Z
I=
sin3 (x) cos2 (x)dx
Power on sin is odd
Z
sin2 (x) cos2 (x) sin(x)dx
Z
(1 − cos2 (x)) cos2 (x) sin(x)dx
=
Let u = cos(x) ⇒ du = − sin(x)dx
Z
− (1 − u 2 )u 2 du
=
−
=
−
=
I
Z
=
(u 2 − u 4 )du
Z
u n du =
u n+1
+C
n+1
!
u5
u3
+
+C
3
5
3
(cos(x))
(cos(x))5
−
+
+ C.
3
5
39 / 58
Trigo Integration: Products of Powers of Sin and Cos
Example (Case3)
Z
Z
=
sin2 (x) cos4 (x)dx
Both powers are even
sin2 (x)(cos2 )2 (x)dx
1 − cos(2x) 1 + cos(2x) 2
)(
) dx
2
2
Z
1 − cos(2x) (1 + cos(2x))2
(
)
)dx
2
4
Z
1
(1 − cos(2x))(1 + 2cos(2x) + cos2 (2x))dx
8
Z
1
1 − cos(2x) + 2cos(2x) − 2cos2 (2x) + cos2 (2x) − cos3 (2x)dx
8
Z
1
1 + cos(2x) − cos2 (2x) − cos3 (2x)dx
8
"
#
Z
Z
1
1
1 + cos(4x)
dx − cos3 (2x)dx
x + sin(2x) −
8
2
2
Z
=
=
=
=
=
=
(
40 / 58
Trigo Integration: Products of Powers of Sin and Cos
Example (Case3)
=
=
=
=
=
"
1
x
8
"
1
x
8
"
1
x
8
"
1
x
8
"
1
x
8
1
+ sin(2x) −
2
Z
+
1
x
sin(2x) −
2
2
+
1
x
sin(2x) −
2
2
+
1
x
sin(2x) −
2
2
+
1
x
sin(2x) −
2
2
#
Z
1 + cos(4x)
3
dx − cos (2x)dx
2
#
Z
sin(4x)
2
−
dx − (1 − sin (2x))cos(2x)dx
8
#
Z
sin(4x)
1
2
−
dx −
(1 − u )du
8
2
#
sin(4x)
x
u3
−
dx − +
+C
8
2
6
#
sin(4x)
x
sin(2x)3
−
dx − +
+C
8
2
6
41 / 58
Trigo Integration: Products of Powers of Sin and Cos
Example (Case4)
=
=
=
=
=
=
Z
I=
sin3 (x) cos3 (x)dx
Z
sin3 (x) cos2 (x) cos(x)dx
Z
sin3 (x)(1 − sin2 (x)) cos(x)dx
Let u = sin(x) ⇒ du = cos(x)dx
Z
u 3 (1 − u 2 )du
Z
(u 3 − u 5 )du
u6
u4
−
+C
4
6
(sin(x))4
(sin(x))6
−
+ C.
4
6
42 / 58
Eliminating Square Roots
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
43 / 58
Eliminating Square Roots
Rule
To eliminate square roots we need to write the expression in the form of the square of
1+cos(2x)
something. Example: you may use: cos2 (x) =
.
2
Example
Z
π
4
p
1 + cos(4x)dx
0
Z
=
π
4
0
=
=
=
π
4
√ Z
2
√
√ q
2 cos2 (2x)dx
0
π
4
√ Z
2
| cos(2x)|dx
0
=
π
4
Z
q
2 cos2 (2x)dx =
0 ≤ 2x ≤
π
2
cos(2x)dx
0
2
sin(2x)
2
√
2/2.
π
4
0
√
2
π
2
sin( ) −
sin(0)
2
2
2
√
=
44 / 58
Integration by Partial Fractions
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
45 / 58
Integration by Partial Fractions
Case 1.
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
46 / 58
Integration by Partial Fractions
Case 1.
Rule
P(x)
dx
Q(x)
If degP < degQ we factorize the denominator and decompose f into partial fractions
according to the following forms:
Z
Given the integral:
Z
f (x)dx =
Form of the denominator
Decomposition of the function
A
x−1
(x − 1)(x − 2)(x − 3)
(x − 1)2 (x − 2)3
A
x−1
+
B
(x−1)2
(x + 1)3 (x 2 + 1)
+
Ax+B
x 2 +3
+
A
x+1
Cx+D
(x 2 +3)2
+
B
x−2
C
x−2
Ax+B
x 2 +3
(x 2 + 3)(x 2 + x + 1)
(x 2 + 3)2 (x 2 + x + 1)2
+
+
+
C
x−3
D
(x−2)2
+
E
(x−2)3
+
Cx+D
(x 2 +x+1)
+
Ex+F
x 2 +x+1
+
Gx+H
(x 2 +x+1)2
C
(x+1)3
+
Dx+E
x 2 +1
B
(x+1)2
+
Where A, B, C, · · · , are constants to be determined.
47 / 58
Integration by Partial Fractions
Case 1.
Example (1)
Z
6x + 7
Evaluate
dx :
(x + 2)2
Solution. Decomposition of f :
6x + 7
A
B
=
+
.
(x + 2)2
x +2
(x + 2)2
Take common denominator :
6x + 7
2 =
(x + 2)
(
⇔
A(x + 2) + B
2 ⇔ 6x + 7 = Ax + 2A + B
(x + 2)
A=6
2A + B = 7
⇔ A = 6,
B = −5.
6
5
6x + 7
=
−
. Integrate both sides:
(x + 2)2
x +2
(x + 2)2
Z
Z
Z
6x + 7
6dx
5dx
5
dx
=
−
= 6 ln |x + 2| +
+ C.
(x + 2)2
x +2
(x + 2)2
x +2
Replacing A, B, we get
48 / 58
Integration by Partial Fractions
Example (2)
Z
Evaluate
Case 1.
1
dx :
x(x 2 + 1)2
Solution. Decomposition of f :
x(x 2
1
A
Bx + C
Dx + E
= + 2
+ 2
.
x(x 2 + 1)2
x
x +1
(x + 1)2
A(x 2 + 1)2 + (Bx + C)(x 2 + 1)x + (Dx + E)x
1
=
2
+ 1)
x(x 2 + 1)2
⇔
1 = Ax 4 + 2Ax 2 + A + Bx 4 + Bx 2 + Cx 3 + Cx + Dx 2 + Ex
⇔
1 = (A + B)x 4 + Cx 3 + (2A + B + D)x 2 + (C + E)x + A

A + B = 0 ⇒ B = −1





C = 0
2A + B + D = 0 ⇒ 2 − 1 + D = 0 ⇒ D = −1


C + E = 0 ⇒ E = −C = 0



A=1
⇔
49 / 58
Integration by Partial Fractions
Case 1.
Example (2)
A = 1,
B = −1,
C = 0,
1
x(x 2 +1)2
1
x
D = −1,
x
x 2 +1
E = 0.
x
(x 2 +1)2
= −
Replacing A, B, C, D and E, we get:
−
Z
Z
Z
Z
dx
xdx
xdx
1
dx =
−
−
.
2
2
2
x(x + 1)
x
x +1
(x 2 + 1)2
Let u = x 2 + 1, du = 2xdx, so
Z
Z
xdx
1
du
1
1
=
= ln |u| + C = ln(x 2 + 1),
x2 + 1
2
u
2
2
Z
Z
1
du
11
1
xdx
=
=−
=−
(x 2 + 1)2
2
u2
2u
2(x 2 + 1)
Finally
Z
1
1
1
dx = ln |x| − ln(x 2 + 1) +
+ C.
x(x 2 + 1)2
2
2(x 2 + 1)
50 / 58
Integration by Partial Fractions
Case 2.
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
51 / 58
Integration by Partial Fractions
Case 2.
Rule
Given the integral:
Z
Z
f (x)dx =
P(x)
dx
Q(x)
If degP ≥ degQ, we perform an Euclidean division to decompose f as:
f (x) = E(x) +
R(x)
Q(x)
where:
E(x) is the polynomial part
R(x)
with deg(R) < deg(Q), so back to case 1.
Q(x)
52 / 58
Integration by Partial Fractions
Case 2.
Example
Z
Evaluate
2x 3 − 4x 2 − x − 3
dx :
x 2 − 2x − 3
Solution:
We perform an Euclidean division, we get
We decompose:
2x 3 −2x 2 +1
x 2 −x
= 2x +
5x−3
.
x 2 −2x−3
5x − 3
5x − 3
A
B
Ax − 3A + Bx + B
=
+
=
.
=
x 2 − 2x − 3
(x + 1)(x − 3)
x +1
x −3
(x + 1)(x − 3))
So A + B = 5, −3A + B = −3.
Hence A = 2, B = 3.
Z
2x 3 − 4x 2 − x − 3
dx
x 2 − 2x − 3
Z =
=
2x +
2
3
+
x +1
x −3
dx
x 2 + 2 ln |x + 1| + 3 ln |x − 3| + C.
53 / 58
Heaviside cover-up function
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
54 / 58
Heaviside cover-up function
Heaviside cover-up function
Rule
Let f (x) =
P(x)
Q(x)
with deg(P) < deg(Q) and
Q(x) = (x − α1 )(x − α2 ) · · · (x − αn ).
If, for example, Q(x) = (x − 1)(x − 2)(x − 3), we decompose
f (x) =
B
C
A
+
+
x −1
x −2
x −3
To find A, multiply f (x) by (x − 1) and replace x = 1,
To find B, multiply f (x) by (x − 2) and replace x = 2
To find C, cover the factor (x − 3) and replace x = 3.
55 / 58
Heaviside cover-up function
Example
x2 + 1
dx :
(x − 1)(x − 2)(x − 3)
2
A
B
C
x +1
=
+
+
Solution. We have
(x − 1)(x − 2)(x − 3)
x −1
x −2
x −3
Z
Evaluate
Then A =
12 + 1
= 1,
(1 − 2)(1 − 3)
C=
B=
22 + 1
= −5
(2 − 1)(2 − 3)
32 + 1
= 5.
(3 − 1)(3 − 2)
x2 + 1
dx
(x − 1)(x − 2)(x − 3)
Z
Z
Z
1
5
5
dx −
dx +
dx
x −1
x −2
x −3
ln |x − 1| − 5 ln |x − 2| + 5 ln |x − 3| + C.
Z
So
=
=
56 / 58
Practice More
1 Derivatives
2 Rules of Derivatives
3 Indefinite Integrals
4 Rules of Anti-derivatives
5 Definite Integral
6 Inverse Trigonometric Functions
7 Logarithm and Exponential
8 Practice More
9 Integration by Parts
10 Integration by Parts Twice (or More)
11 Tabular Integration or American Way
12 Trigo Integration: Products of Powers of Sin and Cos
13 Eliminating Square Roots
14 Integration by Partial Fractions
Case 1.
Case 2.
15 Heaviside cover-up function
16 Practice More
57 / 58
Practice More
Evaluate:
Z
x 2 sinx dx
1
π
4
Z
8
xsin(2x) dx
0
Z
2 4x
x e
2
dx
Z
x 2 ln(x) dx
Z
(x + 1)2 ln(3x) dx
3
4
x +3
dx
2x 3 − 8x
Z
9
1
Z
10
0
Z
e2x cos(x) dx
5
Z
e−x sin(4x) dx
6
1
2
Z
7
0
xe2x dx
Z
2x + 2
dx
(x 2 + 1)(x − 1)3
Z
9x 3 − 3x + 1
dx
x3 − x2
11
12
dx
(x + 1)(x 2 + 1)
58 / 58