Integration list
Integration by part
 Integration by parts is a technique for simplifying the
integration.
 The integration by parts formula is used to find the
integral of the product of two different types of
functions such as logarithmic, inverse trigonometric,
algebraic, trigonometric, and exponential functions.
 For example:
Integration by Parts
Integration by parts is a technique to solve an integration in the form of
product of two functions such as:
2
[x
 sin(5x)]dx
f (x)  x 2
g(x)  sin(5x)
The main interest in integration by parts is to transform an integral into
a new integral that is easier to solve than the original.
Indefinite integrals:
 dv 
 du 
 u dx  dx  uv   v dx  dx
Don‟t try to
understand this yet.
Wait for the
examples that follow
Definite integrals:
b  du 
 dv 
b
dx
u

v
dx


uv





a
a  dx 
a  dx 
b
For convenience, this can be memorized as:
 u dv  uv   v du
Integration by part
 Integration by parts formula-Differentiate Version:
 what is u? what is v?
 U & V cross method.
Integration by Parts – Guideline of Selecting U
Choose u by the following sequence:
1
2
3
4
Logarithmic (log x, lnx)
Algebraic (x, x2 x3,…)
Trigonometry (sin x, cos x, tan x,…)
Exponential (ex , e4x,…)
and the next function automatically becomes dv.
If the new integral is more difficult than the original, change the
choice of u and dv.
Example
Evaluate
 x cos xdx
 x cos x dx
u
dv
Why choose x as u instead of cos x?
x is algebraic function. Meanwhile, cos x is a trigonometric function.
Hence, algebraic function comes first before trigonometric functions.
So � is chosen as �.
Differentiate u
u  x,
dv  cos xdx
du  1,
dx
v  sin x (omit c)
Plug everything into the formula
udv  uv  vdu
 x sin x   sin xdx
Integrating
 sin xdx   cos x  c .
Therefore,
 x cos xdx  x sin x   sin xdx
 x sin x  (cos x)  c
 x sin x  cos x  c
Integrate dv
Example
Evaluate
 3x
2
ln xdx
ln x 3x dx
2
u
dv
Why ln x is a u?
In guideline of choosing „u‟, we refer LATE in which Logarithmic (L)
function comes first in the list. Hence, ln � is chosen as �
u  ln x,
dv  3x2 dx
du  1 ,
dx x
Differentiate u
v  x3 (omit c)
Plug everything into the formula
 udv  uv  vdu
 x3 ln x   x31  dx
x 
Integrating  x 2 dx  1 x3  c , we obtained,
3
 3x
2
ln xdx  x 3 ln x   x 2 dx
 x 3 ln x 
x3
3
c
Integrate dv
Example
Evaluate  2xe x dx
u  2x,
du  2,
dx
dv  e x dx
v  e x (omit c)
By using integration by parts
 udv  uv  vdu
 2xe   e 2dx
x
x
 2xe x  2ex  c
 Sometimes, integration by parts must be applied several
times to evaluate a given integral. See example below.
Example
Evaluate e x cos x dx
u  cos x,
dv  e x dx
du  sin x,
dx
v  ex
By using integration by parts
 udv  uv  vdu
 e x cos x  e x  sin xdx
 e x cos x  e x sin xdx  c
1
Another integration by parts applied to the last integral, i.e.
e
x
sin x dx will complete the solution. Hence, by doing by
parts once again, we obtain
u  sin x,
dv  e x dx
du  cos x,
dx
v  ex
Substitute into the formula
e
x
sin xdv  uv  vdu
 e sin x  e cos xdx
x
x
Substitute result in (2) into Equation (1)
2
e
x
cos xdx  e x cos x  e x sin xdx
 e x cos x   e x sin x   e x cos xdx 


Notice that the last term is similar to the original problem. Hence, by
moving the last term into the left hand side equation, we get
e
x
cos xdx  e x cos xdx  e x cos x  e x sin x  c
2 e x cos xdx  e x cos x  e x sin x  c
We want to find e x cos xdx , therefore

e x cos x  e x sin x
c
e cos xdx 
2
x
Exercise
Please solve
• Try it with the integration by part formular.
• Use the U & V method.
Exercise
Please solve
• Try it with the integration by part formular.
• Use the U & V method.
Exercise
Please solve
• Try it with the integration by part formular.
• Use the U & V method.
Integration by Partial Fractions
 If the integrand is in the form of an algebraic fraction
and the integral cannot be evaluated by simple methods,
the fraction needs to be expressed in partial fraction.
 Definition – Proper Fraction Any rational function of x,
Q(x)
where the P(x) is less thanthe
degree of
P(x)
Q(x) could
be expressed as sum of relatively simpler rational
functions, called partial fractions.
1. Linear Factor
Q(x)  (a1x  b1)(a2 x  b2 )
Partial fraction :
(an x  bn )
A1
A2

(a1x  b1) (a2 x  b2 )
An
(an x  bn )
2. Repeated Linear Factor
Q(x)  (ax  b)n
Partial fraction :
A1
A2

(ax  b) (ax  b)2
3. Quadratic Factor
Q(x)  ax 2  bx  c 
Partial fraction :
Ax  B
(ax2  bx  c)
An
(ax  b)n
Example
Evaluate
Step 1
�
��
(� + �)(� + �)
Factor the denominator
x2  4x  3  (x  3)(x 1)
Step 2
Break up the fraction into sum of “partial fractions”
B
6
A


(x  3)(x 1) x  3 x 1
Step 3
Multiply both sides of the equation by the left side
denominator
6  A(x 1)  B(x  3)
Step 4
Take the roots of the linear factors and plug them,
one at a time, into x on the equation from step 3,
and solve
If
x  1, B  3
x  3, A  3
Split up the original integral and integrate
Step 5
 
x  36dx(x 1)
3
 3 dx
x  3 x 1
 3ln x  3  3ln x 1  c
Example
7x  6 dx

Evaluate
x  3x
3
2
The function can be written as
7x  6
7x  6

x3  3x 2 x 2 (x  3)
The denominator is a combination of linear and repeated linear case,
therefore we have
7x  6  Ax  B  C
x2
x3  3x 2
x3
We have three unknown A, B and C. Multiply both sides by the left
side of denominator,
7x  6  Ax  Bx  3 Cx2
 Ax2  3Ax  Bx  3B  Cx2
 A  C x 2  (3A  B)x  3B
Equating the coefficient of the polynomial, where
x
2
:
A

C

0
x :  3A  B  7
x0 :  3B  6
Solving the simultaneous equation, we obtain A  3,
B  2, C  3.
Alternatively, A, B and C can be found using the another approach.
If
x  3, 9C  27  C  3
x  0 , 6  3B  B  2
x  1, 13  2( A  2)  3  A  3
Hence,
7x  6  3x  2  3
x2
x3  3x 2
x3
The integration becomes

7x  6 dx  3x  2  3
dx
 x2
x3  3x 2
x3
3
3x
dx
  2 dx   22 dx 

x
x
x3
Simplify whenever necessary and then integrate

7x  6 dx 
x3  3x 2
3
3 dx  2x 2 dx 
x

 x  3 dx
 3ln x  2  3ln x  3  c
x
Solve
�+3
��
(�+6)2
Solution:
�+3
2 =
(�+6)
�
(�+6)
+
�
(�+6)2
� + 3 = � (� + 6) + �
� + 3 = �� + 6� + �
Compare the coefficient:
A=1, B= -3
�+3
��
(�+6)2
=
= ln � + 6 +
1
(�+6)
3
(�+6)
−
3
��
(�+6)2
+�
Trigonometric Integrals
 The general idea is to use identities to transform the
integrals we have to find into integrals that are easier
to work with.
 Solve the integral which include the product of
trigonometric.
Trigonometric: Basic identities
Trigonometric: Basic identities
Trigonometric Integrals
 Solve the
 where m and n are nonnegative integers (positive or
zero).
 We can divide the appropriate substitution into three
cases according to m and n being odd or even.
Trigonometric Integrals
Exercise
Please solve
• Try solve it with the trigo identities.
Exercise
Please solve
• Try solve it with the trigo identities.
Trigonometric Substitutions
Trigonometric Substitutions
Trigonometric Substitutions
Trigonometric Substitutions
Exercise
Please evaluate
Exercise
Please evaluate
Numerical Integration
 Trapezoidal Approximations
Numerical Integration
 Trapezoidal Approximations
 Procedure to solve:
1.
2.
Find out all y (table form)
Solve it with the Trapezoidal Rule.
Exercise
Numerical Integration
 Simpson’s rule
Step to use simpson’s theory
Find the x-intervals
2. FInd the value of y for respestive x intervals
3. Apply the simpson’s theory.
4. Sub all value of y and calculate the answer.
1.
Exercise
Improper Integration
Improper Integration
* This integration is finite
Exercise: