3.3 Integration by parts

MOOC MAT438/ UiTM

Integration Techniques

 Integration of Rational Function using Partial
Fractions

 Integration of Trigonometric Function
(Powers of sine cosine and secant tangent)

 Integration by Trigonometric Substitution
 Integration by Parts

MOOC MAT438/ UiTM

At the end of this session, the students should be able to
❶ Recognize when to use Integration by Parts
❷ Memorize the Integration by Parts formula : ‫ ׬‬ = − ‫ ׬‬
❸ Use the Integration by Parts formula to solve integration problems.
➍ Memorize the ILATE acronym (Inverse trigonometry, Logarithm,

Algebra, Trigonometry, and Exponential functions)
➎ Recognize the potential function as u and dv
➏ Recognize special cases that allow the use of Tabular Integrals

MOOC MAT438/ UiTM

Integration by Integration
u-substitution by Parts

න 2 = 2 න 2 = 2

= න ∙ = 2 = න ∙ = 2
2 2

= 1 න = 2 = 2
2
න = + Can’t simplify !
= 1 + the correct
2 න = + method is..
Use Integration
= 1 2 + # by Parts
2

MOOC MAT438/ UiTM

What is ILATE??

ILATE is,

I Inverse Trigonometric functions tan−1x, sin−1x, etc NR

L Logarithmic functions ln x NR

A Algebraic functions x, 3x2, x3, etc NR

T Trigonometric functions sinx, cos3x, tan2x, etc R

E Exponential functions ex, e2x, e3x, etc R

• Using ILATE rule, whichever function comes first in the table Where,
NR : non repeatable
list, it should be u (easy to differentiate) .
• Note that dx always included in dv R: repeatable
• There are two (2) methods to solve integration by parts

This method can use for all

 Using formula : ‫ ׬‬ = − ‫ ׬‬ types of question involving
Integration by Parts

 Tabular Integrals We only can use this method if the

MOOC MAT438/ UiTM function has repeatable function

Example 1 : ‫ ׬‬

MOOC MAT438/ UiTM

 Using formula : ‫ ׬‬ = − ‫ ׬‬

(logarithmic function (NR) – doesn’t has repeatable function – never/can’t use tabular integral)
Evaluate න

Solution = = Be careful
1 =
න = න Common mistake!!
= − න =
1 1
= − න ∙ To check whether the answer is correct or wrong,
= − න 1 න = +
= − + # differentiate back the final answer. If the result (after actually,

MOOC MAT438/ UiTM differentiate) is the same as the question given, this ✓1

න ≠ +
But…

✓ 1

=

means your solution is correct. So, to check the answer,

you can try to differentiate back

= − +

TRY THIS!
=

Example 2 : ‫ ׬‬ −

MOOC MAT438/ UiTM

 Using formula : ‫ ׬‬ = − ‫ ׬‬

(logarithmic function (NR) – doesn’t has repeatable function – never/can’t use tabular integral)

Evaluate න − To evaluate ‫׬‬ 1 , we can use either
−3

Solution  When the numerator is a derivative of the denominator,
→ therefore, the answer is ln |denominator| + C.

න − 3 = න = or
= − න =
= − 3  Use formula ‫׬‬ + −1 = ‫׬‬ 1 = 1 + +
1 +

= − 3
or
 Use u-substitution, where = − 3

1 quotient
= − 3 − න ∙ − 3
divisor 1
degree 1 − 3 + 0

− 3 =
= − 3 − න − 3 degree 1
− − 3
Separating the 3 Improper Fraction :
integrals = − 3 − න 1 + − 3 The degree of the top is Need long 3
greater than or equal to division!
the degree of the bottom remainder
1
= − 3 − න 1 + 3 න − 3
→ − 3 = +
Integrate in bracket, one by one
3
= − 3 − + 3 − 3 + → − 3 = 1 + − 3

Expand (to eliminate bracket)

= − 3 − − 3 − 3 + #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

 Using formula : ‫ ׬‬ = − ‫ ׬‬

(Inverse Trigonometric (NR) and algebraic functions (NR) – doesn’t has repeatable function – never/can’t use tabular integral)

Use integration by parts to solve න −

Solution ILATE = −1 −
= +
= 2
න 2 −1 = න −1 ∙ 2 2 2 − +
න + =
AI I A = 2
== නන

== −− නන 1 = 2
= 1 + 2
quotient

= 2 −1 − න 2 ∙ 1 divisor
+ 2
1 2 = 1
1 + 2 2 + 0 + 1 2 + 0 + 0
= 2 −1 2
− න 1 + 2 − 2 + 0 + 1

1 Improper Fraction : Need long −1
1 − 1 + 2 The degree of the top is greater division!
= 2 −1 − න
than or equal to the degree of remainder

Integrate in bracket, one by one the bottom

= 2 −1 − − −1 + 2
→ 1 + 2 = +

Expand (to elimate bracket) 2 1
→ 1 + 2 = 1 − 1 + 2
= 2 −1 − + −1 + #

MOOC MAT438/ UiTM

Example 4 : ‫ ׬‬ −

MOOC MAT438/ UiTM

 Using formula : ‫ ׬‬ = − ‫ ׬‬

(logarithmic function (NR) – doesn’t has repeatable function – never/can’t use tabular integral)

Use integration by parts to solve න − integrate

Solution differentiate

= −1 3 = න 3 3
= න ∙ −18
1 − 9 2
1
න −1 3 = න = ∙ 3
= − න 1 − 3 2 11
= −6න
3

= 1 − 9 2 = 1 −12
6
= 1 − 9 2 = − න

= −1 3 − න ∙ 3 Integrate = −18 1
1 − 9 2 3
1 2
= −1 3 − න 3 න = −18 = − 6 1 +
1 − 9 2 1 − 9 2
2
using u-subs. 12
= − 6 ∙ 1 +
= −1 3 − 1 1 − 9 2 +
−3

න −1 3 = −1 3 + 1 1 − 9 2 + # න 3 1 1 − 9 2 +
3 1 − 9 2 = − 3

Or trigo subs with Radical 2 − 2

Trigo-subs : =

MOOC MAT438/ UiTM

Example 5 : ‫ ׬‬ +

MOOC MAT438/ UiTM

 Tabular integrals

(algebraic (NR) and exponential functions (R) – has repeatable function – can use tabular integral)

Use integration by parts to solve න + Product function : න 2 = 1 2 +
never use product 2

Column u : differentiate rule to evaluate න 1 2 = 1 න 2 = 1 ∙ 1 2 + = 1 2 +
Column dv : integrate the integration 2 2 2 2 4
Solution
න 1 2 = 1 න 2 = 1 ∙ 1 2 + = 1 2 +
න 2 2 + 3 4 4 4 2 8

EA

sign u dv Using ILATE rule, A come
first. So A should be u, and
+ multiply 2 + 3 2 the other function (E)
becomes dv
begin − multiply 1
with +ve + multiply 2 2 2
−
sign, 1 න = +
then multiply 4
alternate 2 2
with the
sign 1
8
0 2 න 0 =

න

∴ න 2 2 + 3 = 2 + 3 ∙ 1 2 − 2 ∙ 1 2 + 2 ∙ 1 2 − න 0 (and we put + C)
2 4 8
Q : How many rows should because,
Q&A we differentiate? simplify = 1 2 2 + 3 − 1 2 + 1 2 + #
2 2 4 = 0
A : until the derivative of
algebraic function gives
the answer 0

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න 2 2 + 3 ∶ 1 2 2 + 3 − 1 2 + 1 2 +
2 2 4

න 2 2 + 3 = න 2 + 3 2 Note : dx always included in dv

EA u differentiate integrate
= න = 2
dv = 2 + 3
= 1 2
න = 1 +
2
= 2

= − න

= 2 + 3 ∙ 1 2 − 1 2 ∙ 2
2 න2

= 1 2 2 + 3 − න 2
2

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න 2 2 + 3 ∶ 1 2 2 + 3 − 1 2 + 1 2 +
2 2 4

න 2 2 + 3 = 1 2 2 + 3 − න 2 Notes : dx always included in dv
2 Au Edv

= 1 2 2 + 3 − න differentiate integrate
2 = = 2

= 1 2 න = 1 +

1 = 2
= 2 2 2 + 3 − − න

= 1 2 2 + 3 − 1 2 − 1 2
2 ∙2 න2

= 1 2 2 + 3 − 1 2 + 1 න 2 න = 1 +
2 2 2

= 1 2 2 + 3 − 1 2 + 11 2 + 1 2 2 + 3 − 1 2 + 1 2 + #
2 2 2∙2 =2 2 4

MOOC MAT438/ UiTM

Example 6 : ‫ ׬‬ −

MOOC MAT438/ UiTM

 Tabular integrals

(algebraic (NR) and exponential functions (R) – has repeatable function – can use tabular integral)

Use integration by parts to solve න − 1
න 2 = 2 2 +

Column u : differentiate Product function : 11 11 1
Solution Column dv : integrate never use product න 2 2 = 2 න 2 = 2 ∙ − 2 2 + = − 4 2 +
rule to evaluate the
න 2 − 1 2 11 11 1
integration න − 4 2 = − 4 න 2 = − 4 ∙ 2 2 + = − 8 2 +
AT

sign u dv Using ILATE rule, A come
first. So A should be u, and
+ multiply 2 − 1 2 the other function (T)
becomes dv
begin − multiply 1 න 0 =
with +ve + multiply 2 2 2
− න = + (and we put + C)
sign, 2 1 because,
then multiply − 4 2
alternate න = − +
with the 1 = 0
sign

0 − 8 2

න

∴ න 2 − 1 2 = 2 − 1 1 1 1 + න 0
∙ 2 2 − 2 ∙ − 4 2 + 2 ∙ − 8 2
Q : How many rows should

Q&A we differentiate? simplify = 1 2 − 1 2 + 1 1 + #
A : until the derivative of 2 2 2 − 4 2

algebraic function gives

the answer 0

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න 2 − 1 2 ∶ 1 2 − 1 2 + 1 1 +
2 2 2 − 4 2

න 2 − 1 2 = න differentiate integrate 1
Au Tdv = − න = 2 − 1 = 2 න = +

= 2 1
= 2 2

Note : dx always included in dv 11
∙ 2 2 − න 2 2 ∙ 2
= 2 − 1

1 2 − 1 2 − න 2
=2
Au T differentiate integrate
1 dv =
=2
2 − 1 2 − න = = 2 1
න = − +
Note : dx always included in dv 1

1 2 − 1 2 − − න = − 2 2
=2

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න 2 − 1 2 1 2 − 1 2 + 1 1 +
∶ 2 2 2 − 4 2

න 2 − 1 2 1 2 − 1 2 − න = = 2
=2 =
1
1 2 − 1 2 − − න = − 2 2
=2

1 2 − 1 2 − 11
=2 ∙ − 2 2 − න − 2 2

1 2 − 1 2 + 1 − 1 1
=2 2 2 2 න 2 න = +

1 2 −1 2 + 1 − 11 +
=2 2 2 2 ∙ 2 2

1 2 − 1 2 + 1 1 #
MOOC MAT438/ UiTM = 2 2 2 − 4 2 +

MOOC MAT438/ UiTM

 Tabular integrals

(algebraic (NR) and exponential functions (R) – has both repeatable function – can use tabular integral)

Use integration by parts to solve න −3 2 + 1 Product function :
never use product
Solution Column u : differentiate rule to evaluate the
Column dv : integrate
integration

න −3 2 + 1 dv Using ILATE rule, T come
first. So T should be u, and
ET the other function (E)
becomes dv
sign u

+ 2 + 1 −3 න = +

begin multiply
with +ve
− 2 2 + 1 − 1 −3 1
sign, 3 −3
then multiply න −3 = −3 +
alternate
with the + −4 2 + 1 1 −3 1 1 1 1 1
sign 9 3 3 3 3 9
multiply න න − −3 = − න −3 = − ∙ − −3 + = −3 +

∴ න −3 2 + 1 = 2 + 1 ∙ − 1 −3 − 2 2 + 1 ∙ 1 −3 + න −4 2 + 1 ∙ 1 −3
3 9 9

Q&A Q : How many rows should
we differentiate?

A : until the derivative of Trigo

gives the same function as

the given question

MOOC MAT438/ UiTM

 Tabular integrals

(algebraic (NR) and exponential functions (R) – has both repeatable function – can use tabular integral)

Use integration by parts to solve න −3 2 + 1

Solution

Integration by parts

∴ න −3 2 + 1 = 2 + 1 1 − 2 2 + 1 1 න −4 2 + 1 1 again??
∙ − −3 ∙ −3 + ∙ −3
39 9

1 න −3 2 + 1 = − 1 −3 2 + 1 − 2 −3 2 +1 − 4 න −3 2 +1
3 9 9

1 න −3 2 + 1 + 4 න −3 2 + 1 = − 1 −3 2 + 1 − 2 −3 2 +1 Same term
9 3 9 with LHS!
So, we can
13 න −3 2 + 1 = − 1 −3 2 + 1 + 2 + 1 move to LHS
9 9

න −3 2 + 1 = 9 ∙ − 1 −3 2 + 1 + 2 + 1 +
13 9

→ න −3 2 + 1 = − 1 −3 3 2 + 1 + 2 2 + 1 + #
13

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න −3 2 + 1 ∶ − 1 −3 3 2 + 1 + 2 2 + 1 +
13

න −3 2 + 1 = න 2 + 1 ∙ −3 Note : dx always included in dv

ET u dv differentiate integrate
= න
= (2 +1) = −3 න 1
= +

= − න = 2 cos (2 +1) = − 1 −3
3

= (2 +1) ∙ − 1 −3 − න − 1 −3 ∙ 2 cos (2 +1)
3 3

= − 1 −3 (2 +1) + 2 න −3 cos (2 +1)
3 3

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න −3 2 + 1 ∶ − 1 −3 3 2 + 1 + 2 2 + 1 +
13

න −3 2 + 1 = − 1 −3 (2 +1) + 2 න −3 cos (2 +1)
3 3 E T

= − 1 −3 (2 +1) 2 ∙ −3 Notes : dx always included in dv
3 + 3 න 2 + 1

1 2 u dv
3 + 3 න
= − −3 (2 +1) differentiate integrate
= −3
= cos (2 +1)

= − 1 −3 (2 +1) 2 − න = −2 sin (2 +1) = − 1 −3
3 +3 3

න = +

= − 1 −3 (2 +1) 2 cos (2 +1) ∙ − 1 −3 − න − 1 −3 ∙ −2 sin (2 +1)
3 +3 3 3

MOOC MAT438/ UiTM

‫׬‬ = − ‫׬‬

න −3 2 + 1 ∶ − 1 −3 3 2 + 1 + 2 2 + 1 +
13

න −3 2 + 1 = − 1 −3 (2 +1) 2 cos (2 +1) ∙ − 1 −3 − න − 1 −3 ∙ −2 sin (2 +1)
න −3 2 + 1 3 +3 3 3

= − 1 −3 (2 +1) − 2 −3 cos (2 +1) − 4 න −3 sin (2 +1) at this point,
3 9 9 it is similar to
using tabular
integrals

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

= 3 Can’t use integration
1 by u-subs..
= 3 ∙ 3

= න ∙ =
= න = න ∙

= +
= 3 + #

MOOC MAT438/ UiTM

න 3 ‫׬‬ = − ‫׬‬

1
∶ 2 3 + 3 +

න 3 = න Note : dx always included in dv differentiate integrate
= ( 3 ) =
u dv
= − න 1 =
= − ( 3 ) ∙ 3 ∙ 3
Can’t use Tabular Integrals
− ( 3 )
=

= 3 − න ∙ − ( 3 )



= 3 + න ( 3 )

MOOC MAT438/ UiTM

න 3 ‫׬‬ = − ‫׬‬

1
∶ 2 3 + 3 +

Note : dx always included in dv

න 3 = 3 + න ( 3 )

u dv differentiate integrate
= 3 + න = ( 3 ) =

= 3 + − න 1 =
= ( 3 ) ∙ 3 ∙ 3

( 3 )
=

= 3 + 3 − න ∙ ( 3 )



= 3 + 3 − න ( 3 )

MOOC MAT438/ UiTM

න 3 ‫׬‬ = − ‫׬‬

1
∶ 2 3 + 3 +

න 3 = 3 + 3 − න ( 3 ) Integration by parts
again??

න ( 3 ) + න ( 3 ) = 3 + 3 Same term
with LHS!
So, we can
move to LHS

න ( 3 ) = 3 + 3

1
න ( 3 ) = 2 3 + 3 + #

MOOC MAT438/ UiTM

= 3 + # 1
= 2 3 + 3 + #

1 +
= 2 3 + 3

= 3 + 1 = 3 + 3
= 2

1 ′ = − 3 1 1
= 3 ∙ 3 ∙ 3 + 0 ∙ 3 ∙ 3 + 3 ∙ 3 ∙ 3

3 1 ′ = 3 3
= ′ = 2 − +



= ′ + ′


= 1 3 + 1 3 + − 1 3 + 1 3 +0
2
2 2 2

MOOC MAT438/ UiTM
= 3

න 3 1
∶ 2 3 + 3 +

2 ( 3 ) 2 − ( 3 ) + 3
න 3 = 3 − 2 ∙ − +න2 ∙
2

න 3 11
u dv න 3 = 3 + 2 ( 3 ) + 2 න − ( 3 ) + 3

11 1
න 3 = 3 + 2 ( 3 ) − 2 න 3 + 2 න 3

sign u dv = ( 3 ) not the same
+ multiply 3 1 term as LHS!

′ = − ( 3 ) ∙ 3 ∙ 3

′ = − ( 3 )

( 3 )

− −multiply = − ( 3 ) = − 3 =
′ = 1
1
′ = − ( 3 ) ∙ 3 ∙ 3
− ( 3 ) + 3 2 ′ − ′ ( 3 )
′ = 2 ′ = −
+
multiply 22 න

′ = − ( 3 ) + 3

MOOC MAT438/ UiTM 2

MOOC MAT438/ UiTM