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LIATE Method and Tabular Integration by Parts, Exams of Differential and Integral Calculus

The liate method is a rule for integration by parts, helping to determine which function should be used as u and dv. Examples and exceptions to the liate rule. Tabular integration by parts is an alternative method for recursive integration by parts.

What you will learn

  • What is the LIATE method and how is it used in integration by parts?
  • What are the exceptions to the LIATE rule and how can they be solved?
  • How does tabular integration by parts simplify the process of recursive integration by parts?

Typology: Exams

2021/2022

Uploaded on 09/27/2022

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“LIATE” AND TABULAR INTERGRATION BY PARTS
1. LIATE
The LIATE method was first mentioned by Herbert E. Kasube in [1]. The
function that appears first in the following list should be uwhen using integration
by parts:
LLogatithmic functions ln(x), log2(x), etc.
IInverse trig. functions tan1(x), sin1(x), etc.
AAlgebraic functions x, 3x2, 5x25, etc.
TTrig. functions cos(x), tan(x), etc.
EExponential functions ex, 2x, etc.
Example 1.
Zxsin(x)dx.
Following the LIATE method, u=xand dv = sin(x)dx since xis an algebraic
function and sin(x) is a trigonometric function. Therefore,
u=x dv = sin(x)dx
du =dx v =cos(x)
and
Zxsin(x)dx =xcos(x)Z(cos(x))dx
=xcos(x) + sin(x) + C.
WARNING: This technique is not perfect!(1.1)
There are exceptions to LIATE. Some of these can be solved using the order
“ILATE” instead. Sometimes, something completely different needs to be con-
sidered.
Example 2.
Zx3ex2
dx.
Following the LIATE rule, u=x3and dv =ex2dx. However, we would actually
set u=x2and dv =xex2.
u=x2dv =xex2dx
du = 2xdx v =1
2ex2
1
pf3

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1. LIATE

The LIATE method was first mentioned by Herbert E. Kasube in [1]. The function that appears first in the following list should be u when using integration by parts:

L Logatithmic functions ln(x), log 2 (x), etc. I Inverse trig. functions tan−^1 (x), sin−^1 (x), etc. A Algebraic functions x, 3x^2 , 5x^25 , etc. T Trig. functions cos(x), tan(x), etc. E Exponential functions ex, 2x, etc.

Example 1. ∫ x sin(x)dx.

Following the LIATE method, u = x and dv = sin(x)dx since x is an algebraic function and sin(x) is a trigonometric function. Therefore,

u = x dv = sin(x)dx du = dx v = − cos(x)

and ∫ x sin(x)dx = −x cos(x) −

(− cos(x))dx

= −x cos(x) + sin(x) + C.

WARNING: This technique is not perfect!

There are exceptions to LIATE. Some of these can be solved using the order “ILATE” instead. Sometimes, something completely different needs to be con- sidered.

Example 2. ∫ x^3 ex

2 dx.

Following the LIATE rule, u = x^3 and dv = ex

2 dx. However, we would actually set u = x^2 and dv = xex

2 . u = x^2 dv = xex

2 dx du = 2xdx v = 12 ex 2 1

and so ∫ x^3 ex

2 dx = x^2

ex

2

ex

2 2 xdx

x^2 ex

2 −

xex

2 dx

x^2 ex

2 −

ex

2

  • C

=

ex

2 (x^2 − 1) + C.

Example 3. (^) ∫

sec^3 (x)dx.

Following the LIATE rule, u = 1 and dv = sec^3 (x)dx. However, we would actually set u = sec(x) and dv = sec^2 (x).

u = sec(x) dv = sec^2 (x)dx du = sec(x) tan(x)dx v = tan(x) and so ∫ sec^3 (x)dx = sec(x) tan(x) −

tan^2 (x) sec(x)dx

= sec(x) tan(x) −

(sec^2 (x) − 1) sec(x)dx

= sec(x) tan(x) −

sec^3 (x) − sec(x)

dx

= sec(x) tan(x) −

sec^3 (x)dx +

sec(x)dx

= sec(x) tan(x) −

sec^3 (x)dx + ln | sec(x) + tan(x)|.

Since the integral we are solving reappears, we need to add it to the left side to get

2

sec^3 (x)dx = sec(x) tan(x) + ln | sec(x) + tan(x)|.

Finally,

∫ sec^3 (x)dx =

sec(x) tan(x) +

ln | sec(x) + tan(x)| + C.

  1. Tabular Integration By Parts When integration by parts is needed more than once you are actually doing integration by parts recursively. This leads to an alternative method which just makes the amount of writing significantly less. I will explain this through the following example.