DEMO CLASS
INTEGRATION BY
PARTS
AGENDA
• REVIEW
• INTRODUCTION
• EXAMPLS
• TRICKY EXAMPLS
• SUMMARY
2
REVIEW
COMMON
DIFFERENTIATION
𝑥𝑛
𝑒𝑥
𝑙𝑛𝑥
𝑎𝑥
sin 𝑥
cos 𝑥
tan 𝑥
sec 𝑥
csc 𝑥
cot 𝑥
INTEGRATION BY SUBSTITUTION
Previously
𝑥 2𝑥 + 5𝑑𝑥
But now
𝑥 sin 𝑥 𝑑𝑥
INTRODUCTION
Suppose we wish to integrate the product of two functions, such as x sin x, where
one of the functions is not related to the derivative of the other.
An expression such as this can be integrated using the method of integration by
parts.
When we differentiate the product of two functions u and v we use the product rule:
d
dv
du
(uv ) = u
+v
dx
dx
dx
where u and v are functions of x.
Integrating throughout with respect to x gives:
uv =  u
dv
du
dx +  v
dx
dx
dx
INTRODUCTION
This can be rearranged to give:
dv
du
 u dx dx = uv   v dx dx
To integrate a product using this formula we let one part equal
u and the other equal dv.
dx
du
We find
by differentiating the part we called u.
dx
dv
We find v by integrating the part we called
.
dx
dv
It is important to choose u and
so that
dx
dv
du
is easier to integrate than  u
.
dx
v
dx
 dx
dx
INTRODUCTION
So, to integrate x sin x with respect to x:
Let
differentiate
So
u=x
du
=1
dx
Now, using the formula
and
and
dv
= sin x
dx
integrate
v =  cos x
dv
du
:
 u dx dx = uv   v dx dx
 x sin x dx =  x cos x    cos x dx
=  x cos x   sin x + c
= sin x  x cos x + c
We don’t need
the “+ c” here.
EXAMPLE 1
Find
2x
xe
 . dx
dv
= e2 x
u=x
dx
2x
du
e
So
and
=1
v=
dx
2
dv
du
Now, apply the formula
:
 u dx dx = uv   v dx dx
e2 x
e2 x
2x
 xe dx = x 2   2 dx
xe 2 x e 2 x
=

+c
2
4
e2 x
=
(2 x  1) + c
4
Let
and
EXAMPLE 2
Find
 8 x ln.x dx
We don’t know the integral of ln x so:
let
so
u = ln x
du 1
=
dx x
and
dv
= 8x
dx
and
v = 4 x2
dv
du
dx = uv:   v
dx
dx
dx
1
2
2
 8 x ln x dx = 4 x ln x   4 x × x dx
= 4 x 2 ln x   4 x dx
Now, using the formula,
u
= 4 x 2 ln x  2 x 2 + c
= 2 x 2 (2ln x  1) + c
INTEGRATION BY PARTS OF
DEFINITE INTEGRALS
To evaluate a definite integral using integration by parts we use:
b du
dv
b
a u dx dx = uv a  a v dx dx
b
Evaluate
Let
So
1
−1 𝑥 𝑑𝑥
tan
0
.
u=tan−1 𝑥
𝑑𝑦
1
=
𝑑𝑥
1+𝑥 2
and
and
v=x
dv
=1
dx
TRICKY ONE: (𝒙 − 𝟐) 𝟖 − 𝒙
TRICKY ONE:
𝒆𝒙 𝐂𝑶𝑺 𝑿
SUMMARY
Review common differentiation
Integral by parts