# 8-2: Integration By Parts

```8-2: Integration By Parts
Objectives:
1. To find an
antiderivative using
integration by parts
Assignment:
• P. 531-533: 9-17 odd,
23-33 odd, 39, 43, 49,
53, 73-78, 99, 103a-c,
107
• Homework Supplement
Warm Up
Let 𝑢 and 𝑣 be
differentiable
functions of 𝑥.
𝑑
Find
𝑢𝑣 .
𝑑𝑥
𝑑
𝑢𝑣 = 𝑣𝑢′ + 𝑢𝑣′
𝑑𝑥
𝑑
𝑑𝑢
𝑑𝑣
𝑢𝑣 = 𝑣
+𝑢
𝑑𝑥
𝑑𝑥
𝑑𝑥
Product Rule
The product of two differentiable functions 𝑓
and 𝑔 is also differentiable such that
𝑑
𝑓(𝑥) ∙ 𝑔(𝑥) = 𝑓 ′ 𝑥 ∙ 𝑔 𝑥 + 𝑓(𝑥) ∙ 𝑔′ (𝑥)
𝑑𝑥
Derivative of the first times the second + first times the derivative of the second
Objective 1
You will be able to find an antiderivative using
integration by parts
Working Backwards
Given that 𝑢 and 𝑣
are differentiable
functions of 𝑥, let’s
work the Product
Rule backwards.
Solve for 𝑢 𝑑𝑣
𝑑
𝑑𝑢
𝑑𝑣
𝑢𝑣 = 𝑣
+𝑢
𝑑𝑥
𝑑𝑥
𝑑𝑥
Multiply by 𝑑𝑥
Integrate
both sides
𝑑 𝑢𝑣 = 𝑣 𝑑𝑢 + 𝑢 𝑑𝑣
𝑑 𝑢𝑣 =
𝑢𝑣 =
𝑣 𝑑𝑢 + 𝑢 𝑑𝑣
𝑣 𝑑𝑢 +
𝑢 𝑑𝑣 = 𝑢𝑣 −
𝑢 𝑑𝑣
𝑣 𝑑𝑢
Integration By Parts
If 𝑢 and 𝑣 are functions of 𝑥 and have
continuous derivatives, then
𝑢 𝑑𝑣 = 𝑢𝑣 −
Try letting 𝑢 be
the part whose
derivative is
simpler than 𝑢.
𝑣 𝑑𝑢
Try letting 𝑑𝑣 be
the more
complicated part.
Notice that you
are exchanging
𝑢 𝑑𝑣 for 𝑣 𝑑𝑢,
which should be
simpler and
easier to
integrate.
Mnemonic Device
If 𝑢 and 𝑣 are functions of 𝑥 and have
continuous derivatives, then
𝑢 𝑑𝑣
Try letting 𝑢 be
the part whose
derivative is
simpler than 𝑢.
𝑢←L I
on
g v
= 𝑢𝑣 − 𝑣 𝑑𝑢
ae
r r
i s
t e
Try letting 𝑑𝑣 be
h
the more
mt
complicated part.
i r
c i
g
P
o
w
e
r
E
x
p
o
n
e
n
t
i
a
l
T → 𝑑𝑣
r
i
g
o
n
o
m
e
t
r
i
c
We use
integration by
parts when the
integrand is a
product, usually
of an algebraic
times a
transcendental
function.
Organizational Box
If 𝑢 and 𝑣 are functions of 𝑥 and have
continuous derivatives, then
Original
New
Differentiate
𝑢=
𝑣 𝑑𝑢
𝑣=
𝑑𝑥
𝑑𝑣 =
Integrate
𝑢 𝑑𝑣 = 𝑢𝑣 −
Exercise 1
𝑥𝑒 𝑥 𝑑𝑥 =
𝑢=𝑥
𝑣 = 𝑒𝑥
New
𝑑𝑢 = 𝑑𝑥
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
Original
𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 −
𝑒 𝑥 𝑑𝑥
𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝐶
Exercise 2
Find the average value of 𝑦 = ln 𝑥 on the
closed interval 1, 𝑒 .
Exercise 3
1
arcsin 𝑥 𝑑𝑥
0
=
Exercise 4
Find the particular solution of
given the initial condition
𝑑𝑦
𝑑𝑥
𝜋
,0
2
.
= 𝑥 sin 𝑥
Exercise 5
𝑥 2 sin 𝑥 𝑑𝑥 =
Exercise 6
sec 3 𝑥 𝑑𝑥 =
Exercise 7: AP FRQ
Let 𝑓 be the function defined for 𝑥 &gt; 0, with
𝑓 𝑒 = 2 and 𝑓′, the first derivative of 𝑓, given by
𝑓′ 𝑥 = 𝑥 2 ln 𝑥.
a) Write an equation for the line tangent to the
graph of 𝑓 at the point 𝑒, 2 .
b) Is the graph of 𝑓 concave up or concave down
on the interval 1 &lt; 𝑥 &lt; 3? Give a reason for
c) Use antidifferentiation to find 𝑓 𝑥 .
8-2: Integration By Parts
Objectives:
1. To find an
antiderivative using
integration by parts
Assignment:
• P. 531-533: 9-17 odd,
23-33 odd, 39, 43, 49,
53, 73-78, 99, 103a-c,
107
• Homework Supplement
```