Cul. 7

Integration
by parts
mc-TY-parts-2009-1
A special rule, integration by parts, is available for integrating products of two functions. This
unit derives and illustrates this rule with a number of examples.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• state the formula for integration by parts
• integrate products of functions using integration by parts
Contents
1. Introduction 2
2. Derivation of the formula for integration by parts
Z u
dv
dx
dx = u v ? Z v
du
dx
dx 2
3. Using the formula for integration by parts 5
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1. Introduction
Functions often arise as products of other functions, and we may be required to integrate these
products. For example, we may be asked to determine
Z x cos x dx .
Here, the integrand is the product of the functions x and cos x. A rule exists for integrating
products of functions and in the following section we will derive it.
2. Derivation of the formula for integration by parts
We already know how to differentiate a product: if
y = u v
then
dy
dx
=
d(uv)
dx
= u
dv
dx
+ v
du
dx
.
Rearranging this rule:
u
dv
dx
=
d(uv)
dx ? v
du
dx
.
Now integrate both sides:
Z u
dv
dx
dx = Z d(uv)
dx
dx ? Z v
du
dx
dx .
The first term on the right simplifies since we are simply integrating what has been differentiated.
Z u
dv
dx
dx = u v ? Z v
du
dx
dx .
This is the formula known as integration by parts.
Key Point
Integration by parts
Z u
dv
dx
dx = u v ? Z v
du
dx
dx
The formula replaces one integral (that on the left) with another (that on the right); the intention
is that the one on the right is a simpler integral to evaluate, as we shall see in the following
examples.
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3. Using the formula for integration by parts
Example
Find Z x cos x dx.
Solution
Here, we are trying to integrate the product of the functions x and cos x. To use the integration
by parts formula we let one of the terms be
dv
dx
and the other be u. Notice from the formula that
whichever term we let equal u we need to differentiate it in order to find
du
dx
. So in this case, if
we let u equal x, when we differentiate it we will find
du
dx
= 1, simply a constant. Notice that
the formula replaces one integral, the one on the left, by another, the one on the right. Careful
choice of u will produce an integral which is less complicated than the original.
Choose
u = x and
dv
dx
= cos x .
With this choice, by differentiating we obtain
du
dx
= 1 .
Also from
dv
dx
= cos x, by integrating we find
v = Z cos x dx = sin x .
(At this stage do not concern yourself with the constant of integration). Then use the formula
Z u
dv
dx
dx = u v ? Z v
du
dx
dx :
Z x cos x dx = x sin x ? Z (sin x) · 1 dx
= x sin x + cos x + c
where c is the constant of integration.
In the next Example we will see that it is sometimes necessary to apply the formula for integration
by parts more than once.
Example
Find Z x2e3x dx.
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Solution
We have to make a choice and let one of the functions in the product equal u and one equal
dv
dx
.
As a general rule we let u be the function which will become simpler when we differentiate it. In
this case it makes sense to let
u = x2 and
dv
dx
= e3x .
Then
du
dx
= 2x and v = Z e3xdx =
1
3
e3x .
Then, using the formula for integration by parts,
Z x2e3x dx =
1
3
e3x · x2 ? Z 1
3
e3x · 2x dx
=
1
3
x2e3x ? Z 2
3
xe3x dx .
The resulting integral is still a product. It is a product of the functions 2
3x and e3x. We can use
the formula again. This time we choose
u =
2
3
x and
dv
dx
= e3x .
Then
du
dx
=
2
3
and v = Z e3xdx =
1
3
e3x .
So
Z x2e3x dx =
1
3
x2e3x ? Z 2
3
xe3x dx
=
1
3
x2e3x ? 2
3
x ·
1
3
e3x ? Z 1
3
e3x ·
2
3
dx
=
1
3
x2e3x ?
2
9
xe3x +
2
27
e3x + c
where c is the constant of integration. So we have done integration by parts twice to arrive at
our final answer.
Remember that to apply the formula you have to be able to integrate the function you call
dv
dx
.
This can cause problems — consider the next Example.
Example
Find Z x ln |x| dx.
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Solution
Remember the formula:
Z u
dv
dx
dx = u v ? Z v
du
dx
dx .
It would be natural to choose u = x so that when we differentiate it we get
du
dx
= 1. However
this choice would mean choosing
dv
dx
= ln |x| and we would need to be able to integrate this.
This integral is not a known standard form. So, in this Example we will choose
u = ln |x| and
dv
dx
= x
from which
du
dx
=
1
x
and v = Z x dx =
x2
2
.
Then, applying the formula
Z x ln |x| dx =
x2
2
ln |x| ? Z x2
2 ·
1
x
dx
=
x2
2
ln |x| ? Z x
2
dx
=
x2
2
ln |x| ?
x2
4
+ c
where c is the constant of integration.
Example
Find Z ln |x|dx.
Solution
We can use the formula for integration by parts to find this integral if we note that we can write
ln |x| as 1 · ln |x|, a product. We choose
dv
dx
= 1 and u = ln |x|
so that
v = Z 1 dx = x and
du
dx
=
1
x
.
Then,
Z 1 · ln |x|dx = x ln |x| ? Z x ·
1
x
dx
= x ln |x| ? Z 1 dx
= x ln |x| ? x + c
where c is a constant of integration.
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Example
Find Z ex sin x dx.
Solution
Whichever terms we choose for u and
dv
dx
it may not appear that integration by parts is going
to produce a simpler integral. Nevertheless, let us make a choice:
dv
dx
= sin x and u = ex
so that
v = Z sin x dx = ?cos x and
du
dx
= ex .
Then,
Z ex sin x dx = ex · ?cos x ? Z ?cos x · exdx
= ?cos x · ex + Z ex cos xdx .
We now integrate by parts again choosing
dv
dx
= cos x and u = ex
so that
v = Z cos x dx = sin x and
du
dx
= ex .
So
Z ex sin x dx = ?cos x · ex + ex sin x ? Z sin x · exdx
= ?ex cos x + ex sin x ? Z ex sin xdx .
Notice that the integral we have ended up with is exactly the same as the one we started with.
Let us call this I. That is I = Z ex sin x dx.
So
I = ex sin x ? ex cos x ? I
from which
2I = ex sin x ? ex cos x
and
I =
1
2
(ex sin x ? ex cos x) .
So
Z ex sin x dx =
1
2
(ex sin x ? ex cos x) + c
where c is the constant of integration.
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Exercises
1. Evaluate the following integrals:
(a) Z x sin x dx (b) Z x cos 4x dx (c) Z xe?xdx (d) Z x2 cos x dx
(e) Z 2x2exdx (f) Z x2 ln |x| dx (g) Z tan?1 x dx (h) Z sin?1 x dx
(i) Z ex cos x dx (j) Z sin3 x dx (Hint: write sin3 x as sin2 x sin x.)
2. Calculate the value of each of the following:
(a) Z 
0
x cos 1
2x dx (b) Z 1
0
x2exdx (c) Z 2
1
x3 ln |x| dx (d) Z /4
0
x2 sin 2x dx
(e) Z 1
0
x tan?1 x dx
Answers
1.
(a) ?x cos x + sin x + C (b) 1
4x sin 4x + 1
16 cos 4x + C
(c) ?xe?x ? e?x + C (d) x2 sin x + 2x cos x ? 2 sin x + C
(e) 2x2ex ? 4xex + 4ex + C (f) 1
3x3 ln |x| ? 1
9x3 + C
(g) x tan?1 x ? 1
2 ln |1 + x2| + C (h) x sin?1 x + p1 ? x2 + C
(i) 1
2ex(cos x + sin x) + C (j) ?1
3 (cos x sin2 x + 2 cos x) + C
2.
(a) 2 ? 4 (b) e ? 2 (c) 4 ln 2 ?
15
16
(d)

8 ?
1
4
(e)

4 ?
1
2
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