8.10
Integration by parts
Introduction
The technique known as integration by parts is used to integrate a product of two functions,
for example
Z
Z 1
2x
e sin 3x dx
and
x3 e−2x dx
0
This leaflet explains how to apply this technique.
1. The integration by parts formula
We need to make use of the integration by parts formula which states:
Z
!
dv
u
dx = uv −
dx
Z
!
du
v
dx
dx
Note that the formula replaces one integral, the one on the left, with a different integral, that
on the right. The intention is that the latter is simpler to evaluate. Note also that to apply the
formula we must let one function in the product equal u. We must be able to differentiate this
du
dv
function to find
. We let the other function in the product equal
. We must be able to
dx
dx
integrate this function, to find v. Consider the the following example:
Example
Z
Find 3x sin x dx.
Solution
Compare the required integral with the formula for integration by parts: we see that it makes
sense to choose
dv
u = 3x
and
= sin x
dx
It follows that
du
=3
dx
and
v=
Z
sin x dx = − cos x
dv
to find v there is no need to include a constant of integration. When you
dx
become confident with the method, you may like to think about why this is the case.) Applying
(When integrating
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the formula we obtain
Z
!
du
dx
v
dx
Z
3x sin x dx = uv −
= 3x(− cos x) −
= −3x cos x + 3
Z
Z
(− cos x).(3) dx
cos x dx
= −3x cos x + 3 sin x + c
2. Dealing with definite integrals
When dealing with definite integrals (those with limits of integration) the corresponding formula
is
Z
b
a
!
dv
u
dx = [uv]ba −
dx
Z
b
a
!
du
v
dx
dx
Example
Z
2
Find
xex dx.
0
Solution
du
dv
= ex . Then
= 1 and v = ex . Using the formula for integration by
We let u = x and
dx
dx
parts we obtain
Z
2
x
xe dx =
[xex ]20
−
0
Z
2
ex .1dx
0
= (2e2 ) − (0e0 ) − [ex ]20
= 2e2 − [e2 − 1]
= e2 + 1
(or 8.389 to 3d.p.)
Exercises
R
1. Find a) x sin(2x)dx,
b)
R
te3t dt,
c)
2. Evaluate the following definite integrals:
a)
R1
0
x cos 2x dx,
b)
R π/2
0
x sin 2x dx, c)
R1
−1
R
x cos x dx.
te2t dt.
(Remember to set your calculator to radian mode for evaluating the trigonometric functions.)
3. Find
R2
0
x2 ex dx. (You will need to apply the integration by parts formula twice.)
Answers
1. a) sin42x −
x cos 2x
2
2. a) 0.1006,
+ c,
b) 0.7854,
b) e3t ( 3t − 19 ) + c,
c) cos x + x sin x + c.
c) 1.9488
3. 12.778 (3dp).
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