Chapter 44
Reduction formulae
44.1 Introduction
When using integration by parts in Chapter 43, an
integral such as
x 2 e x dx requires integration by
parts twice. Similarly,
x 3 e x dx requires integration by parts three times. Thus, integrals such as
x 5 e x dx,
x 6 cos x dx and
x 8 sin 2x dx for example,
would take a long time to determine using integration by parts. Reduction formulae provide a quicker
method for determining such integrals and the method
is demonstrated in the following sections.
44.2 Using reduction formulae for
integrals of the form
x n e x dx
To determine
x n e x dx using integration by parts,
let
u = x n from which,
du
dx
= nx n−1 and du =nx n−1 dx
and
dv = e x dx from which,
v =
e
x dx = e
x
Thus,
x
n e
x dx = x n e x −
e
x nx
n−1 dx
using the integration by parts formula,
= x n e x − n
x
n−1 e
x dx
The integral on the far right is seen to be of the same
form as the integral on the left-hand side, except that n
has been replaced by n −1.
Thus, if we let,
x
n e
x dx = I n ,
then
x
n−1 e
x dx = I n−1
Hence
x
n e
x dx = x
n e
x
− n
x
n−1 e
x dx
can be written as:
I n = x
n e
x
− nI n−1
(1)
Equation (1) is an example of a reduction formula since
it expresses an integral in n in terms of the same integral
in n −1.
Problem 1. Determine
x 2 e x dx using a
reduction formula.
Using equation (1) with n = 2 gives:
x
2 e
x dx = I 2 = x 2 e x − 2I 1
and
I 1 = x 1 e x − 1I 0
I 0 =
x
0 e
x dx =
e
x dx = e
x
+ c 1
Hence
I 2 = x 2 e x − 2[xe x − 1I 0 ]
= x 2 e x − 2[xe x − 1(e x + c 1 )]
i.e.
x
2 e
x dx = x 2 e x − 2xe x + 2e x + 2c 1
= e x (x 2 − 2x +2) + c
(where c =2c 1 )
As with integration by parts, in the following examples
the constant of integration will be added at the last step
with indefinite integrals.
Problem 2. Use a reduction formula to determine
x
3 e
x dx.