Chapter 44
Reduction formulae
44.1 Introduction
When using integration by parts in Chapter 43, an
integral such as
∫
x^2 exdx requires integration by
parts twice. Similarly,
∫
x^3 exdx requires integra-
tion by parts three times. Thus, integrals such as∫
x^5 exdx,
∫
x^6 cosxdxand
∫
x^8 sin2xdxfor example,
would take a long time to determine using integra-
tion by parts.Reduction formulaeprovide 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
∫
xnexdx
To determine
∫
xnexdxusing integration by parts,
let u=xnfrom which,
du
dx
=nxn−^1 and du=nxn−^1 dx
and dv=exdxfrom which,
v=
∫
exdx=ex
Thus,
∫
xnexdx=xnex−
∫
exnxn−^1 dx
using the integration by parts formula,
=xnex−n
∫
xn−^1 exdx
The integral on the far right is seen to be of the same
form as the integral on the left-hand side, except thatn
has been replaced byn−1.
Thus, if we let,
∫
xnexdx=In,
then
∫
xn−^1 exdx=In− 1
Hence
∫
xnexdx=xnex−n
∫
xn−^1 exdx
can be written as:
In=xnex−nIn− 1 (1)
Equation (1) is an example of a reduction formula since
it expresses an integral innin terms of thesame integral
inn−1.
Problem 1. Determine
∫
x^2 exdxusing a
reduction formula.
Using equation (1) withn=2gives:
∫
x^2 exdx=I 2 =x^2 ex− 2 I 1
and I 1 =x^1 ex− 1 I 0
I 0 =
∫
x^0 exdx=
∫
exdx=ex+c 1
Hence I 2 =x^2 ex−2[xex− 1 I 0 ]
=x^2 ex−2[xex− 1 (ex+c 1 )]
i.e.
∫
x^2 exdx=x^2 ex− 2 xex+2ex+ 2 c 1
=ex(x^2 − 2 x+2)+c
(wherec=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 exdx.