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.