Higher Engineering Mathematics

Integral calculus

44

Reduction formulae

44.1 Introduction

When using integration by parts in Chapter 43, an

integral such as

∫

x^2 exdxrequires integration by

parts twice. Similarly,

∫

x^3 exdxrequires integra-

tion by parts three times. Thus, integrals such as∫

x^5 exdx,

∫

x^6 cosxdxand

∫

x^8 sin 2xdxfor exam-

ple, would take a long time to determine using

integration 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 that

nhas 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 the

same integral inn−1.

Problem 1. Determine

∫

x^2 exdxusing a reduc-

tion formula.

Using equation (1) withn=2 gives:

∫

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 exam-

ples the constant of integration will be added at the

last step with indefinite integrals.

Problem 2. Use a reduction formula to deter-

mine

∫

x^3 exdx.

From equation (1),In=xnex−nIn− 1

Hence

∫

x^3 exdx=I 3 =x^3 ex− 3 I 2

I 2 =x^2 ex− 2 I 1

I 1 =x^1 ex− 1 I 0