Higher Engineering Mathematics

Integral calculus

43

Integration by parts

43.1 Introduction

From the product rule of differentiation:

d

dx

(uv)=v

du

dx

+u

dv

dx

,

whereuandvare both functions ofx.

Rearranging gives:u

dv

dx

=

d

dx

(uv)−v

du

dx

Integrating both sides with respect toxgives:

∫

u

dv

dx

dx=

∫

d

dx

(uv)dx−

∫

v

du

dx

dx

i.e.

∫

u

dv

dx

dx=uv−

∫

v

du

dx

dx

or

∫

udv=uv−

∫

vdu

This is known as the integration by parts for-

mulaand provides a method of integrating such

products of simple functions as

∫

xexdx,

∫

∫ tsintdt,

eθcosθdθand

∫

xlnxdx.

Given a product of two terms to integrate the ini-

tial choice is: ‘which part to make equal tou’ and

‘which part to make equal tov’. The choice must

be such that the ‘upart’ becomes a constant after

successive differentiation and the ‘dvpart’ can be

integrated from standard integrals. Invariable, the

following rule holds: If a product to be integrated

contains an algebraic term (such asx,t^2 or 3θ) then

this term is chosen as theupart. The one exception

to this rule is when a ‘lnx’ term is involved; in this

case lnxis chosen as the ‘upart’.

43.2 Worked problems on integration

by parts

Problem 1. Determine

∫

xcosxdx.

From the integration by parts formula,

∫

udv=uv−

∫

vdu

Letu=x, from which

du

dx

=1, i.e. du=dxand let

dv=cosxdx, from whichv=

∫

cosxdx=sinx.

Expressions foru,duandvare now substituted

into the ‘by parts’ formula as shown below.

i.e.

∫

xcosxdx=xsinx−(−cosx)+c

=xsinx+cosx+c

[This result may be checked by differentiating the

right hand side,

i.e.

d

dx

(xsinx+cosx+c)

=[(x)(cosx)+(sinx)(1)]−sinx+ 0

using the product rule

=xcosx, which is the function

being integrated]

Problem 2. Find

∫

3 te^2 tdt.

Letu= 3 t, from which,

du

dt

=3, i.e. du=3dtand

let dv=e^2 tdt, from which,v=

∫

e^2 tdt=

1

2

e^2 t

Substituting into

∫

udv=uv−

∫

vdugives:

∫

3 te^2 tdt=(3t)

(

1

2

e^2 t

)

−

∫ (

1

2

e^2 t

)

(3 dt)