Chapter 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 theintegration by parts formula
and provides a method of integrating such prod-
ucts of simple functions as

∫

xexdx,

∫
∫ tsintdt,
eθcosθdθand

∫
xlnxdx.
Given a product of two terms to integrate the initial
choice is: ‘which part to make equal tou’ and ‘which
part tomakeequal tov’.Thechoicemust besuch that the
‘upart’ becomes a constant after successive differenti-
ation and the ‘dvpart’ can be integrated from standard
integrals. Invariably, the followingrule holds: If a prod-
uct to be integrated contains an algebraic term (such as
x,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

Let u=x, from which

du

dx

=1, i.e. du=dx and let

dv=cosxdx, from whichv=

∫

cosxdx=sinx.

Expressions foru,duandvare now substituted into

the ‘by parts’ formula as shown below.









u

x

u vv

(x)

dv du

cos x dx (sin x) (sin x) (dx)

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]