Understanding Engineering Mathematics

∫

( 2 x− 4 )^3 dx=

1

2

∫

( 2 x− 4 )^3 d( 2 x− 4 )

=

1

2

1

4

( 2 x− 4 )^4 +C

=

1

8

( 2 x− 4 )^4 +C

(d( 2 x− 4 )= 2 dx)

There is a useful ‘intuitive’ way of looking at linear substitution. When integrating a
function ofax+b,f(ax+b), we think ‘Well,ax+bis really little different tox.So
treat it likex– if the integral of f(x)isF(x), take the integral off(ax+b)to be
F(ax+b). But if we differentiated this with respect toxwe would get a multiplier “a”
coming in from the function of a function rule. So we need to divide the integral byato

cancel this multiplier. The integral off(ax+b)is thus

1

a

F(ax+b).’ For the example

above this goes through as follows:( 2 x− 4 )^3 is likex^3 , so take its integral to be

( 2 x− 4 )^4

4

.

But if we differentiate this we get a 2 coming from differentiating the bracketed term by

function of a function. So remove this by multiplying by

1

2

so that the final form of the

integral is

1

2

×

( 2 x− 4 )^4

4

=

( 2 x− 4 )^4

8

, as we found above.

Another important point is that the linear substitution is very special – it is theonly
casein whichduanddxare proportional. Suppose we substitute

u=g(x)

so
du
dx

=g′(x)

and therefore

du=g′(x)dx

In the linear substitution case this gives

du=adx

and we could simply divide byato replacedxby

du

a

But in any other substitution wecan’t do this, we would get

dx=

du

g′(x)

and we have the complication of substituting forxin terms ofuing′(x). So, greater care
is needed with other types of substitutions. In general, simple substitutions require special
forms of the integrand – see Section 9.2.6.