Understanding Engineering Mathematics

Solution to review question 9.1.5

These questions all involve linear substitutions. In no case is it necessary

to do anything to the function before substitution.

(i) If you did something like

‘

∫

dx

x− 1

=

∫ (

1

x

− 1

)

dx=...’

stop now and go back to review your basic algebra. It is of course

wrong. You cannot ‘simplify’

1

x− 1

any further and must work with

it as it is, and for this we use a substitution.

The closeness of

1

x− 1

to

1

x

suggests making the substitution

u=x−1sodu=dxand

∫

dx

x− 1

=

∫

du

u

=lnu=ln(x− 1 )+C

or, quicker

∫

dx

x− 1

=

∫

d(x− 1 )

x− 1

=ln(x− 1 )+C

(ii) Here you may have been tempted to use the compound angle formula

to expand the cos and get simple sine and cosine integrals. Again,

while correct, this is unnecessarily messy – simply substitute

u= 3 x+ 2 , dx=

du

3

so

∫

cos( 3 x+ 2 )dx=

∫

cosu

du

3

=

1

3

sinu+C

=

1

3

sin( 3 x+ 2 )+C

or

∫

cos( 3 x+ 2 )dx=

1

3

∫

cos( 3 x+ 2 )d( 3 x+ 2 )

=

1

3

sin( 3 x+ 2 )+C

(iii) By now, you will probably be happy with:

∫

e^2 x−^1 dx=

1

2

∫

e^2 x−^1 d( 2 x− 1 )

=

1

2

e^2 x−^1 +C