Understanding Engineering Mathematics

(iii) When we make a substitution, we can change the limits too – then

we don’t have to return to the original variables.

In

∫ 1

0

x^2

x^3 + 1

dxthe numerator is the derivative of the denomi-

nator (almost), so putu=x^3 +1.

Thendu= 3 x^2 dxorx^2 dx=

du

3

The limits onuare obtained from the substitution

whenx= 0 ,u= 03 + 1 = 1

whenx= 2 ,u= 23 + 1 = 9

So ∫ 2

0

x^2

x^3 + 1

dx=

1

3

∫ 9

1

du

u

=

1

3

[lnu]^91

=

1

3

(ln 9−ln 1)

=

1

3

ln 9

remembering that ln 1=0.

(iv) This sort of integral is very important in the theory of Fourier

series (see Chapter 17). Using the double angle formula we have

∫ 2 π

0

cosxsinxdx=

1

2

∫ 2 π

0

sin 2xdx

=−

1

4

[cos 2x]^20 π= 0

B.

∫ 2

0

dx

x^2 − 1

=

∫ 2

0

dx

(x− 1 )(x+ 1 )

The integrand doesn’t exist (‘becomes singular’) atx=1, which

is within the range of integration.

9.3 Reinforcement

9.3.1 Definition of integration

➤➤

251 253

➤

A.Differentiate the following functions:

(i) 3x^3 (ii)

√

3 x (iii)

2

x^4

(iv) x^2 /^5