Higher Engineering Mathematics

H

Integral calculus

39

Integration using algebraic

substitutions

39.1 Introduction

Functions which require integrating are not always in
the ‘standard form’ shown in Chapter 37. However,
it is often possible to change a function into a form
which can be integrated by using either:

(i) an algebraic substitution (see Section 39.2),

(ii) a trigonometric or hyperbolic substitution (see

Chapter 40),

(iii) partial fractions (see Chapter 41),

(iv) thet=tanθ/2 substitution (see Chapter 42),

(v) integration by parts (see Chapter 43), or

(vi) reduction formulae (see Chapter 44).

39.2 Algebraic substitutions

Withalgebraic substitutions, the substitution usu-
ally made is to letube equal tof(x) such thatf(u)du
is a standard integral. It is found that integrals of the
forms,

k

∫

[f(x)]nf′(x)dxandk

∫

f′(x)

[f(x)]n

dx

(wherekandnare constants) can both be integrated
by substitutinguforf(x).

39.3 Worked problems on integration

using algebraic substitutions

Problem 1. Determine

∫

cos(3x+7) dx.

∫
cos (3x+7) dxis not a standard integral of the
form shown in Table 37.1, page 368, thus an alge-
braic substitution is made.

Letu= 3 x+7 then

du

dx

=3 and rearranging gives

dx=

du

3

. Hence,

∫

cos(3x+7) dx=

∫

(cosu)

du

3

=

∫

1

3

cosudu,

which is a standard integral

=

1

3

sinu+c

Rewritinguas (3x+7) gives:

∫

cos(3x+7) dx=

1

3

sin(3x+7)+c,

which may be checked by differentiating it.

Problem 2. Find

∫

(2x−5)^7 dx.

(2x−5) may be multiplied by itself 7 times and

then each term of the result integrated. However, this

would be a lengthy process, and thus an algebraic

substitution is made.

Letu=(2x−5) then

du

dx

=2 and dx=

du

2

Hence

∫

(2x−5)^7 dx=

∫

u^7

du

2

=

1

2

∫

u^7 du

=

1

2

(

u^8

8

)

+c=

1

16

u^8 +c

Rewritinguas (2x−5) gives:

∫

(2x−5)^7 dx=

1

16

(2x−5)^8 +c

Problem 3. Find

∫

4

(5x−3)

dx.