The Chemistry Maths Book, Second Edition

168 Chapter 6Methods of integration

EXAMPLE 6.4Integrate.

Letu 1 = 12 x

2

1 + 13 x 1 + 11. Thendu 1 = 1 (4x 1 + 1 3)dx,and

0 Exercises 15, 16

There are no all-embracing rules for finding the correct change of variable that will

transform an integral to standard form; proficiency in the art of integration is the

result of a lot of practice. Some of the simpler types of substitution are summarized in

Table 6.2.

Table 6.2

Type Substitution Result

1. u 1 = 1 f(x), du 1 = 1 f′(x) 1 dx


2. u 1 = 1 f(x), du 1 = 1 f′(x) 1 dx


3. u 1 = 1 sin 1 x, du 1 = 1 cos 1 x 1 dx


4. u 1 = 1 cos 1 x, du 1 = 1 −sin 1 x 1 dx

EXAMPLES 6.5Integrals of type 1 :.

(i).

In this case,cos 1 axis proportional to the derivative ofsin 1 ax(and vice versa);

f(x) 1 = 1 sin 1 ax, f′(x) 1 = 1 a 1 cos 1 ax

Therefore, puttingu 1 = 1 sin 1 ax, du 1 = 1 a 1 cos 1 ax 1 dx,

We note that this integral is identical to case 3 in Table 6.1, and this example

demonstrates that there are often several ways of evaluating a particular integral.

I

a

udu

a

uC

a

==+= +ax C

11

2

1

2

22

Z sin

Iaxaxd=Zsin cos x

Zfxf xdx() ()′

Zfxxdx()cos sin −Zfudu()

Zfudu()

Zfx xdx()sin cos

Z

du

u

Z =+lnuC

fx′

fx

dx

()

()

Zudu u C=+

1

2

2

Zfxf xdx() ()′

ZZexdxedueCe

−++xx −u −u −x

+= =−+=−

() (

()

231 2

2 2

43

+++

• 
  

31 x

C

)

Zexdx

−++xx

• 
  

()

()

231

2

43