The Chemistry Maths Book, Second Edition

178 Chapter 6Methods of integration

Solving forI

n

then gives

0 Exercises 52–54

EXAMPLE 6.15Determine a reduction formula for the definite integral

.

From the result of Example 6.14,

Becausesin 0 1 = 10 andcos

n− 1

(π 2 2) 1 = 10 ifn 1 > 11 , it follows that

with

For example,

0 Exercises 55, 56

EXAMPLE 6.16Determine a reduction formula for.

Because , the integrand is written as.

Then, by parts,

Now asr 1 → 1 ∞and, ifn 1 > 1 1,r

n− 1

e 1 = 10 whenr 1 = 10. Then, forn 1 ≥ 12 ,

−ar

→

2

0

Z

0

0

1

22

1

2

1

∞ ∞

erdr

a

er

n

−−ar n ar n−





















=− +

( − ))

2

0

2

2

a

erdr

ar n

Z

∞

−−

−−













−−

1

2

2

2

1

a

are r

ar n

d

dr

eare

−−ar ar

=−

22

2

Ierdr

n

ar n

=

−

Z

0

2

∞

II I II I

42 0 53 1

3

4

3

4

1

2

3

16

4

5

4

5

2

3

8

15

==×=, ==×=

π

IxdxIdx

1

0

2

0

0

2

1

2

==ZZ,==

ππ

π

cos

I

n

n

In

nn

=

−

,≥

−

1

2

2

I

n

xx

n

n

I

n

n

n

=

















• 
  

−

−

−

0

2

1

2

11

π

cos sin

Ixdx

n

n

=Z

0

π 2

cos

I

n

xx

n

n

I

n

n

n

=+

−

−

−

11

1

2

cos sin