Cambridge Additional Mathematics

Integration (Chapter 15) 429

Likewise if n 6 =¡ 1 ,

d

dx

μ

1

a(n+1)

(ax+b)n+1

¶

=

1

a(n+1)

(n+ 1)(ax+b)n£a,

=(ax+b)n

)

Z

(ax+b)ndx=

1

a

(ax+b)n+1

(n+1)

+c for n 6 =¡ 1

We can perform the same process for the circular functions:

d

dx

(sin(ax+b)) =acos(ax+b)

)

Z

acos(ax+b)dx= sin(ax+b)+c

) a

Z

cos(ax+b)dx= sin(ax+b)+c

So,

Likewise we can show

Z

cos(ax+b)dx=

1

a

sin(ax+b)+c for a 6 =0.

Z

sin(ax+b)dx=¡

1

a

cos(ax+b)+c for a 6 =0.

Fora,bconstants with a 6 =0, we have: Function Integral

eax+b

1

a

eax+b+c

(ax+b)n, n 6 =¡ 1

1

a

(ax+b)n+1

n+1

+c

cos(ax+b)

1

a

sin(ax+b)+c

sin(ax+b) ¡

1

a

cos(ax+b)+c

Example 11 Self Tutor

Find: a

Z

(2x+3)^4 dx b

Z

1

p

1 ¡ 2 x

dx

a

Z

(2x+3)^4 dx

=^12 £

(2x+3)^5

5

+c

= 101 (2x+3)^5 +c

b

Z

1

p

1 ¡ 2 x

dx

=

Z

(1¡ 2 x)

¡^12

dx

=¡^12 £

(1¡ 2 x)

1

2

1

2

+c

=¡

p

1 ¡ 2 x+c

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_15\429CamAdd_15.cdr Monday, 7 April 2014 3:59:33 PM BRIAN