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1
CHAPTER:
INTEGRATION
Contents
1
Integration as the Reverse of Differentiation
2
Standard Integration Formulae
2.1
Integration of trigonometric functions
2.2
Integration of exponential function
2.3
Integration of Inverse Trigonometric Functions
3
Computation of Definite Integrals
4
Integration by Partial Fractions
5
Integration by Substitution
6
Integration By Parts
7
Miscellaneous Examples
 1 Integration as the Reverse of Differentiation
The process of finding an expression for y in terms of x from the gradient function,
dy
, is called integration.
dx
It reverses the operation of differentiation.
1
We know that if
y = x2 + C where C is an arbitrary constant
2
dy
Then
=x
dx
1
And if
y = x2
2
dy
Then, again,
=x
dx
1
Hence, to integrate x with respect to x, we write,  x dx = x2 + C where C is an arbitrary constant.
2
n+1
x
In general,
 xn dx =
+C
provided n+ 1  0 and C is an arbitrary constant.
n+1
Remark : Geometrical Interpretation of Integration
y
y = x3 + 15
dy
Consider graphs with gradient function,
= 3x2.
y = x3
dx
15
d
Since
(x3 + C) = 3x2 where C is an arbitrary constant,
y = x3 - 10
dx
dy
thus the equation
= 3x2 represents the family of
x
10
dx
curves y = x3 + C, some of which are shown in the diagram.
A particular member of this family of curves is specified
if we are given one point on the curve. Then the value of C
can be calculated using the given point, and thus the equation of the curve obtained.
 2 Standard Integration Formulae
 Adx
= Ax + C
A
 Axn dx
=
x n+1 + C
n+1
(Ax + B)n + 1
 (Ax + B)n dx =
+C
A(n + 1)
Example 2.1 [a)
a)  3x dx
b)
 13 dx
x
3 2
1
1
3 1 1
x + C b) - 2 + C c)
(3x – 8)7 + C d) - ( - x )8 + C]
2
2x
21
8 2 3
=
=
6
c) 
(3x - 8) dx =
1 1
d)  ( - x)7 dx =
2 3
2
 2.1 Integration of trigonometric functions
As integration is the reverse of differentiation, we have the following results.
d
Since
sin x = cos x
then,
 cos x dx = sin x + C
dx
d
cos x =  sin x
 sin x dx = - cos x + C
dx
d
tan x = sec2 x
 sec2 x dx = tan x + C
dx
d
sec x = sec x tan x
 sec x tan x dx = sec x + C
dx
d
cosec x =  cosec x cot x
 cosec x cot x dx =  cosec x + C
dx
d
cot x =  cosec2 x
 cosec2 x dx =  cot x + C
dx
In general,
 f ‘ (x) sin [f(x)] dx =  cos [f(x)] + C
 f ‘ (x) cos[f(x)] dx = sin [f(x)] + C
 sec x dx
= ln sec x + tan x + C.
1
1
3
1
sin3x
Example 2.2 [a) 3 tan x + C b) - 4 cos 4 + C c) - 2 cot (2x + 4 ) + C d) - 5 cosec 5 + C e) 3 – 3cosx + C]
a)  3 sec2 x dx =
b)  sin 4 d
c)
=
 cosec2 (2x +
3
) dx =
4
d)  cosec 5 cot 5 d =
e)
 (cos 3x + 3 sin x) dx =
 2.2 Integration of exponential function
In general,
 f ‘ (x) e f(x) dx = ef(x) + C
 ax dx =

Specifically,
 ex dx = ex + C,
 Aex dx = Aex + C
1
 e (Ax + B) dx = e(Ax + B) + C
A
ax
+C
ln a
f ' (x)
dx = ln f(x) + C
f(x)
 (f(x)) n f ‘ (x) dx =
(f(x))n + 1
+C
n+ 1
 2 1 2 dx = 1 ln a  x + C
2a a + x
a  x
1
dx = lnx + C
x
1
1

dx = ln  Ax + B  + C
Ax + B
A

3
2
3 2x
1
3
Example 2.3 [a) - e – 5x + C b)
+ C c) 4 ln x + C d) ln 2x + 5 + C e) - ln 4 – 2x+ C]
5
2 ln 3
2
2
a)  2e –5x dx
=
b)  32x dx
=
c) 
4
dx
x
=
d) 
1
dx
2x + 5
=
e) 
3
dx
4 - 2x
=
 2.3 Integration of Inverse Trigonometric Functions
d
f(x)
f ' (x)
Since
(sin-1
)= 2
, x< a
then,
dx
a
a -[f(x)]2
d
f(x)
a f ' (x)
(tan -1
)= 2
dx
a
a + [f(x)]2
 2f ' (x) 2 dx = sin-1 f(x) + C
a
 a -[f(x)]
 2a f ' (x) 2 dx = 1 tan –1 f(x) + C
a
a
a + [f(x)]
Variations of above formulae
 1 2 dx = sin – 1 x + C
 1-x
 1 2 dx = tan –1 x + C
1 + x
 21 2 dx = sin-1 x + C
a
 a -x
 2 1 2 dx = 1 tan –1 x + C
a
a
a + x
 f ' (x) 2 dx = sin – 1 [f(x)] + C , x < 1
 1 - [f(x)]
 f ' (x) 2 dx = tan –1 [f(x)] + C
1 + [f(x)]
Example 2.4 [a) sin
a) 
1
dx =
16 - x2
b) 
1
dx =
9 - 4x2


c) 
1
25 + 16x2
dx =
-1
x
1
2x
1
4x
+ C b) sin –1 ( ) + C c)
tan –1 ( )+ C ]
4
2
3
20
5
4
Example 2.5 [a)
a)
7x4
3
4
+ 2x2 + 4 + C b) – 3 cos x + x 3/2 + C c) 4 ln x - 3tan x + C ]
4
4x
3
( 7x3 + 4x - 35 ) dx =
x

b) 
( 3 sin x + 2 x ) dx =
c)
2
4 - 3x sec x dx =
x

Example 2.6 [a) tan x – x + C b)
a)  tan2 x dx =
b)  cos2 x dx =
c)  sin 3x cos 2x dx =
1
1
1
1
x + sin 2x + C c) cos 5x - cos x + C ]
2
4
10
2
5
Example 2.7
d
Show that
(x sin x) = x cos x + sin x. Hence find
dx
Solution
 x cos x dx.
[x sin x + cos x + C]
Example 2.8
Find  sin2 x cos x dx and hence
Solution
 cos3 x dx.
1
1
[ sin3 x + C, sin x - sin3 x + C ]
3
3
6
(f(x))n + 1
+C ]
n+ 1
2
(ln x)2
c) (1 + ex)3/2 + C d)
+C ]
3
2
Example 2.9 [Using formula :  (f(x)) n f ‘ (x) dx =
(1 + x4)4
tan4 x
+ C b)
+C
16
4
a)  x3 (1 + x4)3 dx =
[a)
b)  sec2 x tan3 x dx =
c)  ex 1 + ex dx =
d) 
ln x
dx =
 x
Example 2.10 [Using formula : 
[a) ln 1 + sin x + C b) a)
 cos x dx =
 1 + sin x
b)
 x 2 dx =
1-x
c)
 xe dx =
e + 4
d)
tan x dx =

x
f ' (x)
dx = ln f(x) + C ]
f(x)
1
ln 1 – x2 + C c) ln ex + 4 + C d) – ln cos x + C ]
2
7
 3 Computation of Definite Integrals
Suppose f(x) is the integrand and F(x) is the anti-derivative of f(x). Then, the definite integral of f(x) between
b
b
two limits x = a and x = b is given by:

f(x)dx = F(x)a = F(b)  F(a).
a
Some properties of Definite Integrals
a
a) 
f(x)dx = 0
a
b
b)
a

f(x)dx = - 
f(x)dx
a
b
c)
b
b
kf(x)dx
=
k

f(x)dx where k is a constant


a
b
d)
a
b
b
{f(x) + g(x)}dx = 
f(x)dx + 
g(x)dx

a
a
b
c
c
e) 
f(x)dx + 
f(x)dx = 
f(x)dx
a
b
a
a
Example 3.1
1
Evaluate a) 
x2dx
0
Solution
e
1
b)  dx
x
1
/4
c) 
( 2cos x + sec2 x)dx
0
1
[ a) b) 1 c) 2 ]
3
8
 4 Integration by Partial Fractions
In this section, we shall consider integrals of the form 
f(x)
(x - a)(x - b)
dx.
1
1
a  x
[Recall :  2
2 dx = 2a ln a + x + C from page 2]


a

x

In this case, simplify the expression using partial fractions, then integrate.
Example 4.1
Find
1
1
x-2

dx. [ ln 
+C]
3
x+1
(x - 2)(x + 1)
Solution
Example 4.2
Find
 23x + 5 dx.
x + x - 12
Solution
[ ln (x – 3)2 ( x + 4) + C ]
9
Example 4.3
Find
1

dx.
(x + 1)(x2 + 2x + 2)
[ ln 
x+1
+k]
x2 + 2x + 2
Solution
Example 4.4
Find
x-4
x-1
1

dx [ ln 
+
+C ]
2x + 1
x-1
(x - 1)2(2x + 1)
Solution
10
 5 Integration by Substitution
dy
= f(x).
dx
dy dy dx
dy
dx
By Chain Rule, we know
=
.

= f(x)
du dx du
du
du
Let
y =  f(x) dx.
Then
Integrating this equation with respect to u, we have
Hence, when simplifying an integral
1. Express f(x) in terms of u,
dx
2. Replace dx by
du.
du
 f(x) dx
Note:
When the integrand contains the following function:
a) a2 - x2
then substitute x = a sin  .
b) a2 + x2
then substitute x = a tan  .
2
2
c) x - a
then substitute x = a sec 
1
1
d)
then substitute tan x = t
A sin x + B cos x + C
2
Example 5.1
Find  10 (2x + 4)4 dx
Solution
[(2x + 4)5 + C ]
dx
du.
du
by a change to a new variable u, we must
y=
f(x) .
where a is a constant.
2t
1 - t2
2t
sin x =
, tan x =
,
2 , cos x =
1+t
1 + t2
1 - t2
11
Example 5.2
1
Find  sin3 2x cos 2x dx. [ (sin 2x)4 + C ]
8
Solution
Example 5.3
Find

x 2x - 1 dx.
Solution
[
1
(2x – 1)3/2 (3x + 1) + C ]
15
12
Example 5.4
4
Find 
d .
 3 + 5 cos 

2 + tan
2
[ ln 
+C]

2 - tan
2
Solution
Example 5.5
Find
1

dx.
1 + sin x + cos x
Solution
[ln  1 + tan
x
+C ]
2
13
Example 5.6
Find a)
 1 2 3/2 dx
(9 - x )
Solution
1
b) 
dx
x x2 - 4
[ a)
x
1
x
+ C b) sec-1 + C]
2
2
9 9 - x2
14
Example 5.7
3
9
Show, by using the substitution x = 3 sin , that 
 9 - x2dx = 4 .
0
Solution
Example 5.8
ln 2
1
Find 
dx
1 + ex
0
Solution
[ ln
4
]
3
15
 6 Integration By Parts
If u and v are functions of x, then by product rule for differentiation:
Integrating, we get
Rearranging, we get
d
dv
du
(uv) = u + v .
dx
dx
dx
dv
du
+v
)dx.
dx
dx
dv
du
=  u dx +  v
dx.
dx
dx
dv
du
 u dx = uv -  v
dx
dx
dx
uv =  (u
b
b
dv
b
du
For definite integrals, the rule for integration by parts becomes u dx = [ uv ] a - v dx
 dx
 dx
a
a
Example 6.1
Find  x cos x dx. [x sin x + cos x + C ]
Solution
Example 6.2
Find a)  x2 ex dx
Solution
b)  x ln x dx
[a) x2 ex – 2xex + 2ex + C b)
x2
x2
ln x +C]
2
4
16
Example 6.3
Find  x (ln x)2 dx. [
x2
x2
x2
(ln x)2 - ln x +
+C ]
2
2
4
Solution
Example 6.4
Find  ln x dx.
Solution
[x ln x – x + C]
17
Example 6.5
Find  ex cos x dx.
1
C
[ ex (sin x + cos x) +
]
2
2
Solution
Example 6.6
e
Find a) 
ln x dx
1
Solution
 /3
b) 
x sin 3x dx
0
[ a) 1 b)

]
9
18
 7 Miscellaneous Examples
Example 7.1 (AJC 98/1/11a)
x+2
1
Integrate 2
with respect to x. [ ln x2 + 2x + 2 + tan-1 ( x + 1) + C ]
x + 2x + 2
2
Solution
Example 7.2 (CJC 96/1/7)
Find
Solution
a)  x tan -1 x dx
b) 
x
dx
x +1
2
[a)
1 2
1
(x + 1)tan – 1 x - x + C b)
2
2
x2 + 1 + C ]
19
Example 7.3 (CJC 96/1/14a)
Find  x cos 2x dx. Hence, or otherwise, find  x cos2 x dx.
1
1
1
1
1
[ x sin 2x + cos 2x + C , x sin 2x + cos 2x + x2 + C ]
2
4
4
8
4
Solution
Example 7.4 (NJC 2000/1/15b)
2x - 4
Let f(x) = 2
x -x+4
i)
Prove that x2 – x + 4 is always positive for all real values of x.
ii)
2x - 4
Evaluate  2
dx
x
 -x+4
iii)
2x - 4
Hence, evaluate  2
dx , correct to 3 decimal places.
x
 -x+4
4
1
6
15
2
1
[ii) ln (x2 – x + 4 ) tan –1 (
x)+C
15
15
15
Solution
iii) 0.575]
20
SUMMARY (Integration)
Trigonometric formula,  cos x dx = sin x + C
 sin x dx = - cos x + C
 sec2 x dx = tan x + C
In general,
Exponential and logarithmic formula
In general,
 f ‘ (x) e f(x) dx = ef(x) + C
 ax dx =

 sec x tan x dx = sec x + C
 cosec x cot x dx =  cosec x + C
 cosec2 x dx =  cot x + C
 f ‘ (x) sin [f(x)] dx = - cos [f(x)] + C
 f ‘ (x) cos[f(x)] dx = sin [f(x)] + C
Specifically,
 ex dx = ex + C,
 Aex dx = Aex + C
1
 e (Ax + B) dx = e(Ax + B) + C
A
ax
+C
ln a
f ' (x)
dx = ln f(x) + C
f(x)
 (f(x)) n f ‘ (x) dx =
1
dx = lnx + C
x
1
1

dx = ln  Ax + B  + C
Ax + B
A

(f(x))n + 1
+C
n+ 1
Inverse Trigonometric formula,
 2 1 2 dx = 1 ln a  x + C
2a a + x
a  x
 2f ' (x) 2 dx = sin-1 f(x) + C
a
 a -[f(x)]
 2a f ' (x) 2 dx = 1 tan –1 f(x) + C
a
a
a + [f(x)]
Variations
 1 2 dx = sin – 1 x + C
 1-x
 1 2 dx = tan –1 x + C
1 + x
 21 2 dx = sin-1 x + C
a
 a -x
 2 1 2 dx = 1 tan –1 x + C
a
a
a + x
 f ' (x) 2 dx = sin – 1 [f(x)] + C , x < 1
 1 - [f(x)]
 f ' (x) 2 dx = tan –1 [f(x)] + C
1 + [f(x)]
Integration By Substitution
dx
y=
f(x) .du du.
1. Express f(x) in terms of u,
dx
2. Replace dx by
du.
du
Note:
When the integrand contains the following function:
a) a2 - x2
then substitute x = a sin  .
b) a2 + x2
then substitute x = a tan  .
2
2
c) x - a
then substitute x = a sec 
1
1
d)
then substitute tan x = t
A sin x + B cos x + C
2
where a is a constant.
2t
1 - t2
2t
sin x =
, tan x =
,
2 , cos x =
1+t
1 + t2
1 - t2
Integration By Parts
dv
du
 u dx = uv -  v
dx
dx
dx
“An optimist is a person who sees a green light everywhere,
while the pessimist sees only the red stop-light.
But the truly wise person is colour blind.” Albert Schweitzer