Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Unit 2
Techniques of Integration

Unit 2
...
0
Evaluate the integral
...
 cos  2 x  dx  sin  2 x   C
2
1
2
...
 2sec  5x  dx  tan  5x   C
5
1
4
...
 e dx  e  C
4
1 26 x
6
...
 sech 8x  tanh 8x  dx   sech 8x   C
8
dx
1
8
...

  ln 22  12 x  C
22  12 x
12

End of Unit 2
...
1
Integration by Parts

If u and v are differentiable functions
of x then

 uv  '  uv ' vu '

 uv  ' dx   uv ' dx   vu ' dx
uv   uv ' dx   vu ' dx

uv   uv ' dx   vu ' dx
du  u ' dx and dv  v ' dx
uv   u dv   v du

 u dv  uv   v du

Example 2
...

1
...


 ln x dx
  u dv
 uv   v du
 x ln x   dx
 x ln x  x  C

Let u  ln x
dx
du 
x

dv  dx
vx

3
...


e x cos x dx


Let u  e

  u dv

du  e dx v  sin x
x

 uv   v du
 e sin x   e sin x dx
x

dv  cos x dx

x

x

For

e x sin x dx,


 e sin x dx
  u dv
 uv   v du
x

Let u  e x

 e x cos x   e x cos x dx

du  e x dx

dv  sin x dx
v   cos x

e
e

x
x

cos x dx  e sin x   e sin x dx
x

x

sin x dx  e x cos x   e x cos x dx

Therefore,
e x cos x dx  e x sin x  e x cos x   e x cos x dx

2 e x cos x dx  e x sin x  e x cos x
e x sin x  e x cos x
e x cos x dx 
C

2

5
...


x

4

sin  2 x  dx
u
4
x
4 x3
2
12 x
24 x
24
0








dv
sin  2 x 
 1 cos  2 x 
2
 1 sin  2 x 
4
1
cos  2 x 
8
1
sin  2 x 
16
1
 32 cos  2 x 

x4 sin  2 x  dx   1 x 4 cos  2 x   x3 sin  2 x 
2


 x cos  2 x   3 x sin  2 x   3 cos  2 x   C
2
4
3
2

2

7
...

Exercises 7
...


End of Unit 2
...
2
Integration of Special Powers
of Trigonometric Functions

Example 2
...
1

 cos

Evaluate

3

x dx
...
If m is odd, save one sine factor and
express the remaining factors in terms of
cosine
...
If n is odd, save one cosine factor and
express the remaining factors in terms of
sine
...
If m and n are both even, use the halfangle identities
1  cos 2u
2
sin u 
2
1  cos 2u
2
cos u 
2

Example 2
...
2
Evaluate the integrals
...


Let u  cos x
du   sin x dx
du  sin x dx

 sin x cos x dx
  sin x cos x sin x dx
   sin x  cos x sin x dx
  1  cos x  cos x sin x dx
  1  u  u du
5

2

4

2

2

2

2

2

2

2 2

2

2

  1  u

u

2 2

2

du

  1  2u 2  u 4  u 2 du
   u  2u  u  du
2

4

6

 u3 2u 5 u 7 
  
 C
5
7
3
3
5
7
cos x 2cos x cos x



C
3
5
7

2
...
If m is odd, save a factor of sec u tan u
and express the remaining factors in terms
of sec u
...
If n is even, save a factor of sec u and
express the remaining factors in terms of
tan u
...
If the power of secant is odd and the power
of tangent is even, then express all factors
in terms of sec u then use integration by
parts
...
2
...

1
...


tan 3 x dx

  tan x  sec x  sec x tan x dx
1

2

   sec x  1  sec x  sec x tan x dx
2

1

   u 2  1 u 1 du

1

   u   du
u

2
u
  ln u  C
2
2
sec x

 ln sec x  C
2

Let u  sec x
du  sec x tan x dx

3
...
1
...

Exercises 7
...


End of Unit 2
...
3
Trigonometric Substitution

Preliminary Exercise
Express the following in terms of
sin  and cos
...
tan 
cos
cos
2
...
sec 
cos
1
4
...

1
1
...
cos 
sec
tan 
sin 
3
...
csc 
sec 
tan 
cos

Consider the integral  a  x dx where a
2

2

is a positive constant
...

2

2

For integrals involving a  u where a  0,
if u  a sin  where       then
2
2
2

2

a  u  a  a sin 
2

2

2

2

2

 a 1  sin  
2

2

 a cos 
 a cos
2

2

Trigonometric Substitution
Expression
1
...


a u
2

2

2

u  a tan 
du  a sec2  d
a  u  a sec
2

2

3
...
3
...

4  x2
1
...




dx
x2  25






3

Let x  5tan 
2

dx
x  25
2

dx  5sec2  d



5sec2  d

3
 5sec 
1 d
 
25 sec
1
  cos d
25

3

x2  25  5sec

1
  cos d
25
1
 sin   C
25
1 tan 
 
C
25 sec
1
x
 
C
25 x2  25

x  5tan 
x  25  5sec
2

3
...
3
...




a 2  x2 dx

  a cos  a cos d
 a  cos  d
2

2

1  cos2
a 
d
2
2
a
  1  cos2  d
2
2

Let x  a sin 
dx  a cos d
a  x  a cos
2

2

2

a
  1  cos2  d
2
2
a 
1

   sin 2   C
2
2

a2
   sin  cos   C
2
a2 
x x a2  x2
  Arcsin  
2
a a
a


x  a sin 
a 2  x2  a cos
sin 2  2sin  cos


C



Required Exercises

Answer Exercises in TC7
...
3 (1-33) on page 600
Check your answers on A-160
...
3

Unit 2
...

x 1 x  2
x  2  3 x  1
1
3


x 1 x  2
 x  1 x  2
x  2  3x  3
 2
x x2
4x  1
 2
x x2

1
3
4x  1

 2
x 1 x  2 x  x  2
4x  1
3 
 1
So,  2
dx   

 dx
x x2
 x 1 x  2 
 ln x  1  3ln x  2  C

When to Use Partial Fractions
P
Consider  dx
...
P and Q are polynomials
...
deg P  deg Q
...

3
...

Otherwise, cancel common factors
...
4
...

2 x  3x  1
4x  3
A
B


 x  1 2 x  1 x  1 2 x  1
A 2 x  1  B  x  1
4x  3

 x  1 2 x  1
 x  1 2 x  1
4 x  3  A  2 x  1  B  x  1

4 x  3  A 2 x  1  B  x  1
if x   1 ,
2

4   1   3  B   1  1
2
2

1 1 B
2
B2
if x  1,

4  1  3  A 2  1  1
1   A
A 1

4x  3
A
B


2
2 x  3x  1 x  1 2 x  1
A  1 and B  2
4x  3
1
2
Therefore,



...

r
 ax  b 

Example 2
...
2
Evaluate
Consider
3x  1
2
x  x  1

3x  1
 x3  2x2  x dx
...

3
2
x  2x  x
A
B
C
 

2
x x  1  x  1

A x  1  Bx  x  1  Cx
3x  1

2
2
x  x  1
x  x  1
2

3x  1  A x  1  Bx  x  1  Cx
2

3x  1  A x  1  Bx  x  1  Cx
2

if x  1,
3 1  1  C
2  C
C2
if x  0,
3 0   1  A 0  1
A 1

2

3x  1  A x  1  Bx  x  1  Cx
2

A  1 and C  2
3x  1  1 x  1  Bx  x  1  2 x
2

3x  1  x2  2 x  1  Bx 2  Bx  2 x
x2 :
x:
C:

0 1 B
 B  1
32 B2
11

3x  1
A
B
C
 

2
3
2
x  2 x  x x x  1  x  1
A  1, B  1, and C  2
3x  1
1
1
2
Therefore, 3
 

2
2
x  2 x  x x x  1  x  1
1
3x  1
1
2 
 x3  2x2  x dx    x  x  1   x  12  dx





1
3x  1
1
2 
 x3  2x2  x dx    x  x  1   x  12  dx




2
For 
dx,
2
 x  1

Let

u  x 1
du  dx

2
  x  12 dx
2
  2 du
u
1
 2  C
u
2

C
x 1

1
3x  1
1
2 
 x3  2x2  x dx    x  x  1   x  12  dx




2
2
  x  12 dx   x  1  C
Therefore,
3x  1
2
 x3  2x2  x  ln x  ln x  1  x  1  C1

Case 3: Non Repeating
Quadratic Factors
If Q has quadratic factors that are non repeating,
for each quadratic factor, ax  bx  c where a
and one of b and c are not zero, write a partial
fraction of the form
2

Ax  B
ax2  bx  c

Example 2
...
3
x2  8
dx
...

Consider 4
2
x  4x
A B Cx  D
x2  8
  2 2
2
2
x 4
x  x  4 x x
x 8

2
2
x  x  4
2

Ax  x2  4  B  x 2  4   Cx  D  x 2
x  x  4
2

2

x2  8  Ax  x2  4   B  x 2  4    Cx  D  x 2

x  8  Ax  x  4   B  x  4    Cx  D  x
2

2

2

2

if x  0,
8  4B
B2
x2  8  Ax  x2  4   2  x 2  4    Cx  D  x 2
x2  8  Ax3  4 Ax  2 x 2  8  Cx3  Dx 2

x  8  Ax  4 Ax  2 x  8  Cx  Dx
2

3

2

x3 :

0  AC

 C  A

x2 :
x:
C:

1 2 D
0  4A
88

 D  1
 A0

3

2

C  0

x 8
A B Cx  D
  2 2
4
2
x  4x
x x
x 4
A  0, B  2, C  0, and D  1
2

x2  8
2
1
Therefore, 4
 2 2

...

Exercises 7
...


End of Unit 2
...
5
Rationalizing Substitution

Example 2
...
1
dx
Evaluate 

...


Example 2
...
2
Evaluate the integrals
...
 3
x x
6
u x
5
6u du
 2 3
6u5 du  dx
u u
1
1
5
6 3
2
3
3
6u du
x  x  u   u
 2
u 1  u 
1
1
6 2
x  x 2  u   u3
3
u du
 6
1 u

u  u 1
u  1 u3
2

3

u du
 6
1 u
1 
 2
 6  u  u  1 
 du
u 1


u u
3

2

 u2
u 2  u
u
u 1
1

1 
 2
 6  u  u  1 
 du
u 1

 u3 u 2

 6    u  ln u  1   C
3 2


   

2 x

1

3

6

3 x

1

6

2

1

1

 x  ln x 6  1  C
6

 2 x  3 3 x  6 6 x  6ln 6 x  1  C

2
...

Exercises 7
...


End of Unit 2
...
6
Half-Angle Substitution

Half-Angle Substitution

If an integrand is a rational function of sin x
and cos x, then let
x
z  tan  2 

x
z  tan  2 

x
sec2  2 
dz 
dx
2
2
dx 
dz
2 x
sec  2 

2
dx 
dz
2 x
1  tan  2 
2 dz
dx 
1  z2

x
z  tan  2 

x
sin x  sin  2  2 

x
x
 2sin  2  cos  2 

sin  2u   2sin u cos u

x
cos  2 
x
x
 2sin  2  cos  2  
x
cos  2 
x
2sin  2 
2 x

 cos  2 
x
cos  2 
x
2tan  2 

2 x
sec  2 

x
z  tan  2 

x
2tan  2 
sin x 
2 x
sec  2 

sin x 

x
2tan  2 

1  tan

2z
sin x 
1  z2

2

 
x
2

x
z  tan  2 

cos x  cos  2 


x
 2cos2  2   1
x
2

2

1
2 x
sec  2 
2

1
2 x
1  tan  2 
2

1
2
1 z

cos  2u   2cos u  1
2

2
cos x 
1
2
1 z
2
1 z
cos z 
2
1 z

Theorem
x
If z  tan  2  , then

2 dz
dx 
2
1 z
2z
sin x 
1  z2
1 z
cos x 
2
1 z
2

Example 2
...
1
Evaluate the following integrals
...

1  sin x  cos x
2dz
1  z2

2
2z
1 z
1

2
2
1 z 1 z
2dz
1  z2

2
2
1  z  2z  1  z
2
1 z

x
z  tan  2 

2dz
dx 
1  z2
2z
sin x 
2
1 z
2
1 z
cos x 
2
1 z

2dz
2
1 z

2
2
1  z  2z  1  z
1  z2
dz
 2
2z  2
dz

z 1
 ln z  1  C
x
 ln tan  2   1  C

dx
2
...

Exercises 7
...


End of Unit 2
...

dx
1
...

2
9x  4
 2 x  3 dx
3
...

sin x  2cos x
5
...


x3 cos  2 x  dx


4

3



---