Chapter 3 - Integrations PDF

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CHAPTER 3 INTEGRATION 3 Integration of hyperbolic functions 3 Integration of inverse trigonometric functions 3 Integration of inverse hyperbolic functions Recall: Methods involved: - Substitution of u - By parts - Tabular method - Partial fractions 1 REVISION: Techniques of integration (a) Integrati...

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CHAPTER 3 INTEGRATION

3.1

Integration of hyperbolic functions

3.2

Integration of inverse trigonometric functions

3.3

Integration of inverse hyperbolic functions

Recall: Methods involved: - Substitution of u - By parts - Tabular method - Partial fractions

1

REVISION: Techniques of integration (a) Integration by substitution Example:

sin x dx 1.  1  cos x 2. 3.



sin x cos 4 x dx 2

 x cos x e

sin x

2

dx

(b) Integration by parts Example: 1.

 x cos x dx

2.

 x sin 2 x dx

2

(c) Tabular methods Example: 1.

x sec 2 x dx



2.  e 3x cos 2 x dx

(d) Integration using partial fractions Example: 1.

2.



3x  2

dx

2

x  3x  2

1  x3  2 x2  x dx

3

3.1 Integrals of Hyperbolic Functions

Integral Formulae 1.  sinh

xdx  cosh x  C

2.  cosh

xdx  sinh x  C

3.  sec h

2

4.

xdx  tanh x  C

2 cos ech xdx   coth x  C 

5.  sec hx tanh

xdx   sec hx  C

6.  cos echx coth xdx

  cos echx  C

4

Example 3.1 : Evaluate the following integrals

a)

 sinh x cosh x dx

b)

2 tanh sec x h x dx 

c)

 xcosh x dx

d)

3 x  cosh x dx

5

3.2 Integration of Inverse Trigonometric Functions Integration formulae of the Inverse Trigonometric Functions Differentiation

Integration

1 d (sin 1 x )  dx 1  x2



1 d 1 x  (cos ) dx 1  x2

d dx

(tan

d dx

(cot

d dx

(sec

1

1

x) x) x) 

1 1 x 1

(csc

1

2

2

x 1 1

x)  x

1 x2  dx 1 x2

2

x 1

 sin 1 x  C  cos 1 x  C

dx

1  x2

 tan 1 x  C

 dx

 cot 1 x  C

1  x2

1 x 2 1 x

d dx

1



dx

x

dx

x

 dx

x 2 1

x2 1

 sec 1 x  C

 csc 1 x  C

6

Example 3.2 : Evaluate the following integrals

7

What about

dx

dx



4  x2

 9  x2 ,  x

,

dx 2

…?

x 10

To find the answer for this question, lets try to solve



dx 2

a x

2

.

Solution: Let

x  au , then dx  adu . Hence



dx



a2  x2



du 1u 2

 sin 1 u  C x   sin 1    C a Using the same method, we can find the solution for

dx

 a2  x2 ,  x

dx x2  a2

,…

8

From the above discussions, we obtain the general integration formulae as follows:

1 

x  sin    C a a2 x2 dx



 dx



2

a x

2

 cos

1  x 

 C a 

dx 1 -1  x   a 2  x2  a tan  a   C  dx 1 1  x   a 2  x 2  a cot  a   C

 

1 1  x   sec    C 2 2 a a x a dx

x

dx x

1 1  x   csc    C 2 2 a a x a

9

Example 3.3 : Evaluate the following integrals

10

3.3 Integration involving Inverse Hyperbolic Functions

Integration formulae of the Inverse Hyperbolic Functions: Differentiation

Integration

d 1 (sinh  1 x )  dx 1  x2



d (cosh 1 x )  dx



1 x2 1

1 d 1 (tanh x )  dx 1 x2

dx 1 x2 dx x2 1 dx

 1 x 2

 sinh 1 x  C  cosh 1 x  C

 tanh 1 x  C

11

What about

dx



x

,

4  x2

dx

dx 2

,

8

 25  x 2

…?

To find the answer for this question, lets try to solve

dx



2

a x

2

.

Solution: Let

x  au , then dx  adu . Hence



dx 2

a x

2





du 1 u 2

 sinh 1 u  C  sinh

1 

x   C a

Using the same method, we can find the solution for

dx

 a2  x2  ,

dx x2  a 2

,… 12

From the above discussions, we obtain the general integration formulae as follows:

dx

1  x 



a2  x2



x  cosh    C a x2 a2



1 1  x  tanh    C ,  dx a a  2 2 a x  1 coth 1  x   C ,  a  a



1  x   sec h 1    C a a x a2  x2



1 1  x  cos ech    C  a a x a2  x2

dx

 sinh

  C a 

1 

x a x a

dx

dx

13

Examples 3.4: 1. Solve the following:

14...