sin . a c s o b sin a b si n a b - - PowerPoint PPT Presentation

Trigonometric Relations

 

2

1 cos 1 cos2 2 x x  

2 2

tan sec 1 x x    

2

1 sin 1 cos2 2 x x  

2 2

cot csc 1 x x  

   

cos 1 sin . co i 2 s s n a b a b a b        

   

sin 1 sin . si

• 2

n c s a b a b a b        

   

cos 1 cos . co

• 2

s c s a b a b a b        

Examples

2

( cos 3 ) i x dx



1 1 sin6 2 6 x x C          1 1 cos6 2 x dx  



   

1 sin 7 sin 3 2 x x dx       



   

1 1 1 cos 7 sin 3 2 7 3 x x C             

   

sin 2 .cos 5 ( ) x ii x dx



 

2

1 2sin5 ( ) x i x ii d 



 

1 4sin5 2 1 cos10 x x dx    



2

1 4sin5 4sin 5 x x dx   



4 1 cos5 2 sin10 5 10 x x x x C            

 

2

tan2 3sec2 ( ) x x x iv d 



2 2

tan 2 6sec2 tan2 9sec 2 x x x x dx       



 

2 2

sec 2 1 6sec2 tan2 9sec 2 x x x x dx        



2

6sec2 tan2 10sec 2 1 x x x dx       



6 10 sec2 tan2 2 2 x x x C    

2 2

cos 2 . ) 3 ( sin v x x dx



1 1 cos4 cos6 cos4 .cos6 4 x x x x dx    



 

1 1 1 cos4 cos6 cos2 cos10 4 2 x x x x dx     



1 1 1 1 1 1 sin4 sin6 sin2 sin10 4 4 6 2 2 10 x x x x x C                  

   

1 1 1 cos4 . 1 cos6 2 2 x x dx   



Important Rules

   

'

(1)

n f

x f x dx        



   

'

) (2 f x dx f x



 

1

1

n

f x C n



      

 

ln f x C      

 

4 2 2

3 ( )

x x

e e i dx 



   

4 2 2

1 3 2 2

x x

e e dx  



 

5 2

3 1 2 5

x

e C   

2

( 5 ) 3 x x i x i d 



 

 

1/ 2 2

1 3 5 6 6 x x dx  



 

3/2 2

3 5 1 6 3/2 x C   

Examples

2 3

1 ( 2 5 ) 6 x dx x ii x i   



2 3

1 6 6 6 2 6 5 x dx x x    



3

1 ln 2 6 5 6 x x C    

2 ln ( ) iv dx x x



1/ 2 ln x dx x  

2lnlnx C  

1 2

( ) sin 1 x dx x v







1 2

sin 1 x dx x



 



 

1/ 2 1 2

1 sin 1 x dx x



 



sin3 cos3 5 ( ) x dx x vi 



1 3sin3 3 cos3 5 x dx x    



1ln cos3 5 3 x C    

 

 

3/2 1

sin / 3/2 x C



 

tan ( ) i x dx



sin cos x dx x  

sec ( ) iii x dx



sec tan sec sec tan x x x dx x x   



ln sec tan x x C   

ln cosx C  

Examples

cot ( ) ii x dx



cos sin x dx x  

ln sinx C  

• csc

( ) iv x dx



csc cot csc csc cot x x x dx x x   



ln csc cot x x C   

* For many integrals, a substitution can be used to transform the integrand and make possible the finding of an antiderivative. There are a variety of such substitutions, each depending on the form of the integrand. * If a component of the integrand can be viewed as the derivative of another component of the integrand, a substitution can be made to simplify the integrand.

Integration by Substitution

Example

 

4 5

sin x x dx



5

x u 

1cos 5 u C   

 

5

1cos 5 x C   

sin 5 du u  

4

5x dx du 

Substitute from (2) &(3) into (1)

 

4 5

sin x x dx



(1) (2) (3)

Example

 

 

2

ln 1 ln y dy y y 



ln y u 

2

1 ln 1 2 u C   

 

2

1 ln 1 ln 2 y C   

2

1 u du u  



1 dy du y 

Substitute from (2) &(3) into (1)

 

 

2

ln 1 ln y dy y y 



(1) (2) (3)

Example

2 2tan

sec

x

x e dx



tanx u 

2

1 2

u

e C  

2tan

1 2

x

e C  

2u

e du  

2

sec x dx du 

Substitute from (2) &(3) into (1)

2 2tan

sec

x

x e dx



(1) (2) (3)

Example

2

sec tan 9 4sec x x dx x 



secx u 

2 9 4

1 1 4 du u  



1

1 1 . tan 4 3/2 3/2 u C

 

      

2

1 9 4 du u  



sec tan x x dx du 

Substitute from (2) &(3) into (1)

2

sec tan 9 4sec x x dx x 



(1) (2) (3)

3 2 a 

Example

1 x dx x 



x u 

2

2 1 u du u  



 

1 2 1 1 u du u          



2 1 u udu u  



2 dx udu 

Substitute from (2) &(3) into (1)

1 x dx x 



(1) (2) (3)

2

x u 

2

u

1 u 

u

2

u u 

• -

u 

1  1 u  

+ +

1

 

2

2 ln 1 2 u u u C           

Example

1

x x dx

e e  



x

e u 

2

1 1 du u  



1

tan u C



 

1

1 1du u u u   



1 dx du u 

Substitute from (2) &(3) into (1)

1

x x dx

e e  



(1) (2) (3) ln x u 