SlideShare a Scribd company logo
(6) t results
(6) t results
                           x
                If t  tan
                           2
(6) t results
                           x
                If t  tan
                           2
                              2t              1 t2
                then sin x         , cos x 
                             1 t 2
                                              1 t 2
(6) t results
                           x
                If t  tan
                           2
                              2t              1 t2
                then sin x         , cos x 
                             1 t 2
                                              1 t 2
                         dt 1 2 x
                            sec
                         dx 2       2
                            1          2 x
                            1  tan 
                            2            2

                            1  t 2 
                            1
                            2
(6) t results
                           x
                If t  tan
                           2
                              2t              1 t2
                then sin x         , cos x 
                             1 t 2
                                              1 t 2
                         dt 1 2 x
                            sec
                         dx 2       2
                            1          2 x
                            1  tan 
                            2            2

                            1  t 2 
                            1
                            2
                               dx     2
                                  
                               dt 1  t 2
                                     2dt
                               dx 
                                    1 t 2
dx
e.g. 
       7  6 cos x
dx
e.g.                 t  tan
                              x
       7  6 cos x            2
                           2dt
                     dx 
                          1 t 2
dx
e.g.                       t  tan
                                    x
       7  6 cos x                  2
              2dt                2dt
                           dx 
          1 t 2              1 t 2
              1  t 2 
        7  6         
              1  t 
                     2
dx
e.g.                          t  tan
                                       x
       7  6 cos x                     2
              2dt                   2dt
                              dx 
          1 t 2                 1 t 2
              1  t 2 
        7  6         
              1  t 
                     2


                 2dt
   
        7  7t 2  6  6t 2
          2dt
   
        13  t 2
dx
e.g.                          t  tan
                                       x
       7  6 cos x                     2
              2dt                   2dt
                              dx 
           1 t 2                1 t 2
              1  t 2 
        7  6         
              1  t 
                     2


                 2dt
   
        7  7t 2  6  6t 2
          2dt
   
        13  t 2
        2        1 t
           tan          c
        13           13
dx
e.g.                             t  tan
                                          x
       7  6 cos x                        2
              2dt                      2dt
                                 dx 
           1 t 2                   1 t 2
              1  t 2 
        7  6          
              1  t 
                      2


                 2dt
   
        7  7t 2  6  6t 2
          2dt
   
        13  t 2
        2        1 t
           tan           c
        13            13
                         x
        2           tan 
          tan 1        2c
        13          13 
                          
                          
If t  tan x
If t  tan x   dt
                   sec 2 x
               dx
                   1  tan 2 x
                   1 t2
If t  tan x   dt
                   sec 2 x       dx
                                     
                                        1
               dx                 dt 1  t 2
                   1  tan 2 x         dt
                                  dx 
                   1 t2              1 t2
If t  tan x       dt
                        sec 2 x       dx
                                          
                                             1
                    dx                 dt 1  t 2
                        1  tan 2 x         dt
                                       dx 
                        1 t2              1 t2


In General :
                x
   If t  tan
                a
If t  tan x       dt
                        sec 2 x            dx
                                               
                                                  1
                    dx                      dt 1  t 2
                        1  tan 2 x              dt
                                            dx 
                        1 t2                   1 t2


In General :
                x       dt 1 2 x
   If t  tan              sec
                a       dx a      a
                           1          x
                           1  tan 2 
                           a          a

                           1  t 2 
                           1
                           a
If t  tan x       dt
                        sec 2 x            dx
                                               
                                                  1
                    dx                      dt 1  t 2
                        1  tan 2 x              dt
                                            dx 
                        1 t2                   1 t2


In General :
                x       dt 1 2 x                dx     a
   If t  tan              sec                    
                a       dx a      a             dt 1  t 2
                           1          x             adt
                           1  tan 2         dx 
                           a          a            1 t2
                           1  t 2 
                           1
                           a
If t  tan x       dt
                        sec 2 x            dx
                                               
                                                  1
                    dx                      dt 1  t 2
                        1  tan 2 x              dt
                                            dx 
                        1 t2                   1 t2


In General :
                x       dt 1 2 x                dx     a
   If t  tan              sec                    
                a       dx a      a             dt 1  t 2
                           1          x             adt
                           1  tan 2         dx 
                           a          a            1 t2
                           1  t 2 
                           1
                           a
Exercise 2C; 20 21, 24, 25

More Related Content

Viewers also liked

Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mau
Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mauXay dung he thong xu ly nuic thai du an benh vien da khoa ca mau
Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mauDo Trung
 
Xay dung dong tien p2-E0
Xay dung dong tien p2-E0Xay dung dong tien p2-E0
Xay dung dong tien p2-E0Phạm Nhung
 
Xarxes 1
Xarxes 1 Xarxes 1
Xarxes 1
Manel Apio
 
Xebec digital presentation. Engage. Lead
Xebec digital presentation. Engage. LeadXebec digital presentation. Engage. Lead
Xebec digital presentation. Engage. LeadJulia Dutta
 
XCOM: Enemy Unknown
XCOM: Enemy UnknownXCOM: Enemy Unknown
XCOM: Enemy UnknownAndrew John
 
X-Construction Lite: HCI Evaluation
X-Construction Lite: HCI EvaluationX-Construction Lite: HCI Evaluation
X-Construction Lite: HCI Evaluation
twh
 
Xela key
Xela key Xela key
Xela key xelapa
 
Xact aceds 5-7-14 webcast
Xact aceds 5-7-14 webcastXact aceds 5-7-14 webcast
Xact aceds 5-7-14 webcastRobbie Hilson
 

Viewers also liked (15)

X4 drill practice
X4 drill practiceX4 drill practice
X4 drill practice
 
X Ciclo de Cine " Pobreza y Exclusión Social"
X Ciclo de Cine " Pobreza y Exclusión Social"X Ciclo de Cine " Pobreza y Exclusión Social"
X Ciclo de Cine " Pobreza y Exclusión Social"
 
Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mau
Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mauXay dung he thong xu ly nuic thai du an benh vien da khoa ca mau
Xay dung he thong xu ly nuic thai du an benh vien da khoa ca mau
 
Xay dung dong tien p2-E0
Xay dung dong tien p2-E0Xay dung dong tien p2-E0
Xay dung dong tien p2-E0
 
hojahoja
hojahojahojahoja
hojahoja
 
Xarxes 1
Xarxes 1 Xarxes 1
Xarxes 1
 
Xebec digital presentation. Engage. Lead
Xebec digital presentation. Engage. LeadXebec digital presentation. Engage. Lead
Xebec digital presentation. Engage. Lead
 
Xcix
XcixXcix
Xcix
 
X 10 11-immanuel-djhs
X 10 11-immanuel-djhsX 10 11-immanuel-djhs
X 10 11-immanuel-djhs
 
XCOM: Enemy Unknown
XCOM: Enemy UnknownXCOM: Enemy Unknown
XCOM: Enemy Unknown
 
Xcongressonbookbarcelona
XcongressonbookbarcelonaXcongressonbookbarcelona
Xcongressonbookbarcelona
 
X-Construction Lite: HCI Evaluation
X-Construction Lite: HCI EvaluationX-Construction Lite: HCI Evaluation
X-Construction Lite: HCI Evaluation
 
Xela key
Xela key Xela key
Xela key
 
X.3 pai
X.3 paiX.3 pai
X.3 pai
 
Xact aceds 5-7-14 webcast
Xact aceds 5-7-14 webcastXact aceds 5-7-14 webcast
Xact aceds 5-7-14 webcast
 

More from Nigel Simmons

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
Nigel Simmons
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
Nigel Simmons
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)Nigel Simmons
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)Nigel Simmons
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)Nigel Simmons
 

More from Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 

X2 t04 03 t results (2013)

  • 2. (6) t results x If t  tan 2
  • 3. (6) t results x If t  tan 2 2t 1 t2 then sin x  , cos x  1 t 2 1 t 2
  • 4. (6) t results x If t  tan 2 2t 1 t2 then sin x  , cos x  1 t 2 1 t 2 dt 1 2 x  sec dx 2 2 1 2 x  1  tan  2 2  1  t 2  1 2
  • 5. (6) t results x If t  tan 2 2t 1 t2 then sin x  , cos x  1 t 2 1 t 2 dt 1 2 x  sec dx 2 2 1 2 x  1  tan  2 2  1  t 2  1 2 dx 2  dt 1  t 2 2dt dx  1 t 2
  • 6. dx e.g.  7  6 cos x
  • 7. dx e.g.  t  tan x 7  6 cos x 2 2dt dx  1 t 2
  • 8. dx e.g.  t  tan x 7  6 cos x 2 2dt 2dt dx   1 t 2 1 t 2 1  t 2  7  6  1  t  2
  • 9. dx e.g.  t  tan x 7  6 cos x 2 2dt 2dt dx   1 t 2 1 t 2 1  t 2  7  6  1  t  2 2dt  7  7t 2  6  6t 2 2dt  13  t 2
  • 10. dx e.g.  t  tan x 7  6 cos x 2 2dt 2dt dx   1 t 2 1 t 2 1  t 2  7  6  1  t  2 2dt  7  7t 2  6  6t 2 2dt  13  t 2 2 1 t  tan c 13 13
  • 11. dx e.g.  t  tan x 7  6 cos x 2 2dt 2dt dx   1 t 2 1 t 2 1  t 2  7  6  1  t  2 2dt  7  7t 2  6  6t 2 2dt  13  t 2 2 1 t  tan c 13 13  x 2  tan   tan 1  2c 13  13     
  • 12. If t  tan x
  • 13. If t  tan x dt  sec 2 x dx  1  tan 2 x  1 t2
  • 14. If t  tan x dt  sec 2 x dx  1 dx dt 1  t 2  1  tan 2 x dt dx   1 t2 1 t2
  • 15. If t  tan x dt  sec 2 x dx  1 dx dt 1  t 2  1  tan 2 x dt dx   1 t2 1 t2 In General : x If t  tan a
  • 16. If t  tan x dt  sec 2 x dx  1 dx dt 1  t 2  1  tan 2 x dt dx   1 t2 1 t2 In General : x dt 1 2 x If t  tan  sec a dx a a 1 x  1  tan 2  a a  1  t 2  1 a
  • 17. If t  tan x dt  sec 2 x dx  1 dx dt 1  t 2  1  tan 2 x dt dx   1 t2 1 t2 In General : x dt 1 2 x dx a If t  tan  sec  a dx a a dt 1  t 2 1 x adt  1  tan 2  dx  a a 1 t2  1  t 2  1 a
  • 18. If t  tan x dt  sec 2 x dx  1 dx dt 1  t 2  1  tan 2 x dt dx   1 t2 1 t2 In General : x dt 1 2 x dx a If t  tan  sec  a dx a a dt 1  t 2 1 x adt  1  tan 2  dx  a a 1 t2  1  t 2  1 a
  • 19. Exercise 2C; 20 21, 24, 25