sin a c s o b sin a b si n a b
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Nov 05, 2023 •232 likes •428 views

Trigonometric Relations 1 2 2 2 sin x 1 cos2 x tan x sec x 1 2 1 2 2 2 cos x 1 cos2 x cot x csc x 1 2 1 sin

  2. Trigonometric Relations 1       2 2 2 sin x 1 cos2 x tan x sec x 1 2 1       2 2 2 cos x 1 cos2 x cot x csc x 1 2 1           sin . a c s o b  sin a b si n a b  2 1           sin . s n i cos co s a b  a b a b  2 1           cos . a c s o b cos a b co s a b   2

  3. Examples  1 1 2    ( ) cos 3 i x dx cos6 x dx 2   1 1 sin6      x x C   2 6      ( ) ii sin 2 x .cos 5 x dx 1           sin 7 x sin 3 x dx   2     1 1 1           cos 7 sin 3 x x C    2 7 3

  4.    2  ( ii i ) 1 2sin5 x d x     2 1 4sin5 x 4sin 5 x dx        1 4sin5 x 2 1 cos10 x dx    4 1        x cos5 x 2 x sin10 x C   5 10

  5.    2  ( iv ) tan2 x 3sec2 x d x       2 2 tan 2 x 6sec2 x tan2 x 9sec 2 x dx            2 2 sec 2 x 1 6sec2 x tan2 x 9sec 2 x dx         2 6sec2 x tan2 x 10sec 2 x 1 dx   6 10     sec2 tan2 x x x C 2 2

  6.  2 2 ( v ) cos 2 . x sin 3 x dx 1 1         1 cos4 x . 1 cos6 x dx 2 2 1 1      cos4 x cos6 x cos4 .cos6 x x dx 4 1 1         1 cos4 x cos6 x cos2 x cos10 x dx 4 2     1 1 1 1 1 1          x sin4 x sin6 x sin2 x sin10 x  C     4 4 6 2 2 10

  7. Important Rules    n 1   f x   n f            ' (1)  f x   x dx C   1 n   ' f x        ln f x C (2 ) dx     f x

  8. Examples       4 1 4      2 x 2 x 2 x 2 x ( ) i e 3 e dx e 3 2 e dx 2   5  2 x e 3 1   C 2 5   1 1/ 2        2 2 3 x 5 6 x dx ( i i ) x 3 x 5 d x 6   3/2  2 3 x 5 1   C 6 3/2

  9.  2  2 1 6 x 6 x 1    dx ( i ii ) dx     3 3 6 2 x 6 x 5 2 x 6 x 5 1 ln 2     3 6 5 x x C 6   2 1/ x dx  ( ) iv dx 2 ln x ln x x   2lnln x C

  10.   1 1 sin x dx sin x dx    ( ) v  2 1 x  2 1 x   1 1/ 2    1 sin x dx  2 1 x   3/2      1 sin / 3/2 x C   sin3 x 1 3sin3 x dx    ( ) vi dx   cos3 x 5 3 cos3 x 5  1ln cos3    x 5 C 3

  11. Examples   ( ) i tan x dx ( ) ii cot x dx   sin x dx -   cos x dx - cos x sin x     ln cos x - C ln sin x C   ( iii ) sec x dx ( iv ) csc x dx   sec x tan x csc x cot x     csc x dx sec x dx   csc x cot x sec x tan x       ln csc x cot x C ln sec x tan x C

  12. Integration by Substitution * 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.

  13.    4 5 sin x x dx (1) Example  5 (2) x u  4 5 x dx du (3) Substitute from (2) &(3) into (1)     du  4 5 sin x x dx sin u 5  1cos   u C 5    1cos   5 x C 5

  14. ln y   Example  dy (1)   2  y 1 ln y  ln y u (2) 1 dy  du (3) y Substitute from (2) &(3) into (1) ln y u      dy du    2 2  1 u y 1 ln y 1 ln 1    2 u C 2 1 ln 1   2    ln y C 2

  15.  2 2tan x Example sec x e dx (1)  tan x u (2)  2 (3) sec x dx du Substitute from (2) &(3) into (1)    2 2tan x 2 u sec x e dx e du 1   2 u e C 2 1   2tan x e C 2

  16. sec x tan x dx  Example (1)  2 9 4sec x  sec x u (2)  sec tan x x dx du (3) Substitute from (2) &(3) into (1) sec x tan x dx  1   du   2 2 9 4 u 9 4sec x 3 1 1  a   du  2 2 9 4 u 4    1 1 u   1   . tan C   4 3/2 3/2

  17. x  Example dx (1) x  1  (2) x u  2 x u  dx 2 udu (3) Substitute from (2) &(3) into (1) x u     dx 2 udu u 1  x  u 1 1 u  2 2 1 u u   2 du - -   2 u 1 u u   1          2 u 1 du u    u 1 + +   u 1   2 u          1 2 ln 1 u u C  2 

  18. 1  x dx Example (1) e   x e  x (2) e u  x ln u 1  dx du (3) u Substitute from (2) &(3) into (1) 1 1 1 du    x dx e  u    x 1 u e u 1   du  2 u 1    1 tan u C

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