Tablica Der Int

Vrijednosti sinusa i kosinusa

0 0 1

ϕ

sin ϕ cos ϕ

π 6 1 2 3 2

√

π

π

√42 √22

√33 2 1 2

2

π 2

f (x)

f  (x)

f (x)

1 0

xa

axa−1

ln x

sin x

cos x

loga x

cos x

− sin x

sh x

1 x ln a ch x

ch x

sh x

Adicijski teoremi

sin(x cos(x

± y) ± y) tg(x ± y ) ctg(x ± y )

= sin x cos y = cos x cos y

± cos x sin y  sin x sin y

=

tg x tg y 1 tg x tg y

=

ctg x ctg y 1 ctg y ctg x

±



±



sin2x = 2 si sin x cos x cos2x = cos2 x sin2 x =

ctg2x =

−

2 tg x 1 tg2 x

−

2

1 cos2 x 1 sin2 x 1 1 x2 1 1 x2 1 1 + x2 1 1 + x2

tg x ctg x

Funkcije viˇ sestrukih sestru kih argumenata argumen ata

tg 2x

Tablica derivacija

arccos x arctgx arcctgx

ctg x 1 2 ctg ctg x

−

sin x sin y =

1 2 1 2

y y sin x + sin y = 2 sin sin x+ cos x− 2 2

sin x

cos − sin y = 2 cos

x+y 2

y sin x− 2

y y cos x + cos y = 2 cos cos x+ cos x− 2 2

cos x

− cos y = −2sin

x+y 2

y sin x− 2

Funkcije poloviˇ cnih cnih argumenata argumen ata

sin2 x2 cos2 x2

= =

1 cos x 2 1+cos x 2

−

Neke Nek e vaˇ zne zn e formu fo rmule le 2

sin x

=

cos2 x =

2

tg x 1+tg2 x 1 1+tg2 x 2 tg 2 1+tg2 2 x

sin x cos x

= =

arthx arcthx

a ln a

  dx x

−

x

2

− √ √ − − −

||

 

xα dx =

x +1 α+1

 

ax dx =

a ln a

 

ex dx = ex + C 

 

sin xdx =

 

cos xdx = sin x + C 

 

dx sin2 x

=

 

dx cos2 x

= tg x + C 

 

dx x2 +a2

=

1 a

 

dx x2 a2

=

1 2a

 

√adx−x = arcsin( ax ) + C , a > 0

− 2

α

x

+ C, α

∈ R \ {−1}

+ C 

− cos x + C 

− ctg x + C  arctg( xa ) + C , a > 0

 

 

−a  + C , a > 0 ln  xx+ a

2

√

 

√xdx+A = ln |x + x2 + A| + C , A = 0

 

sh xdx = ch x + C 

 

ch xdx = sh x + C 

 

dx sh2 x

=

 

dx ch2 x

= th x + C 

 

dx sin x

= ln tg x2 + C 

 

dx cos x

2

x

2

1 ch2 x 1 sh2 x 1 1 + x2 1 x2 1 1 1 x2 1 1 x2

= ln x + C 

x

1 tg2 1+tg2

x

Tablica integrala

sin x cos y = 12 (sin(x + y) + sin(x cos x cos y =

archx

x

a

− y)) (cos(x + y) + cos(x − y )) (cos(x − y) − cos(x + y ))

arshx

ex

x

Formule pretvorbe

cthx

−

ex

1

thx

− √ − −√ −

arcsin x

f  (x)

− cth x + C 

| | = ln |tg( + )| + C. x 2

π 4