Fórmulas de Cálculo Diferencial e Integral (Página 1 de 3)                                                                                                                                                                                                        http://www.geocities.com/calculusjrm/                                               Jesús Rubí M.

Fórmulas de                                                                ( a + b ) ⋅ ( a 2 − ab + b 2 ) = a 3 + b3                                                        θ      sin  cos   tg   ctg  sec csc                                  Gráfica 4. Las funciones trigonométricas inversas
                                                                                                                                                                                                                                                 arcctg x , arcsec x , arccsc x :                      sin α + sin β = 2sin
                                                                                                                                                                                                                                                                                                                            1               1
                                                                                                                                                                                                                                                                                                                              (α + β ) ⋅ cos (α − β )
                                                                                                                                                                                                    ∞        ∞
                                                                           ( a + b ) ⋅ ( a3 − a 2 b + ab 2 − b3 ) = a 4 − b 4
                                                                                                                                                                           0        0    1    0          1                                                                                                                  2               2
Cálculo Diferencial                                                                                                                                                        30      12    3 2 1 3     3 2 3 2                                             4                                                                  1               1
                                                                                                                                                                                                                                                                                                       sin α − sin β = 2 sin (α − β ) ⋅ cos (α + β )
                                                                           ( a + b ) ⋅ ( a 4 − a 3b + a 2 b 2 − ab3 + b 4 ) = a 5 + b5
e Integral VER.6.8                                                                                                                                                         45     1 2 1 2      1    1     2   2                                          3
                                                                                                                                                                                                                                                                                                                            2               2
                                                                           ( a + b ) ⋅ ( a5 − a 4 b + a 3b 2 − a 2 b3 + ab 4 − b5 ) = a 6 − b 6
                                                                                                                                                                                                                                                                                                                             1               1
                                                                                                                                                                           60       3 2 12      3 1 3    2 2 3                                                                                         cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β )
Jesús Rubí Miranda (jesusrubim@yahoo.com)                                                                                                                                  90       1    0    ∞     0    ∞   1
                                                                                                                                                                                                                                                         2                                                                   2               2
http://www.geocities.com/calculusjrm/                                                  ⎛ n                     ⎞                                                                                                                                                                                                              1               1
                                                                                                                                                                                                                                                                                                       cos α − cos β = −2 sin (α + β ) ⋅ sin (α − β )
                                                                           ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈
                                                                                                k +1
                                                                                                                                                                                                 ⎡ π π⎤                                                  1
                                                                                                                                                              impar    y = ∠ sin x y ∈ ⎢− , ⎥                                                                                                                                 2               2
                                                                                       ⎝ k =1                  ⎠                                                                       ⎣ 2 2⎦
                             VALOR ABSOLUTO
                                                                                                                                                                                                                                                                                                                                sin (α ± β )
                                                                                                                                                                                                                                                         0

                                                                                           ⎛                         ⎞                                                 y = ∠ cos x y ∈ [ 0, π ]
                                                                                               n
    ⎧a si a ≥ 0                                                            ( a + b ) ⋅ ⎜ ∑ ( −1)
                                                                                                              k +1
                                                                                                      a n − k b k −1 ⎟ = a n − b n ∀ n ∈                      par                                                                                                                                      tg α ± tg β =
 a =⎨                                                                                      ⎝ k =1                    ⎠
                                                                                                                                                                                                                                                         -1
                                                                                                                                                                                                                                                                                       arc ctg x                            cos α ⋅ cos β
    ⎩− a si a < 0                                                                                                                                                                                          π π                                                                         arc sec x

                                                                                                                                                                       y = ∠ tg x           y∈ −            ,                                                                          arc csc x
                                                                                                                                                                                                                                                                                                                      1
                                                                                                  SUMAS Y PRODUCTOS                                                                                                                                                                                    sin α ⋅ cos β =   ⎡sin (α − β ) + sin (α + β ) ⎤
                                                                                                                                                                                                                                                         -2

 a = −a                                                                                                                                                                                                    2 2
                                                                                                                                                                                                                                                                                                                      2⎣                              ⎦
                                                                                                                                                                                                                                                           -5                 0                    5

                                                                                                                    n

a ≤ a y −a ≤ a                                                             a1 + a2 +               + an = ∑ ak                                                         y = ∠ ctg x = ∠ tg
                                                                                                                                                                                          1
                                                                                                                                                                                                            y ∈ 0, π                                      IDENTIDADES TRIGONOMÉTRICAS                                 1
                                                                                                                k =1                                                                      x                                                                                                            sin α ⋅ sin β = ⎡cos (α − β ) − cos (α + β ) ⎤
                                                                                                                                                                                                                                                 sin θ + cos 2 θ = 1                                                  2⎣                              ⎦
                                                                                                                                                                                                                                                     2

 a ≥0 y a =0 ⇔ a=0                                                          n

                                                                           ∑ c = nc
                                                                                                                                                                                           1
                                                                                                                                                                       y = ∠ sec x = ∠ cos   y ∈ [ 0, π ]                                        1 + ctg 2 θ = csc 2 θ                                                 1
                              n                    n
                                                                                                                                                                                           x                                                                                                           cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤
                             ∏a          = ∏ ak                                                                                                                                                                                                                                                                        2⎣                               ⎦
                                                                           k =1
 ab = a b ó                                                                                                                                                                                                                                      tg 2 θ + 1 = sec 2 θ
                                    k                                       n                   n
                                                                                                                                                                                           1       ⎡ π π⎤
                             k =1              k =1
                                                                           ∑ ca           = c ∑ ak                                                                     y = ∠ csc x = ∠ sen    y ∈ ⎢− , ⎥                                                                                                               tg α + tg β
                                                                                                                                                                                                   ⎣ 2 2⎦                                        sin ( −θ ) = − sin θ
                                                                                  k
                                     n                      n              k =1                k =1                                                                                        x                                                                                                           tg α ⋅ tg β =
 a+b ≤ a + b ó                      ∑a                 ≤ ∑ ak               n                           n                n                                                                                                                                                                                            ctg α + ctg β
                                                                                                                                                                                                                                                 cos ( −θ ) = cos θ
                                               k
                                    k =1                k =1
                                                                           ∑ ( ak + bk ) = ∑ ak + ∑ bk                                                                Gráfica 1. Las funciones trigonométricas: sin x ,
                                                                                                                                                                                                                                                                                                                    FUNCIONES HIPERBÓLICAS
                                                                                                                                                                      cos x , tg x :
                                                                                                                                                                                                                                                 tg ( −θ ) = − tg θ
                                                                           k =1                        k =1             k =1
                                    EXPONENTES
                                                                                                                                                                                                                                                                                                                ex − e− x
                                                                                                                                                                                                                                                                                                       sinh x =
                                                                            n
a p ⋅ a q = a p+q                                                          ∑(a
                                                                           k =1
                                                                                  k       − ak −1 ) = an − a0                                                               2
                                                                                                                                                                                                                                                 sin (θ + 2π ) = sin θ                                               2
ap                                                                                                                                                                                                                                                                                                              e x + e− x
   = a p−q
                                                                                                                                                                          1.5
                                                                            n
                                                                                                  n                                                                                                                                              cos (θ + 2π ) = cos θ                                 cosh x =
aq
                                                                           ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤
                                                                                ⎣              ⎦    ⎣                 ⎦
                                                                                                                                                                            1

                                                                                                                                                                                                                                                 tg (θ + 2π ) = tg θ
                                                                                                                                                                                                                                                                                                                     2
(a )
   p q
            =a          pq                                                 k =1                                                                                           0.5

                                                                                                                                                                                                                                                                                                       tgh x =
                                                                                                                                                                                                                                                                                                               sinh x e x − e − x
                                                                                                                                                                                                                                                                                                                       =
                                                                                                 n
                                                                                                = (a + l )                                                                                                                                       sin (θ + π ) = − sin θ                                        cosh x e x + e− x
(a ⋅b)
                                                                                                                                                                            0
                = a ⋅b
            p            p    p
                                                                                                 2                                                                       -0.5
                                                                                                                                                                                                                                                 cos (θ + π ) = − cos θ                                            1       e x + e− x
        p                                                                   n
                                                                                          1 − r n a − rl                                                                                                                                                                                               ctgh x =        =
⎛a⎞  ap
⎜ ⎟ = p                                                                    ∑ ar = a 1 − r = 1 − r
                                                                                  k −1                                                                                     -1
                                                                                                                                                                                                                                                 tg (θ + π ) = tg θ                                             tgh x e x − e − x
⎝b⎠  b                                                                     k =1
                                                                                                                                                                         -1.5                                                        sen x
                                                                                                                                                                                                                                                                                                                    1            2
                                                                                                                                                                                                                                                 sin (θ + nπ ) = ( −1) sin θ                           sech x =          =
                                                                                                                                                                                                                                     cos x                                n

                                                                           ∑ k = 2 ( n2 + n )
                                                                            n
                                                                                     1                                                                                                                                               tg x
a p/q = a p                                                                                                                                                                                                                                                                                                     cosh x e x + e − x
                q
                                                                                                                                                                           -2
                                                                                                                                                                             -8   -6        -4        -2     0   2       4       6           8

                                                                                                                                                                                                                                                 cos (θ + nπ ) = ( −1) cos θ
                                                                                                                                                                                                                                                                          n
                                                                           k =1
                  LOGARITMOS                                                                                                                                                                                                                                                                                       1            2
                                                                                                                                                                                                                                                                                                       csch x =         =
                                                                           ∑ k 2 = 6 ( 2n3 + 3n2 + n )
                                                                            n
                                                                                       1                                                                              Gráfica 2. Las funciones trigonométricas csc x ,
log a N = x ⇒ a x = N                                                                                                                                                                                                                            tg (θ + nπ ) = tg θ                                            sinh x e x − e − x
                                                                                                                                                                      sec x , ctg x :
log a MN = log a M + log a N                                               k =1
                                                                                                                                                                                                                                                                                                       sinh :     →
                                                                                                                                                                                                                                                 sin ( nπ ) = 0
                                                                           ∑ k 3 = 4 ( n 4 + 2n3 + n 2 )
                                                                            n
                                                                                       1
      M                                                                                                                                                                   2.5
                                                                                                                                                                                                                                                                                                       cosh :     → [1, ∞
log a     = log a M − log a N                                              k =1                                                                                             2                                                                    cos ( nπ ) = ( −1)
                                                                                                                                                                                                                                                                      n
      N                                                                                                                                                                                                                                                                                                tgh :    → −1,1
                                                                           ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n )                                                                                                                                  tg ( nπ ) = 0
                                                                            n
                                                                                        1                                                                                 1.5

log a N r = r log a N                                                                                                                                                       1                                                                                                                          ctgh :     − {0} → −∞ , −1 ∪ 1, ∞
                                                                           k =1
                                                                                                                                                                                                                                                     ⎛ 2n + 1 ⎞
                                                                                                                                                                                                                                                              π ⎟ = ( −1)
          log b N ln N
                                                                                                                                                                                                                                                                                                                  → 0 ,1]
                                                                                                                                                                                                                                                                          n
                                                                                                      + ( 2n − 1) = n
                                                                                                                                                                          0.5
log a N =         =                                                        1+ 3 + 5 +                                            2                                                                                                               sin ⎜                                                 sech :
           log b a ln a                                                                                                                                                     0
                                                                                                                                                                                                                                                     ⎝ 2        ⎠
                                                                                      n                                                                                                                                                                                                                csch :     − {0} →             − {0}
                                                                           n! = ∏ k                                                                                                                                                                   ⎛ 2n + 1 ⎞
                                                                                                                                                                         -0.5
log10 N = log N y log e N = ln N                                                                                                                                           -1                                                                    cos ⎜        π⎟=0
                ALGUNOS PRODUCTOS                                                 k =1
                                                                                                                                                                         -1.5                                                                         ⎝ 2        ⎠                                     Gráfica 5. Las funciones hiperbólicas sinh x ,
a ⋅ ( c + d ) = ac + ad                                                    ⎛n⎞         n!                                                                                                                                            csc x
                                                                                                                                                                                                                                                    ⎛ 2n + 1 ⎞
                                                                           ⎜ ⎟=                 , k≤n                                                                                                                                                                                                  cosh x , tgh x :
                                                                                                                                                                           -2
                                                                                                                                                                                                                                                             π⎟=∞
                                                                                                                                                                                                                                     sec x


                                                                           ⎝ k ⎠ ( n − k )!k !
                                                                                                                                                                                                                                     ctg x       tg ⎜
( a + b) ⋅ ( a − b) = a − b         2              2
                                                                                                                                                                         -2.5
                                                                                                                                                                             -8   -6        -4        -2     0   2       4       6           8      ⎝ 2        ⎠                                                    5


                                                                                         n
                                                                                            ⎛n⎞                                                                                                                                                                   π⎞
                                                                                                                                                                                                                                                                                                                    4


( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b 2                        ( x + y ) = ∑ ⎜ ⎟ xn−k y k                                                                                                                                                        ⎛
                                                                                    n
                                 2                                                                                                                                    Gráfica 3. Las funciones trigonométricas inversas
                                                                                                                                                                                                                                                 sin θ = cos ⎜ θ − ⎟                                                3

                                                                                       k =0 ⎝ k ⎠                                                                     arcsin x , arccos x , arctg x :                                                        ⎝    2⎠
( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2
                                                                                                                                                                                                                                                                                                                    2
                                 2

                                                                                                                                      n!                                                                                                                     ⎛    π⎞
                                                                           ( x1 + x2 +                + xk ) = ∑
                                                                                                                                                                                                                                                                                                                    1

                                                                                                                                              x1n1 ⋅ x2 2                                                                                        cos θ = sin ⎜ θ + ⎟
                                                                                                                n
( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd
                                                                                                                                                      n
                                                                                                                                                            xknk            4

                                                                                                                                                                                                                                                             ⎝    2⎠                                                0
                                                                                                                               n1 ! n2 ! nk !
( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd
                                                                                                                                                                                                                                                                                                                    -1

                                                                                                                                                                                                                                                 sin (α ± β ) = sin α cos β ± cos α sin β
                                                                                                                                                                            3
                                                                                           CONSTANTES                                                                                                                                                                                                               -2


( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd                                  π = 3.14159265359…                                                                               2
                                                                                                                                                                                                                                                 cos (α ± β ) = cos α cos β ∓ sin α sin β                           -3
                                                                                                                                                                                                                                                                                                                                                      senh x
                                                                                                                                                                                                                                                                                                                                                      cosh x
                                                                                                                                                                                                                                                                                                                                                      tgh x


( a + b ) = a3 + 3a 2b + 3ab 2 + b3
         3                                                                 e = 2.71828182846…                                                                               1
                                                                                                                                                                                                                                                                tg α ± tg β
                                                                                                                                                                                                                                                                                                                    -4
                                                                                                                                                                                                                                                                                                                      -5                          0            5


                                                                                         TRIGONOMETRÍA                                                                                                                                           tg (α ± β ) =                                                  FUNCIONES HIPERBÓLICAS INV
( a − b ) = a 3 − 3a 2b + 3ab 2 − b3
         3                                                                                                                                                                                                                                                     1 ∓ tg α tg β
                                                                                                                                                                                                                                                                                                                           (                  )
                                                                                                                                                                            0


                                                                           sen θ =
                                                                                    CO
                                                                                               cscθ =
                                                                                                        1
                                                                                                                                                                                                                                                 sin 2θ = 2sin θ cos θ                                 sinh −1 x = ln x + x 2 + 1 , ∀x ∈
( a + b + c ) = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc
               2
                                                                                   HIP                sen θ
                                                                                                                                                                                                                                                                                                                            (                 )
                                                                                                                                                                           -1

                                                                                                                                                                                                                                                 cos 2θ = cos 2 θ − sin 2 θ
                                                                                                                                                                                                                             arc sen x

                                                                                                                                                                                                                                                                                                       cosh −1 x = ln x ± x 2 − 1 , x ≥ 1
                                                                                                                                                                                                                             arc cos x


( a − b ) ⋅ ( a + ab + b ) = a − b
                                                                                    CA                  1                                                                                                                    arc tg x
                         2               2              3       3
                                                                           cosθ =              secθ =                                                                      -2
                                                                                                                                                                                                                                                            2 tg θ
                                                                                                      cosθ
                                                                                                                                                                             -3        -2        -1          0       1       2               3
                                                                                   HIP                                                                                                                                                           tg 2θ =                                                             1 ⎛1+ x ⎞
( a − b ) ⋅ ( a 3 + a 2 b + ab 2 + b3 ) = a 4 − b 4                               sen θ CO             1                                                                                                                                                  1 − tg 2 θ                                   tgh −1 x =     ln ⎜   ⎟,            x <1
                                                                           tgθ =       =       ctgθ =                                                                                                                                                                                                                2 ⎝1− x ⎠
( a − b ) ⋅ ( a 4 + a 3b + a 2 b 2 + ab3 + b 4 ) = a 5 − b5                       cosθ CA             tgθ                                                                                                                                                  1
                                                                                                                                                                                                                                                 sin 2 θ = (1 − cos 2θ )                                          1 ⎛ x +1⎞
                                                                                                                                                                                                                                                           2                                           ctgh −1 x = ln ⎜   ⎟,                   x >1
                    ⎛   n
                                           ⎞                               π radianes=180                                                                                                                                                                                                                         2 ⎝ x −1⎠
( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n                        ∀n ∈                                                                                                                                                                                    1
                                                                                                                                                                                                                                                 cos 2 θ = (1 + cos 2θ )
                    ⎝ k =1                 ⎠                                                                                                                                                                                                                2                                                         ⎛ 1 ± 1 − x2 ⎞
                                                                                                                                                                                                                                                          1 − cos 2θ                                   sech −1 x = ln ⎜             ⎟, 0 < x ≤ 1
                                                                                                                                                                                                                                                 tg 2 θ =                                                             ⎜     x       ⎟
                                                                                                                                                                                                                                                                                                                      ⎝             ⎠
                                                                                                                     HIP
                                                                                                                                                   CO                                                                                                     1 + cos 2θ
                                                                                                                                                                                                                                                                                                                      ⎛1      x2 + 1 ⎞
                                                                                                                                                                                                                                                                                                       csch −1 x = ln ⎜ +             ⎟, x ≠ 0
                                                                                                            θ                                                                                                                                                                                                         ⎜x       x ⎟
                                                                                                                                                                                                                                                                                                                      ⎝               ⎠
                                                                                                                        CA
Fórmulas de Cálculo Diferencial e Integral (Página 2 de 3)                                                                                                                                                                 http://www.geocities.com/calculusjrm/                                                                          Jesús Rubí M.

                                                                                                                                                                                                                                                                ∫ tgh udu = ln cosh u
          IDENTIDADES DE FUNCS HIP                           d            dv       du                                 DERIVADA DE FUNCS HIPERBÓLICAS                                                     INTEGRALES DE FUNCS LOG & EXP
                                                                ( uv ) = u + v
cosh 2 x − sinh 2 x = 1                                                                                            d                   du
                                                                                                                                                                                                 ∫ e du = e
                                                                                                                                                                                                   u               u
                                                             dx           dx       dx                                 sinh u = cosh u
1 − tgh 2 x = sech 2 x                                       d                dw       dv      du                  dx                  dx                                                                                                                       ∫ ctgh udu = ln sinh u
                                                                ( uvw ) = uv + uw + vw                                                                                                                     au ⎧a > 0
                                                                                                                                                                                                 ∫ a du = ln a ⎨a ≠ 1                                           ∫ sech udu = ∠ tg ( sinh u )
                                                                                                                                                                                                    u
                                                             dx               dx       dx      dx                  d                   du
ctgh 2 x − 1 = csch 2 x                                                                                               cosh u = sinh u
                                                                                                                                                                                                               ⎩
                                                             d ⎛ u ⎞ v ( du dx ) − u ( dv dx )                     dx                  dx
sinh ( − x ) = − sinh x                                                                                                                                                                                                                                         ∫ csch udu = − ctgh ( cosh u )
                                                                                                                                                                                                                                                                                                   −1
                                                                ⎜ ⎟=                                               d                   du                                                                              au ⎛      1 ⎞
                                                             dx ⎝ v ⎠               v2                                tgh u = sech 2 u                                                           ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟
                                                                                                                                                                                                       u

cosh ( − x ) = cosh x                                                                                              dx                  dx                                                                         ⎝          ⎠                                                        1
                                                                                                                                                                                                                                                                             = ln tgh u
tgh ( − x ) = − tgh x
                                                             d n
                                                             dx
                                                                ( u ) = nu dx
                                                                           n −1 du
                                                                                                                   d
                                                                                                                      ctgh u = − csch 2 u
                                                                                                                                          du
                                                                                                                                                                                                 ∫ ue du = e ( u − 1)
                                                                                                                                                                                                       u               u                                                              2
                                                                                                                   dx                     dx                                                                                                                                   INTEGRALES DE FRAC
sinh ( x ± y ) = sinh x cosh y ± cosh x sinh y               dF dF du                                                                                                                            ∫ ln udu =u ln u − u = u ( ln u − 1)
                                                                =   ⋅ (Regla de la Cadena)                         d
                                                                                                                      sech u = − sech u tgh u
                                                                                                                                               du                                                                                                                    du       1      u
cosh ( x ± y ) = cosh x cosh y ± sinh x sinh y               dx du dx                                              dx                          dx                                                               1                    u                          ∫ u 2 + a 2 = a ∠ tg a
                                                             du   1                                                                                                                              ∫ log         udu =( u ln u − u ) = ( ln u − 1)
               tgh x ± tgh y
                                                                                                                                                                                                           a
                                                                =                                                  d                            du                                                             ln a                 ln a
tgh ( x ± y ) =                                                                                                       csch u = − csch u ctgh u                                                                                                                                1       u
                                                             dx dx du                                                                                                                                                                                                     = − ∠ ctg
              1 ± tgh x tgh y                                                                                      dx                           dx                                                                u2
                                                                                                                                                                                                 ∫ u log a udu = 4 ⋅ ( 2log a u − 1)
                                                                                                                                                                                                                                                                              a       a
                                                             dF dF du                                                                                                                                                                                                          1 u−a
sinh 2 x = 2sinh x cosh x                                                                                                  DERIVADA DE FUNCS HIP INV
                                                                =
                                                                                                                                                                                                                                                                ∫ u 2 − a 2 = 2a ln u + a ( u > a )
                                                                                                                                                                                                                                                                     du                      2   2
                                                             dx dx du                                              d                 1      du                                                                 u2
cosh 2 x = cosh 2 x + sinh 2 x
                                                             dy dy dt f 2′ ( t )                  ⎪ x = f1 ( t )
                                                                                                  ⎧
                                                                                                                      senh −1 u =         ⋅
                                                                                                                                   1 + u 2 dx                                                    ∫ u ln udu = 4 ( 2ln u − 1)                                                   1 a+u
                                                                                                                                                                                                                                                                ∫ a 2 − u 2 = 2a ln a − u ( u < a )
                                                                                                                   dx                                                                                                                                                du                      2   2
tgh 2 x =
            2 tgh x                                              =           =             donde ⎨
          1 + tgh 2 x                                        dx dx dt           f1′( t )          ⎪ y = f2 (t )
                                                                                                  ⎩                d                  ±1     du                         ⎧+ si cosh -1u > 0
                                                                                                                                                                        ⎪                                 INTEGRALES DE FUNCS TRIGO
                                                                                                                      cosh −1 u =           ⋅ , u >1                    ⎨
           1                                                        DERIVADA DE FUNCS LOG & EXP                    dx                u 2 − 1 dx                         ⎪− si cosh u < 0
                                                                                                                                                                        ⎩
                                                                                                                                                                                   -1
                                                                                                                                                                                                 ∫ sin udu = − cos u                                                                   INTEGRALES CON
sinh 2 x = ( cosh 2 x − 1)
           2                                                 d
                                                                ( ln u ) =
                                                                            du dx 1 du
                                                                                     = ⋅
                                                                                                                   d
                                                                                                                      tgh −1 u =
                                                                                                                                    1 du
                                                                                                                                         ⋅ , u <1                                                ∫ cos udu = sin u                                              ∫
                                                                                                                                                                                                                                                                          du
                                                                                                                                                                                                                                                                                       = ∠ sin
                                                                                                                                                                                                                                                                                                   u
           1                                                 dx               u          u dx                      dx            1 − u 2 dx                                                                                                                           a2 − u2                      a
cosh 2 x = ( cosh 2 x + 1)                                                                                                                                                                       ∫ sec udu = tg u
                                                                                                                                                                                                           2
                                                             d               log e du                              d                 1 du
           2                                                    ( log u ) =         ⋅                                      −1
                                                                                                                      ctgh u =            ⋅ , u >1                                                                                                                                     = −∠ cos
                                                                                                                                                                                                                                                                                                        u
          cosh 2 x − 1                                                                                                            1 − u 2 dx                                                     ∫ csc udu = − ctg u
                                                                                                                                                                                                           2
                                                             dx                u dx                                dx                                                                                                                                                                                   a
tgh 2 x =
          cosh 2 x + 1                                                                                                                        du ⎧− si                  sech −1 u > 0, u ∈ 0,1
                                                                                                                                                                                                                                                                                             (                         )
                                                             d                log e du
                                                                ( log a u ) = a ⋅ a > 0, a ≠ 1                     d
                                                                                                                      sech −1 u =
                                                                                                                                      ∓1
                                                                                                                                             ⋅ ⎨
                                                                                                                                                 ⎪
                                                                                                                                                                                                 ∫ sec u tg udu = sec u                                         ∫
                                                                                                                                                                                                                                                                          du
                                                                                                                                                                                                                                                                                       = ln u + u 2 ± a 2
                                                             dx                  u        dx                                                                            sech −1 u < 0, u ∈ 0,1                                                                       u 2 ± a2
tgh x =
          sinh 2 x                                                                                                 dx             u 1 − u 2 dx ⎪ + si
                                                                                                                                                 ⎩
        cosh 2 x + 1
                                                             d u
                                                                ( e ) = eu ⋅ du                                    d                  1       du                                                 ∫ csc u ctg udu = − csc u                                                    du1         u
                                                             dx               dx                                      csch −1 u = −          ⋅ , u≠0                                                                                                            ∫ u a 2 ± u 2 = a ln a + a 2 ± u 2
e x = cosh x + sinh x                                                                                              dx               u 1 + u 2 dx                                                 ∫ tg udu = − ln cos u = ln sec u
e − x = cosh x − sinh x
                                                             d u
                                                                ( a ) = au ln a ⋅ du
                                                                                                                                                                                                 ∫ ctg udu = ln sin u
                                                                                                                                                                                                                                                                      du        1       a
                                                             dx                     dx                               INTEGRALES DEFINIDAS, PROPIEDADES
                                                                                                                                                                                                                                                                ∫ u u 2 − a 2 = a ∠ cos u
ax 2 + bx + c = 0
                              OTRAS                          d v
                                                             dx
                                                                ( u ) = vu v−1 du + ln u ⋅ u v ⋅ dv
                                                                                dx                dx
                                                                                                                   Nota. Para todas las fórmulas de integración deberá
                                                                                                                   agregarse una constante arbitraria c (constante de                            ∫ sec udu = ln sec u + tg u                                                    1       u
                                                                                                                                                                                                                                                                              = ∠ sec
       ⇒ x=
               −b ± b 2 − 4ac
                                                                    DERIVADA DE FUNCIONES TRIGO                    integración).
                                                                                                                                                                                                 ∫ csc udu = ln csc u − ctg u                                                   a       a
                                                                                                                   ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx
                                                                                                                    b                                           b          b
                                                             d                       du                                                                                                                                                                                                       a2
                      2a                                        ( sin u ) = cos u                                                                                                                            u 1                                                    2     2      u 2
                                                                                                                                                                                                                                                                ∫ a − u du = 2 a − u + 2 ∠ sen a
                                                                                                                                                                                                                                                                                          2        u
                                                                                                                                                                                                 ∫ sin         udu =
                                                                                                                                                                                                                − sin 2u
                                                                                                                    a                                          a          a                                2
                                                             dx                      dx
       b 2 − 4ac = discriminante
                                                                                                                   ∫ cf ( x ) dx = c ⋅ ∫ f ( x ) dx c ∈
                                                                                                                    b                                    b
                                                                                                                                                                                                             2 4
                                                                                                                                                                                                                                                                                                                                  (                     )
                                                             d                          du
                                                                ( cos u ) = − sin u
                                                                                                                                                                                                                                                                                                2
                                                                                                                                                                                                                                                                                 u 2          a
exp (α ± i β ) = e   α
                         ( cos β ± i sin β )   si α , β ∈                                                           a                                    a
                                                                                                                                                                                                              u 1                                               ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a
                                                                                                                                                                                                                                                                    2     2               2          2 2

                                                                                                                                                                                                 ∫ cos udu = 2 + 4 sin 2u
                                                                                                                                                                                                       2
                                                             dx                         dx
                                                                                                                   ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
                                                                                                                    b                        c                      b
                            LÍMITES                          d                       du
                                                                ( tg u ) = sec2 u                                                                                                                                                                                               MÁS INTEGRALES
                                                                                                                    a                    a                          c

                                                                                                                                                                                                 ∫ tg udu = tg u − u
                                                                                                                                                                                                     2

                                                                                                                   ∫ f ( x ) dx = − ∫ f ( x ) dx
             1                                                                                                      b                                a
lim (1 + x ) x = e = 2.71828...                              dx                      dx                                                                                                                                                                                                       e au ( a sin bu − b cos bu )
                                                                                                                                                                                                                                                                ∫e        sin bu du =
                                                                                                                                                                                                                                                                     au
                                                                                                                                                                                                               udu = − ( ctg u + u )
x →0                                                                                                                a                            b
                                                             d
                                                                ( ctg u ) = − csc2 u
                                                                                         du
                                                                                                                                                                                                 ∫ ctg
                                                                                                                                                                                                           2
                                                                                                                                                                                                                                                                                                           a 2 + b2
                                                                                                                   ∫ f ( x ) dx = 0
                 x                                                                                                   a
     ⎛ 1⎞                                                    dx                           dx
lim ⎜1 + ⎟ = e                                                                                                      a                                                                                                                                                                            e au ( a cos bu + b sin bu )
x →∞
     ⎝ x⎠                                                    d
                                                                ( sec u ) = sec u tg u
                                                                                           du
                                                                                                                   m ⋅ ( b − a ) ≤ ∫ f ( x ) dx ≤ M ⋅ ( b − a )
                                                                                                                                         b                                                       ∫ u sin udu = sin u − u cos u                                  ∫e
                                                                                                                                                                                                                                                                     au
                                                                                                                                                                                                                                                                          cos bu du =
                                                                                                                                                                                                                                                                                           a2 + b2
      sen x                                                  dx                            dx
                                                                                                                                                                                                 ∫ u cos udu = cos u + u sin u
                                                                                                                                         a
            =1
                                                                                                                     ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈
lim                                                                                                                                                                                                                                                                          1              1
                                                                                                                                                                                                                                                                ∫ sec u du = 2 sec u tg u + 2 ln sec u + tg u
                                                                                                                                                                                                                                                                          3
x →0                                                         d                                 du
        x
                                                                ( csc u ) = − csc u ctg u                                                                                                                  INTEGRALES DE FUNCS TRIGO INV
     1 − cos x                                               dx                                dx
                                                                                                                   ∫ f ( x ) dx ≤ ∫ g ( x ) dx
                                                                                                                    b                        b
lim            =0                                                                                                                                                                                                                                                               ALGUNAS SERIES
                                                                                                                                                                                                 ∫ ∠ sin udu = u∠ sin u + 1 − u
                                                                                                                                                                                                                                         2
x →0                                                         d                         du
                                                                ( vers u ) = sen u
                                                                                                                    a                    a
          x
                                                                                                                     ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a , b ]                                                                                                                                                                           f '' ( x0 )( x − x0 )
                                                                                                                                                                                                                                                                                                                                                            2
      ex −1                                                  dx                        dx                                                                                                                                                                       f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) +
             =1                                                                                                                                                                                  ∫ ∠ cos udu = u∠ cos u − 1 − u
                                                                                                                                                                                                                                          2
lim                                                                                                                                                                                                                                                                                                                                          2!
                                                                                                                    ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b
x →0    x                                                             DERIV DE FUNCS TRIGO INVER                     b                           b


                                                                                                                                                                                                 ∫ ∠ tg udu = u∠ tg u − ln 1 + u                                                            f ( n ) ( x0 )( x − x0 )
                                                                                                                                                                                                                                                                                                                           n
      x −1                                                   d                      1         du                     a                       a                                                                                            2
lim         =1                                                  ( ∠ sin u ) =               ⋅                                                                INTEGRALES                                                                                                            +    +                                      : Taylor
 x →1 ln x                                                   dx                   1 − u 2 dx                                                                                                                                                                                                                n!
                                                                                                                                                                                                 ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u
                                                                                                                                                                                                                                               2
                      DERIVADAS                              d                      1       du                     ∫ adx =ax                                                                                                                                                                                     f '' ( 0 ) x 2
                                                                ( ∠ cos u ) = −                                                                                                                                                                                 f ( x ) = f (0) + f ' ( 0) x +
                                                                                                                                                                                                 ∫ ∠ sec udu = u∠ sec u − ln ( u + u                        )
                                                                                          ⋅
                       f ( x + ∆x ) − f ( x )          ∆y                                                                                                                                                                                          2
                                                                                                                                                                                                                                                       −1
                                                                                                                   ∫ af ( x ) dx = a ∫ f ( x ) dx
              df                                                                  1 − u 2 dx
Dx f ( x ) =                                                 dx                                                                                                                                                                                                                                                       2!
                 = lim                        = lim
              dx ∆x →0          ∆x              ∆x → 0 ∆x
                                                             d                  1     du                                                                                                                               = u∠ sec u − ∠ cosh u                                                f ( n) ( 0 ) x n
d                                                               ( ∠ tg u ) =        ⋅
                                                                                                                   ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ±                                                                                                                  +    +                         : Maclaurin
                                                                                                                                                                                                 ∫ ∠ csc udu = u∠ csc u + ln ( u +                          )
   (c) = 0                                                   dx              1 + u 2 dx
                                                                                                                                                                                                                                               u2 − 1
                                                                                                                                                                                                                                                                                  n!
                                                                                                                   ∫ udv = uv − ∫ vdu ( Integración por partes )
dx                                                           d                     1     du                                                                                                                                                                                 x 2 x3     xn
d                                                               ( ∠ ctg u ) = −        ⋅                                                                                                                                                                        ex = 1 + x +   + + +      +
   ( cx ) = c                                                dx                 1 + u 2 dx
                                                                                                                                 u n+1
                                                                                                                                                                                                                     = u∠ csc u + ∠ cosh u                                  2! 3!      n!
dx
                                                                                             du ⎧+ si u > 1        ∫ u du = n + 1                n ≠ −1
                                                                                                                         n
                                                             d                       1                                                                                                                             INTEGRALES DE FUNCS HIP                                 x 3 x5 x 7                   x 2 n −1
                                                                ( ∠ sec u ) = ±             ⋅ ⎨                                                                                                                                                                 sin x = x − +      −  + + ( −1)
                                                                                                                                                                                                                                                                                                n −1
d
   ( cx n ) = ncx n−1                                        dx                 u u 2 − 1 dx ⎩− si u < −1            du                                                                          ∫ sinh udu = cosh u                                                       3! 5! 7!                  ( 2n − 1)!
dx
                                                                                           du ⎧− si u > 1          ∫u        = ln u
                                                                                                                                                                                                                                                                                                                                           x 2 n− 2
d                      du dv dw
   (u ± v ± w ± ) = ± ± ±
                                                             d
                                                                ( ∠ csc u ) = ∓
                                                                                    1
                                                                                          ⋅ ⎨                                                                                                    ∫ cosh udu = sinh u                                            cos x = 1 −
                                                                                                                                                                                                                                                                                       x2 x4 x6
                                                                                                                                                                                                                                                                                         +  −   +                    + ( −1)
                                                                                                                                                                                                                                                                                                                                  n −1

dx                     dx dx dx                              dx                 u u 2 − 1 dx ⎩+ si u < −1                                                                                                                                                                              2! 4! 6!                                          ( 2n − 2 ) !
                                                                                                                                                                                                 ∫ sech udu = tgh u
                                                                                                                                                                                                               2

d              du                                            d                     1       du                                                                                                                                                                                 x2 x3 x 4                       n
   ( cu ) = c                                                   ( ∠ vers u ) =           ⋅                                                                                                                                                                      ln (1 + x ) = x −
                                                                                                                                                                                                                                                                                + −      + + ( −1)
                                                                                                                                                                                                                                                                                                     n −1 x

                                                                                                                                                                                                 ∫ csch udu = − ctgh u
                                                                                                                                                                                                               2
dx             dx                                            dx                  2u − u 2 dx                                                                                                                                                                                  2   3    4                     n
                                                                                                                                                                                                                                                                                                         2 n −1
                                                                                                                                                                                                                                                                            x3 x 5 x 7
                                                                                                                                                                                                 ∫ sech u tgh udu = − sech u                                    ∠ tg x = x − + −
                                                                                                                                                                                                                                                                            3   5   7
                                                                                                                                                                                                                                                                                       + + ( −1)
                                                                                                                                                                                                                                                                                                 n −1 x

                                                                                                                                                                                                                                                                                                      2n − 1
                                                                                                                                                                                                 ∫ csch u ctgh udu = − csch u
Fórmulas de Cálculo Diferencial e Integral (Página 3 de 3)       http://www.geocities.com/calculusjrm/   Jesús Rubí M.
                 ALFABETO GRIEGO
               Mayúscula Minúscula Nombre          Equivalente
                                                    Romano
      1           Α             α         Alfa         A
      2           Β             β         Beta         B
      3           Γ             γ       Gamma          G
      4           ∆             δ         Delta        D
      5           Ε             ε       Epsilon        E
      6           Ζ             ζ         Zeta         Z
      7           Η             η          Eta         H
      8           Θ         θ       ϑ     Teta         Q
      9           Ι             ι         Iota          I
      10          Κ             κ        Kappa         K
      11          Λ             λ       Lambda         L
      12          Μ             µ          Mu          M
      13          Ν             ν          Nu          N
      14          Ξ             ξ          Xi          X
      15          Ο             ο       Omicron        O
      16          Π         π ϖ             Pi         P
      17          Ρ          ρ            Rho          R
      18          Σ         σ  ς         Sigma         S
      19          Τ          τ             Tau         T
      20          Υ          υ           Ipsilon       U
      21          Φ         φ ϕ            Phi         F
      22          Χ          χ              Ji         C
      23          Ψ          ψ             Psi         Y
      24          Ω          ω          Omega          W

                        NOTACIÓN
sin       Seno.
cos       Coseno.
tg        Tangente.
sec       Secante.
csc       Cosecante.
ctg       Cotangente.
vers Verso seno.
arcsin θ =      sin θ   Arco seno de un ángulo θ .
u = f ( x)
sinh Seno hiperbólico.
cosh Coseno hiperbólico.
tgh       Tangente hiperbólica.
ctgh Cotangente hiperbólica.
sech Secante hiperbólica.
csch Cosecante hiperbólica.
u, v, w      Funciones de x , u = u ( x ) , v = v ( x ) .

          Conjunto de los números reales.
  = {…, −2, −1,0,1, 2,…}            Conjunto de enteros.

          Conjunto de números racionales.
  c
          Conjunto de números irracionales.

   = {1, 2,3,…} Conjunto de números naturales.

          Conjunto de números complejos.

Calculo

  • 1.
    Fórmulas de CálculoDiferencial e Integral (Página 1 de 3) http://www.geocities.com/calculusjrm/ Jesús Rubí M. Fórmulas de ( a + b ) ⋅ ( a 2 − ab + b 2 ) = a 3 + b3 θ sin cos tg ctg sec csc Gráfica 4. Las funciones trigonométricas inversas arcctg x , arcsec x , arccsc x : sin α + sin β = 2sin 1 1 (α + β ) ⋅ cos (α − β ) ∞ ∞ ( a + b ) ⋅ ( a3 − a 2 b + ab 2 − b3 ) = a 4 − b 4 0 0 1 0 1 2 2 Cálculo Diferencial 30 12 3 2 1 3 3 2 3 2 4 1 1 sin α − sin β = 2 sin (α − β ) ⋅ cos (α + β ) ( a + b ) ⋅ ( a 4 − a 3b + a 2 b 2 − ab3 + b 4 ) = a 5 + b5 e Integral VER.6.8 45 1 2 1 2 1 1 2 2 3 2 2 ( a + b ) ⋅ ( a5 − a 4 b + a 3b 2 − a 2 b3 + ab 4 − b5 ) = a 6 − b 6 1 1 60 3 2 12 3 1 3 2 2 3 cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β ) Jesús Rubí Miranda (jesusrubim@yahoo.com) 90 1 0 ∞ 0 ∞ 1 2 2 2 http://www.geocities.com/calculusjrm/ ⎛ n ⎞ 1 1 cos α − cos β = −2 sin (α + β ) ⋅ sin (α − β ) ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈ k +1 ⎡ π π⎤ 1 impar y = ∠ sin x y ∈ ⎢− , ⎥ 2 2 ⎝ k =1 ⎠ ⎣ 2 2⎦ VALOR ABSOLUTO sin (α ± β ) 0 ⎛ ⎞ y = ∠ cos x y ∈ [ 0, π ] n ⎧a si a ≥ 0 ( a + b ) ⋅ ⎜ ∑ ( −1) k +1 a n − k b k −1 ⎟ = a n − b n ∀ n ∈ par tg α ± tg β = a =⎨ ⎝ k =1 ⎠ -1 arc ctg x cos α ⋅ cos β ⎩− a si a < 0 π π arc sec x y = ∠ tg x y∈ − , arc csc x 1 SUMAS Y PRODUCTOS sin α ⋅ cos β = ⎡sin (α − β ) + sin (α + β ) ⎤ -2 a = −a 2 2 2⎣ ⎦ -5 0 5 n a ≤ a y −a ≤ a a1 + a2 + + an = ∑ ak y = ∠ ctg x = ∠ tg 1 y ∈ 0, π IDENTIDADES TRIGONOMÉTRICAS 1 k =1 x sin α ⋅ sin β = ⎡cos (α − β ) − cos (α + β ) ⎤ sin θ + cos 2 θ = 1 2⎣ ⎦ 2 a ≥0 y a =0 ⇔ a=0 n ∑ c = nc 1 y = ∠ sec x = ∠ cos y ∈ [ 0, π ] 1 + ctg 2 θ = csc 2 θ 1 n n x cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤ ∏a = ∏ ak 2⎣ ⎦ k =1 ab = a b ó tg 2 θ + 1 = sec 2 θ k n n 1 ⎡ π π⎤ k =1 k =1 ∑ ca = c ∑ ak y = ∠ csc x = ∠ sen y ∈ ⎢− , ⎥ tg α + tg β ⎣ 2 2⎦ sin ( −θ ) = − sin θ k n n k =1 k =1 x tg α ⋅ tg β = a+b ≤ a + b ó ∑a ≤ ∑ ak n n n ctg α + ctg β cos ( −θ ) = cos θ k k =1 k =1 ∑ ( ak + bk ) = ∑ ak + ∑ bk Gráfica 1. Las funciones trigonométricas: sin x , FUNCIONES HIPERBÓLICAS cos x , tg x : tg ( −θ ) = − tg θ k =1 k =1 k =1 EXPONENTES ex − e− x sinh x = n a p ⋅ a q = a p+q ∑(a k =1 k − ak −1 ) = an − a0 2 sin (θ + 2π ) = sin θ 2 ap e x + e− x = a p−q 1.5 n n cos (θ + 2π ) = cos θ cosh x = aq ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤ ⎣ ⎦ ⎣ ⎦ 1 tg (θ + 2π ) = tg θ 2 (a ) p q =a pq k =1 0.5 tgh x = sinh x e x − e − x = n = (a + l ) sin (θ + π ) = − sin θ cosh x e x + e− x (a ⋅b) 0 = a ⋅b p p p 2 -0.5 cos (θ + π ) = − cos θ 1 e x + e− x p n 1 − r n a − rl ctgh x = = ⎛a⎞ ap ⎜ ⎟ = p ∑ ar = a 1 − r = 1 − r k −1 -1 tg (θ + π ) = tg θ tgh x e x − e − x ⎝b⎠ b k =1 -1.5 sen x 1 2 sin (θ + nπ ) = ( −1) sin θ sech x = = cos x n ∑ k = 2 ( n2 + n ) n 1 tg x a p/q = a p cosh x e x + e − x q -2 -8 -6 -4 -2 0 2 4 6 8 cos (θ + nπ ) = ( −1) cos θ n k =1 LOGARITMOS 1 2 csch x = = ∑ k 2 = 6 ( 2n3 + 3n2 + n ) n 1 Gráfica 2. Las funciones trigonométricas csc x , log a N = x ⇒ a x = N tg (θ + nπ ) = tg θ sinh x e x − e − x sec x , ctg x : log a MN = log a M + log a N k =1 sinh : → sin ( nπ ) = 0 ∑ k 3 = 4 ( n 4 + 2n3 + n 2 ) n 1 M 2.5 cosh : → [1, ∞ log a = log a M − log a N k =1 2 cos ( nπ ) = ( −1) n N tgh : → −1,1 ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n ) tg ( nπ ) = 0 n 1 1.5 log a N r = r log a N 1 ctgh : − {0} → −∞ , −1 ∪ 1, ∞ k =1 ⎛ 2n + 1 ⎞ π ⎟ = ( −1) log b N ln N → 0 ,1] n + ( 2n − 1) = n 0.5 log a N = = 1+ 3 + 5 + 2 sin ⎜ sech : log b a ln a 0 ⎝ 2 ⎠ n csch : − {0} → − {0} n! = ∏ k ⎛ 2n + 1 ⎞ -0.5 log10 N = log N y log e N = ln N -1 cos ⎜ π⎟=0 ALGUNOS PRODUCTOS k =1 -1.5 ⎝ 2 ⎠ Gráfica 5. Las funciones hiperbólicas sinh x , a ⋅ ( c + d ) = ac + ad ⎛n⎞ n! csc x ⎛ 2n + 1 ⎞ ⎜ ⎟= , k≤n cosh x , tgh x : -2 π⎟=∞ sec x ⎝ k ⎠ ( n − k )!k ! ctg x tg ⎜ ( a + b) ⋅ ( a − b) = a − b 2 2 -2.5 -8 -6 -4 -2 0 2 4 6 8 ⎝ 2 ⎠ 5 n ⎛n⎞ π⎞ 4 ( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b 2 ( x + y ) = ∑ ⎜ ⎟ xn−k y k ⎛ n 2 Gráfica 3. Las funciones trigonométricas inversas sin θ = cos ⎜ θ − ⎟ 3 k =0 ⎝ k ⎠ arcsin x , arccos x , arctg x : ⎝ 2⎠ ( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2 2 2 n! ⎛ π⎞ ( x1 + x2 + + xk ) = ∑ 1 x1n1 ⋅ x2 2 cos θ = sin ⎜ θ + ⎟ n ( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd n xknk 4 ⎝ 2⎠ 0 n1 ! n2 ! nk ! ( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd -1 sin (α ± β ) = sin α cos β ± cos α sin β 3 CONSTANTES -2 ( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd π = 3.14159265359… 2 cos (α ± β ) = cos α cos β ∓ sin α sin β -3 senh x cosh x tgh x ( a + b ) = a3 + 3a 2b + 3ab 2 + b3 3 e = 2.71828182846… 1 tg α ± tg β -4 -5 0 5 TRIGONOMETRÍA tg (α ± β ) = FUNCIONES HIPERBÓLICAS INV ( a − b ) = a 3 − 3a 2b + 3ab 2 − b3 3 1 ∓ tg α tg β ( ) 0 sen θ = CO cscθ = 1 sin 2θ = 2sin θ cos θ sinh −1 x = ln x + x 2 + 1 , ∀x ∈ ( a + b + c ) = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc 2 HIP sen θ ( ) -1 cos 2θ = cos 2 θ − sin 2 θ arc sen x cosh −1 x = ln x ± x 2 − 1 , x ≥ 1 arc cos x ( a − b ) ⋅ ( a + ab + b ) = a − b CA 1 arc tg x 2 2 3 3 cosθ = secθ = -2 2 tg θ cosθ -3 -2 -1 0 1 2 3 HIP tg 2θ = 1 ⎛1+ x ⎞ ( a − b ) ⋅ ( a 3 + a 2 b + ab 2 + b3 ) = a 4 − b 4 sen θ CO 1 1 − tg 2 θ tgh −1 x = ln ⎜ ⎟, x <1 tgθ = = ctgθ = 2 ⎝1− x ⎠ ( a − b ) ⋅ ( a 4 + a 3b + a 2 b 2 + ab3 + b 4 ) = a 5 − b5 cosθ CA tgθ 1 sin 2 θ = (1 − cos 2θ ) 1 ⎛ x +1⎞ 2 ctgh −1 x = ln ⎜ ⎟, x >1 ⎛ n ⎞ π radianes=180 2 ⎝ x −1⎠ ( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n ∀n ∈ 1 cos 2 θ = (1 + cos 2θ ) ⎝ k =1 ⎠ 2 ⎛ 1 ± 1 − x2 ⎞ 1 − cos 2θ sech −1 x = ln ⎜ ⎟, 0 < x ≤ 1 tg 2 θ = ⎜ x ⎟ ⎝ ⎠ HIP CO 1 + cos 2θ ⎛1 x2 + 1 ⎞ csch −1 x = ln ⎜ + ⎟, x ≠ 0 θ ⎜x x ⎟ ⎝ ⎠ CA
  • 2.
    Fórmulas de CálculoDiferencial e Integral (Página 2 de 3) http://www.geocities.com/calculusjrm/ Jesús Rubí M. ∫ tgh udu = ln cosh u IDENTIDADES DE FUNCS HIP d dv du DERIVADA DE FUNCS HIPERBÓLICAS INTEGRALES DE FUNCS LOG & EXP ( uv ) = u + v cosh 2 x − sinh 2 x = 1 d du ∫ e du = e u u dx dx dx sinh u = cosh u 1 − tgh 2 x = sech 2 x d dw dv du dx dx ∫ ctgh udu = ln sinh u ( uvw ) = uv + uw + vw au ⎧a > 0 ∫ a du = ln a ⎨a ≠ 1 ∫ sech udu = ∠ tg ( sinh u ) u dx dx dx dx d du ctgh 2 x − 1 = csch 2 x cosh u = sinh u ⎩ d ⎛ u ⎞ v ( du dx ) − u ( dv dx ) dx dx sinh ( − x ) = − sinh x ∫ csch udu = − ctgh ( cosh u ) −1 ⎜ ⎟= d du au ⎛ 1 ⎞ dx ⎝ v ⎠ v2 tgh u = sech 2 u ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟ u cosh ( − x ) = cosh x dx dx ⎝ ⎠ 1 = ln tgh u tgh ( − x ) = − tgh x d n dx ( u ) = nu dx n −1 du d ctgh u = − csch 2 u du ∫ ue du = e ( u − 1) u u 2 dx dx INTEGRALES DE FRAC sinh ( x ± y ) = sinh x cosh y ± cosh x sinh y dF dF du ∫ ln udu =u ln u − u = u ( ln u − 1) = ⋅ (Regla de la Cadena) d sech u = − sech u tgh u du du 1 u cosh ( x ± y ) = cosh x cosh y ± sinh x sinh y dx du dx dx dx 1 u ∫ u 2 + a 2 = a ∠ tg a du 1 ∫ log udu =( u ln u − u ) = ( ln u − 1) tgh x ± tgh y a = d du ln a ln a tgh ( x ± y ) = csch u = − csch u ctgh u 1 u dx dx du = − ∠ ctg 1 ± tgh x tgh y dx dx u2 ∫ u log a udu = 4 ⋅ ( 2log a u − 1) a a dF dF du 1 u−a sinh 2 x = 2sinh x cosh x DERIVADA DE FUNCS HIP INV = ∫ u 2 − a 2 = 2a ln u + a ( u > a ) du 2 2 dx dx du d 1 du u2 cosh 2 x = cosh 2 x + sinh 2 x dy dy dt f 2′ ( t ) ⎪ x = f1 ( t ) ⎧ senh −1 u = ⋅ 1 + u 2 dx ∫ u ln udu = 4 ( 2ln u − 1) 1 a+u ∫ a 2 − u 2 = 2a ln a − u ( u < a ) dx du 2 2 tgh 2 x = 2 tgh x = = donde ⎨ 1 + tgh 2 x dx dx dt f1′( t ) ⎪ y = f2 (t ) ⎩ d ±1 du ⎧+ si cosh -1u > 0 ⎪ INTEGRALES DE FUNCS TRIGO cosh −1 u = ⋅ , u >1 ⎨ 1 DERIVADA DE FUNCS LOG & EXP dx u 2 − 1 dx ⎪− si cosh u < 0 ⎩ -1 ∫ sin udu = − cos u INTEGRALES CON sinh 2 x = ( cosh 2 x − 1) 2 d ( ln u ) = du dx 1 du = ⋅ d tgh −1 u = 1 du ⋅ , u <1 ∫ cos udu = sin u ∫ du = ∠ sin u 1 dx u u dx dx 1 − u 2 dx a2 − u2 a cosh 2 x = ( cosh 2 x + 1) ∫ sec udu = tg u 2 d log e du d 1 du 2 ( log u ) = ⋅ −1 ctgh u = ⋅ , u >1 = −∠ cos u cosh 2 x − 1 1 − u 2 dx ∫ csc udu = − ctg u 2 dx u dx dx a tgh 2 x = cosh 2 x + 1 du ⎧− si sech −1 u > 0, u ∈ 0,1 ( ) d log e du ( log a u ) = a ⋅ a > 0, a ≠ 1 d sech −1 u = ∓1 ⋅ ⎨ ⎪ ∫ sec u tg udu = sec u ∫ du = ln u + u 2 ± a 2 dx u dx sech −1 u < 0, u ∈ 0,1 u 2 ± a2 tgh x = sinh 2 x dx u 1 − u 2 dx ⎪ + si ⎩ cosh 2 x + 1 d u ( e ) = eu ⋅ du d 1 du ∫ csc u ctg udu = − csc u du1 u dx dx csch −1 u = − ⋅ , u≠0 ∫ u a 2 ± u 2 = a ln a + a 2 ± u 2 e x = cosh x + sinh x dx u 1 + u 2 dx ∫ tg udu = − ln cos u = ln sec u e − x = cosh x − sinh x d u ( a ) = au ln a ⋅ du ∫ ctg udu = ln sin u du 1 a dx dx INTEGRALES DEFINIDAS, PROPIEDADES ∫ u u 2 − a 2 = a ∠ cos u ax 2 + bx + c = 0 OTRAS d v dx ( u ) = vu v−1 du + ln u ⋅ u v ⋅ dv dx dx Nota. Para todas las fórmulas de integración deberá agregarse una constante arbitraria c (constante de ∫ sec udu = ln sec u + tg u 1 u = ∠ sec ⇒ x= −b ± b 2 − 4ac DERIVADA DE FUNCIONES TRIGO integración). ∫ csc udu = ln csc u − ctg u a a ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx b b b d du a2 2a ( sin u ) = cos u u 1 2 2 u 2 ∫ a − u du = 2 a − u + 2 ∠ sen a 2 u ∫ sin udu = − sin 2u a a a 2 dx dx b 2 − 4ac = discriminante ∫ cf ( x ) dx = c ⋅ ∫ f ( x ) dx c ∈ b b 2 4 ( ) d du ( cos u ) = − sin u 2 u 2 a exp (α ± i β ) = e α ( cos β ± i sin β ) si α , β ∈ a a u 1 ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a 2 2 2 2 2 ∫ cos udu = 2 + 4 sin 2u 2 dx dx ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx b c b LÍMITES d du ( tg u ) = sec2 u MÁS INTEGRALES a a c ∫ tg udu = tg u − u 2 ∫ f ( x ) dx = − ∫ f ( x ) dx 1 b a lim (1 + x ) x = e = 2.71828... dx dx e au ( a sin bu − b cos bu ) ∫e sin bu du = au udu = − ( ctg u + u ) x →0 a b d ( ctg u ) = − csc2 u du ∫ ctg 2 a 2 + b2 ∫ f ( x ) dx = 0 x a ⎛ 1⎞ dx dx lim ⎜1 + ⎟ = e a e au ( a cos bu + b sin bu ) x →∞ ⎝ x⎠ d ( sec u ) = sec u tg u du m ⋅ ( b − a ) ≤ ∫ f ( x ) dx ≤ M ⋅ ( b − a ) b ∫ u sin udu = sin u − u cos u ∫e au cos bu du = a2 + b2 sen x dx dx ∫ u cos udu = cos u + u sin u a =1 ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈ lim 1 1 ∫ sec u du = 2 sec u tg u + 2 ln sec u + tg u 3 x →0 d du x ( csc u ) = − csc u ctg u INTEGRALES DE FUNCS TRIGO INV 1 − cos x dx dx ∫ f ( x ) dx ≤ ∫ g ( x ) dx b b lim =0 ALGUNAS SERIES ∫ ∠ sin udu = u∠ sin u + 1 − u 2 x →0 d du ( vers u ) = sen u a a x ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a , b ] f '' ( x0 )( x − x0 ) 2 ex −1 dx dx f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) + =1 ∫ ∠ cos udu = u∠ cos u − 1 − u 2 lim 2! ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b x →0 x DERIV DE FUNCS TRIGO INVER b b ∫ ∠ tg udu = u∠ tg u − ln 1 + u f ( n ) ( x0 )( x − x0 ) n x −1 d 1 du a a 2 lim =1 ( ∠ sin u ) = ⋅ INTEGRALES + + : Taylor x →1 ln x dx 1 − u 2 dx n! ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u 2 DERIVADAS d 1 du ∫ adx =ax f '' ( 0 ) x 2 ( ∠ cos u ) = − f ( x ) = f (0) + f ' ( 0) x + ∫ ∠ sec udu = u∠ sec u − ln ( u + u ) ⋅ f ( x + ∆x ) − f ( x ) ∆y 2 −1 ∫ af ( x ) dx = a ∫ f ( x ) dx df 1 − u 2 dx Dx f ( x ) = dx 2! = lim = lim dx ∆x →0 ∆x ∆x → 0 ∆x d 1 du = u∠ sec u − ∠ cosh u f ( n) ( 0 ) x n d ( ∠ tg u ) = ⋅ ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ± + + : Maclaurin ∫ ∠ csc udu = u∠ csc u + ln ( u + ) (c) = 0 dx 1 + u 2 dx u2 − 1 n! ∫ udv = uv − ∫ vdu ( Integración por partes ) dx d 1 du x 2 x3 xn d ( ∠ ctg u ) = − ⋅ ex = 1 + x + + + + + ( cx ) = c dx 1 + u 2 dx u n+1 = u∠ csc u + ∠ cosh u 2! 3! n! dx du ⎧+ si u > 1 ∫ u du = n + 1 n ≠ −1 n d 1 INTEGRALES DE FUNCS HIP x 3 x5 x 7 x 2 n −1 ( ∠ sec u ) = ± ⋅ ⎨ sin x = x − + − + + ( −1) n −1 d ( cx n ) = ncx n−1 dx u u 2 − 1 dx ⎩− si u < −1 du ∫ sinh udu = cosh u 3! 5! 7! ( 2n − 1)! dx du ⎧− si u > 1 ∫u = ln u x 2 n− 2 d du dv dw (u ± v ± w ± ) = ± ± ± d ( ∠ csc u ) = ∓ 1 ⋅ ⎨ ∫ cosh udu = sinh u cos x = 1 − x2 x4 x6 + − + + ( −1) n −1 dx dx dx dx dx u u 2 − 1 dx ⎩+ si u < −1 2! 4! 6! ( 2n − 2 ) ! ∫ sech udu = tgh u 2 d du d 1 du x2 x3 x 4 n ( cu ) = c ( ∠ vers u ) = ⋅ ln (1 + x ) = x − + − + + ( −1) n −1 x ∫ csch udu = − ctgh u 2 dx dx dx 2u − u 2 dx 2 3 4 n 2 n −1 x3 x 5 x 7 ∫ sech u tgh udu = − sech u ∠ tg x = x − + − 3 5 7 + + ( −1) n −1 x 2n − 1 ∫ csch u ctgh udu = − csch u
  • 3.
    Fórmulas de CálculoDiferencial e Integral (Página 3 de 3) http://www.geocities.com/calculusjrm/ Jesús Rubí M. ALFABETO GRIEGO Mayúscula Minúscula Nombre Equivalente Romano 1 Α α Alfa A 2 Β β Beta B 3 Γ γ Gamma G 4 ∆ δ Delta D 5 Ε ε Epsilon E 6 Ζ ζ Zeta Z 7 Η η Eta H 8 Θ θ ϑ Teta Q 9 Ι ι Iota I 10 Κ κ Kappa K 11 Λ λ Lambda L 12 Μ µ Mu M 13 Ν ν Nu N 14 Ξ ξ Xi X 15 Ο ο Omicron O 16 Π π ϖ Pi P 17 Ρ ρ Rho R 18 Σ σ ς Sigma S 19 Τ τ Tau T 20 Υ υ Ipsilon U 21 Φ φ ϕ Phi F 22 Χ χ Ji C 23 Ψ ψ Psi Y 24 Ω ω Omega W NOTACIÓN sin Seno. cos Coseno. tg Tangente. sec Secante. csc Cosecante. ctg Cotangente. vers Verso seno. arcsin θ = sin θ Arco seno de un ángulo θ . u = f ( x) sinh Seno hiperbólico. cosh Coseno hiperbólico. tgh Tangente hiperbólica. ctgh Cotangente hiperbólica. sech Secante hiperbólica. csch Cosecante hiperbólica. u, v, w Funciones de x , u = u ( x ) , v = v ( x ) . Conjunto de los números reales. = {…, −2, −1,0,1, 2,…} Conjunto de enteros. Conjunto de números racionales. c Conjunto de números irracionales. = {1, 2,3,…} Conjunto de números naturales. Conjunto de números complejos.