Fórmulas de Cálculo Diferencial e Integral (Página 1 de 3)                                                                                                                                                                                                               http://www.geocities.com/calculusjrm/                                       Jesús Rubí M.
Fórmulas de                                                             ( a + b ) ⋅ ( a − ab + b ) = a + b
                                                                                           2                       2           3      3                                                                                                         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
                                                                                                                                                                                                                                                                                                                           2               2
Cálculo Diferencial                                                                                                                                                                                   HIP
                                                                                                                                                                                                                                       CO               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 ACTUALIZADO AGO-2007                                                                                                                                                                 θ                                                       3
                                                                                                                                                                                                                                                                                                                           2               2
                                                                        ( a + b ) ⋅ ( a5 − a 4 b + a 3b 2 − a 2 b3 + ab 4 − b5 ) = a 6 − b 6
                                                                                                                                                                                                                                                                                                                            1               1
                                                                                                                                                                                                          CA                                                                                          cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β )
Jesús Rubí Miranda (jesusrubim@yahoo.com)                                                                                                                                                                                                               2                                                                   2               2
Móvil. Méx. DF. 044 55 13 78 51 94                                                                                                                                                                                                                                                                                           1               1
                                                                                       ⎛                                          ⎞                                                                                                                                                                   cos α − cos β = −2 sin (α + β ) ⋅ sin (α − β )
                                                                                           n
                                                                        ( a + b ) ⋅ ⎜ ∑ ( −1)
                                                                                                          k +1
                                                                                                                   a n − k b k −1 ⎟ = a n + b n ∀ n ∈        impar         θ      sin  cos   tg   ctg  sec csc
                                                                                                                                                                                                                                                        1

                                                                                                                                                                                                                                                                                                                             2               2
                                                                                       ⎝ k =1                                     ⎠
1. VALOR ABSOLUTO                                                                                                                                                         0        0    1    0     ∞    1   ∞
                                                                                                                                                                                                                                                                                                                               sin (α ± β )
                                                                                                                                                                                                                                                        0

                                                                                       ⎛   n
                                                                                                     ⎞                                                                                                                                                                                                tg α ± tg β =
                                                                        ( a + b ) ⋅ ⎜ ∑ ( −1)
                                                                                                          k +1                                                                    12
    ⎧a si a ≥ 0                                                                       a n − k b k −1 ⎟ = a n − b n ∀ n ∈                                     par          30            3 2 1 3     3 2 3 2
 a =⎨                                                                         ⎝ k =1                 ⎠
                                                                                                                                                                                                                                                        -1
                                                                                                                                                                                                                                                                                      arc ctg x                            cos α ⋅ cos β
                                                                                                                                                                          45     1 2 1 2                 2   2
    ⎩− a si a < 0
                                                                                                                                                                                              1    1                                                                                  arc sec x
                                                                                                                                                                                                                                                                                      arc csc x
                                                                                                                                                                                                                                                                                                                     1
                                                                                                                                                                                                                                                                                                                       ⎡sin (α − β ) + sin (α + β ) ⎤
                                                                        5. SUMAS Y PRODUCTOS                                                                              60       3 2 12      3 1 3    2 2 3                                           -2
                                                                                                                                                                                                                                                                                                      sin α ⋅ cos β =
 a = −a                                                                                                                                                                                                                                                                                                              2⎣                             ⎦
                                                                                                                                                                                                                                                          -5                      0               5


                                                                                                                                                                                             ∞          ∞
                                                                                                               n
                                                                                               + a n = ∑ ak
                                                                                                                                                                          90       1    0          0        1
a ≤ a y −a ≤ a                                                          a1 + a2 +                                                                                                                                                               8. IDENTIDADES TRIGONOMÉTRICAS                                       1
                                                                                                            k =1                                                                      ⎡ π π⎤                                                                                                          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                                                                                           y = ∠ sin x y ∈ ⎢− , ⎥
                                                                        ∑ c = nc                                                                                                      ⎣ 2 2⎦
                                                                                                                                                                                                                                                1 + ctg 2 θ = csc 2 θ
                                                                                                                                                                                                                                                                                                      cos α ⋅ cos β =
                                                                                                                                                                                                                                                                                                                                1
                                                                                                                                                                                                                                                                                                                                  ⎡cos (α − β ) + cos (α + β ) ⎤
                                                                                                                                                                      y = ∠ cos x y ∈ [ 0, π ]
                                 n                    n

                                ∏a          = ∏ ak                                                                                                                                                                                                                                                                              2⎣                             ⎦
                                                                         k =1
 ab = a b ó                            k                                  n                 n                                                                                                                                                   tg 2 θ + 1 = sec 2 θ
                                k =1              k =1
                                                                        ∑ ca         = c ∑ ak
                                                                                                                                                                      y = ∠ tg x           y∈ −
                                                                                                                                                                                                          π π                                                                                                   tg α + tg β
                                                                                                                                                                                                                                                sin ( −θ ) = − sin θ
                                                                                k
                                        n                  n             k =1              k =1
                                                                                                                                                                                                           ,                                                                                          tg α ⋅ tg β =
 a+b ≤ a + b ó                         ∑a             ≤ ∑ ak              n                         n                   n
                                                                                                                                                                                                          2 2                                                                                                  ctg α + ctg β
                                                                                                                                                                                                                                                cos ( −θ ) = cos θ
                                                  k
                                       k =1               k =1
                                                                        ∑(a     k    + bk ) = ∑ ak + ∑ bk
                                                                                                                                                                      y = ∠ ctg x = ∠ tg
                                                                                                                                                                                                     1
                                                                                                                                                                                                           y ∈ 0, π                                                                                   9. FUNCIONES HIPERBÓLICAS
                                                                                                                                                                                                                                                tg ( −θ ) = − tg θ
                                                                         k =1                      k =1                k =1
2. EXPONENTES                                                                                                                                                                                        x                                                                                                         ex − e− x
                                                                                                                                                                                                                                                                                                      sinh x =
                                                                          n
 a p ⋅ a q = a p+q                                                      ∑(a
                                                                         k =1
                                                                                k    − ak −1 ) = an − a0                                                              y = ∠ sec x = ∠ cos
                                                                                                                                                                                          1
                                                                                                                                                                                            y ∈ [ 0, π ]                                        sin (θ + 2π ) = sin θ                                               2
 ap                                                                                                                                                                                       x
    = a p−q                                                                                                                                                                                                                                                                                                    e x + e− x
                                                                          n
                                                                                                                   n                                                                      1       ⎡ π π⎤                                        cos (θ + 2π ) = cos θ                                 cosh x =
 aq
                                                                        ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤
                                                                          ⎣                ⎦     ⎣                 ⎦                                                  y = ∠ csc x = ∠ sen    y ∈ ⎢− , ⎥
                                                                                                                                                                                                  ⎣ 2 2⎦                                        tg (θ + 2π ) = tg θ
                                                                                                                                                                                                                                                                                                                    2
(a )
   p q
            =a          pq                                               k =1                                                                                                             x
                                                                                                                                                                                                                                                                                                      tgh x =
                                                                                                                                                                                                                                                                                                              sinh x e x − e − x
                                                                                                                                                                                                                                                                                                                      =
                                                                                             n
                                                                                               (a + l )                                                              Gráfica 1. Las funciones trigonométricas: sin x ,                          sin (θ + π ) = − sin θ                                        cosh x e x + e− x
(a ⋅b)          = a ⋅b                                                                                    =
            p               p    p
                                                                                             2                                                                       cos x , tg x :                                                             cos (θ + π ) = − cos θ                                            1       e x + e− x
        p                                                                n
                                                                                     1 − r n a − rl                                                                                                                                                                                                   ctgh x =        =
⎛a⎞  ap
⎜ ⎟ = p                                                                 ∑ ar k −1 = a 1 − r = 1 − r                                                                        2                                                                    tg (θ + π ) = tg θ                                             tgh x e x − e − x
⎝b⎠  b                                                                  k =1
                                                                                                                                                                                                                                                                                                                   1            2
                                                                                                                                                                                                                                                sin (θ + nπ ) = ( −1) sin θ                           sech x =          =
                                                                                                                                                                         1.5                                                                                              n

                                                                        ∑ k = 2 ( n2 + n )
                                                                         n
 a = a                                                                          1
                                                                                                                                                                                                                                                                                                               cosh x e x + e − x
  p/q           q       p
                                                                                                                                                                           1

                                                                                                                                                                                                                                                cos (θ + nπ ) = ( −1) cos θ
                                                                                                                                                                                                                                                                          n
3. LOGARITMOS                                                           k =1                                                                                                                                                                                                                                      1            2
                                                                                                      n ( n + 1)( 2n + 1)
                                                                                                                                                                         0.5
                                                                                                                                                                                                                                                                                                      csch x =         =
                                                                        ∑ k 2 = 6 ( 2n3 + 3n2 + n ) =                                                                                                                                           tg (θ + nπ ) = tg θ                                            sinh x e x − e − x
                                                                          n
 log a N = x ⇒ a x = N                                                          1
                                                                                                                                                                           0


log a MN = log a M + log a N                                            k =1                                   6                                                                                                                                                                                      sinh :     →
                                                                                                                                                                        -0.5
                                                                                                                                                                                                                                                sin ( nπ ) = 0
                                                                        ∑ k = 4 ( n + 2n + n )                                                                                                                                                                                                                   → [1, ∞
                                                                         n
      M                                                                      3  1 4        3    2                                                                                                                                                                                                     cosh :
         = log a M − log a N                                                                                                                                                                                                                    cos ( nπ ) = ( −1)
                                                                                                                                                                          -1
                                                                                                                                                                                                                                                                     n
log a                                                                   k =1
      N                                                                                                                                                                 -1.5                                                        sen x
                                                                                                                                                                                                                                                                                                      tgh :    → −1,1
                                                                                                                                                                                                                                                tg ( nπ ) = 0
                                                                                                                                                                                                                                    cos x


                                                                        ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n )
                                                                         n
log a N = r log a N
       r                                                                         1
                                                                                                                                                                                                                                                                                                                 − {0} → −∞ , −1 ∪ 1, ∞
                                                                                                                                                                                                                                    tg x
                                                                                                                                                                          -2
                                                                                                                                                                            -8   -6        -4        -2     0   2       4       6           8                                                         ctgh :
                                                                        k =1
                                                                                                                                                                                                                                                    ⎛ 2n + 1 ⎞
                                                                                                                                                                                                                                                            π ⎟ = ( −1)
          log b N ln N
                                                                                                                                                                                                                                                                                                                 → 0 ,1]
                                                                                                                                                                                                                                                                        n
log a N =         =                                                     1+ 3 + 5 +                + ( 2n − 1) = n 2                                                  Gráfica 2. Las funciones trigonométricas csc x ,                           sin ⎜                                                 sech :
           log b a ln a                                                                                                                                                                                                                             ⎝ 2       ⎠
                                                                                                                                                                     sec x , ctg x :                                                                                                                  csch :     − {0} →             − {0}
                                                                                                                                                                                                                                                    ⎛ 2n + 1 ⎞
                                                                                 n
log10 N = log N y log e N = ln N                                        n! = ∏ k                                                                                                                                                                cos ⎜       π⎟=0
4. ALGUNOS PRODUCTOS                                                            k =1                                                                                     2.5                                                                        ⎝ 2       ⎠                                       Gráfica 5. Las funciones hiperbólicas sinh x ,
 a ⋅ ( c + d ) = ac + ad                                                ⎛n⎞         n!                                                                                     2
                                                                                                                                                                                                                                                   ⎛ 2n + 1 ⎞                                         cosh x , tgh x :
                                                                        ⎜ ⎟=                 , k≤n                                                                                                                                              tg ⎜       π⎟=∞
                                                                        ⎝ k ⎠ ( n − k ) !k !
                                                                                                                                                                         1.5

( a + b ) ⋅ ( a − b ) = a 2 − b2                                                                                                                                           1
                                                                                                                                                                                                                                                   ⎝ 2      ⎠                                                      5


                                                                                                                                                                                                                                                                                                                   4
                                                                                      n
                                                                                         ⎛n⎞                                                                                                                                                                ⎛    π⎞
( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b 2                     ( x + y ) = ∑ ⎜ ⎟ x n −k y k
                                 2                                               n                                                                                       0.5
                                                                                                                                                                                                                                                sin θ = cos ⎜ θ − ⎟                                                3

                                                                                    k =0 ⎝ k ⎠                                                                                                                                                              ⎝    2⎠
                                                                                                                                                                           0


( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2
                                                                                                                                                                                                                                                                                                                   2
                                 2
                                                                                                                                                                        -0.5

                                                                                                                                                                                                                                                            ⎛    π⎞                                                1


                                                                        ( x1 + x2 +               + xk ) = ∑
                                                                                                                                     n!                                                                                                         cos θ = sin ⎜ θ + ⎟
( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd                                                                                               x1n1 ⋅ x2 2
                                                                                                           n                                                              -1
                                                                                                                                                     n
                                                                                                                                                           xknk
                                                                                                                                                                                                                                                            ⎝    2⎠                                                0
                                                                                                                              n1 ! n2 ! nk !                            -1.5


( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd
                                                                                                                                                                                                                                    csc x                                                                          -1

                                                                                                                                                                                                                                                sin (α ± β ) = sin α cos β ± cos α sin β
                                                                                                                                                                          -2                                                        sec x
                                                                        6. CONSTANTES                                                                                                                                               ctg x                                                                          -2


( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd
                                                                                                                                                                        -2.5

                                                                         π = 3.14159265359…                                                                                                                                                     cos (α ± β ) = cos α cos β ∓ sin α sin β
                                                                                                                                                                            -8   -6        -4        -2     0   2       4       6           8                                                                                                                senh x
                                                                                                                                                                                                                                                                                                                   -3                                        cosh x
                                                                                                                                                                                                                                                                                                                                                             tgh x


( a + b ) = a3 + 3a 2b + 3ab 2 + b3                                      e = 2.71828182846…
         3                                                                                                                                                                                                                                                                                                         -4
                                                                                                                                                                     Gráfica 3. Las funciones trigonométricas inversas                                        tg α ± tg β                                            -5                          0                    5

                                                                                                                                                                                                                                                tg (α ± β ) =                                         10. FUNCIONES HIPERBÓLICAS INV
( a − b ) = a 3 − 3a 2b + 3ab 2 − b3
         3                                                              7. TRIGONOMETRÍA                                                                             arcsin x , arccos x , arctg x :                                                         1 ∓ tg α tg β

( a + b + c ) = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc
               2                                                         sen θ =
                                                                                  CO
                                                                                            cscθ =
                                                                                                     1                                                                     4                                                                    sin 2θ = 2sin θ cos θ                                                     (
                                                                                                                                                                                                                                                                                                      sinh −1 x = ln x + x 2 + 1 , ∀x ∈      )
                                                                                                   sen θ
                                                                                                                                                                                                                                                                                                                           (                 )
                                                                                 HIP
                                                                                                                                                                                                                                                cos 2θ = cos θ − sin θ
                                                                                                                                                                                                                                                                 2            2
                                                                                                                                                                                                                                                                                                      cosh −1 x = ln x ± x 2 − 1 , x ≥ 1
( a − b ) ⋅ ( a 2 + ab + b 2 ) = a 3 − b3
                                                                                                                                                                           3
                                                                                  CA                 1
                                                                         cosθ =             secθ =                                                                                                                                                         2 tg θ
                                                                                 HIP               cosθ                                                                                                                                         tg 2θ =                                                             1 ⎛1+ x ⎞
( a − b ) ⋅ ( a 3 + a 2 b + ab 2 + b3 ) = a 4 − b 4                                                                                                                                                                                                      1 − tg 2 θ
                                                                                                                                                                           2

                                                                                sen θ CO            1                                                                                                                                                                                                 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


                                                                                                                                                                                                                                                                                                                        1 ⎛ x +1⎞
                                                                                                                                                                           0
                                                                                                                                                                                                                                                          2                                           ctgh −1 x =        ln ⎜   ⎟,            x >1
                    ⎛   n
                                              ⎞                                                                                                                                                                                                                                                                         2 ⎝ x −1⎠
( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n                     ∀n ∈   π radianes=180                                                                                                                                                                     1
                                                                                                                                                                                                                                                cos 2 θ = (1 + cos 2θ )
                    ⎝ k =1                    ⎠                                                                                                                           -1
                                                                                                                                                                                                                            arc sen x
                                                                                                                                                                                                                                                           2                                                         ⎛ 1 ± 1 − x2             ⎞
                                                                                                                                                                                                                                                                                                      sech −1 x = ln ⎜                        ⎟, 0 < x ≤ 1
                                                                                                                                                                                                                            arc cos x
                                                                                                                                                                                                                            arc tg x
                                                                                                                                                                                                                                                         1 − cos 2θ                                                  ⎜                        ⎟
                                                                                                                                                                          -2
                                                                                                                                                                            -3        -2        -1          0       1       2               3   tg 2 θ =                                                             ⎝     x                  ⎠
                                                                                                                                                                                                                                                         1 + cos 2θ
                                                                                                                                                                                                                                                                                                                     ⎛1              x2 + 1 ⎞
                                                                                                                                                                                                                                                                                                      csch −1 x = ln ⎜ +                    ⎟, x ≠ 0
                                                                                                                                                                                                                                                                                                                     ⎜x               x ⎟
                                                                                                                                                                                                                                                                                                                     ⎝                      ⎠
Fórmulas de Cálculo Diferencial e Integral (Página 2 de 3)                                                                                                                                             http://www.geocities.com/calculusjrm/                                                  Jesús Rubí M.
                                                                                                                                                                                                                                         ∫ tgh udu = ln cosh u
11. IDENTIDADES DE FUNCS HIP                                 d            dv      du                          18. DERIVADA DE FUNCS HIPERBÓLICAS                        22. 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 x = sech x                                           d               dw        dv      du              dx                  dx                                                                                                    ∫ ctgh udu = ln sinh u
                                                                ( uvw ) = uv + uw + vw
          2           2
                                                                                                                                                                                            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 ⎠                                                                                                   ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟
                                                                                     2
                                                                                                                  tgh u = sech 2 u
                                                                                                                                                                                u
                                                                                   v
cosh ( − x ) = cosh x                                                                                          dx                  dx                                                    ⎝          ⎠                                                           1
                                                                                                                                                                                                                                                       = ln tgh u
tgh ( − x ) = − tgh x
                                                             d n
                                                             dx
                                                                ( u ) = nu n−1 du
                                                                               dx
                                                                                                               d
                                                                                                                  ctgh u = − csch 2 u
                                                                                                                                      du
                                                                                                                                                                        ∫ ue du = e ( u − 1)
                                                                                                                                                                                u               u                                                               2
                                                                                                               dx                     dx                                                                                                 26. 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
                                                                                                                                                                                             2
                                                                                                               dx                           dx                                             u
                                                                                                                                                                         ∫ u log a udu = 4 ⋅ ( 2log a u − 1)
                                                                                                                                                                                                                                                        a       a
                                                             dF dF du                                                                                                                                                                                    1 u−a
sinh 2 x = 2sinh x cosh x                                                                                     19. DERIVADA DE FUNCS HIP INV
                                                                =
                                                                                                                                                                                                                                          ∫ u 2 − a 2 = 2a ln u + a ( u > a )
                                                                                                                                                                                                                                               du                      2   2
                                                             dx dx du                                          d                 1      du                                             u 2
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
                                                                                                                                                 ⎪                      23. INTEGRALES DE FUNCS TRIGO
                                                                                                                 cosh −1 u =           ⋅ , u >1 ⎨
           1                                                 15. DERIVADA DE FUNCS LOG & EXP                  dx                u 2 − 1 dx       ⎪− si cosh u < 0
                                                                                                                                                 ⎩
                                                                                                                                                            -1
                                                                                                                                                                        ∫ sin udu = − cos u                                              27. INTEGRALES CON RAIZ
sinh 2 x =   ( cosh 2 x − 1)                                                                                                                                                                                                                    du
                                                                                                                                                                                                                                          ∫ a 2 − u 2 = ∠ sin a
                                                                                                                                                                                                                                                                 u
           2                                                  d
                                                                 ( ln u ) =
                                                                             du dx 1 du
                                                                                    = ⋅
                                                                                                              d
                                                                                                                 tgh −1 u =
                                                                                                                               1 du
                                                                                                                                    ⋅ , u <1                            ∫ cos udu = sin u
           1                                                  dx               u       u dx                   dx            1 − u 2 dx
cosh x = ( cosh 2 x + 1)                                                                                                                                                ∫ sec udu = tg u
      2                                                                                                                                                                             2
                                                                                                                                                                                                                                                                                 u
                                                              d               log e du                        d                 1 du                                                                                                                            = −∠ cos
           2                                                     ( log u ) =       ⋅                             ctgh −1 u =         ⋅ , u >1                                                                                                                                    a
          cosh 2 x − 1                                                                                                       1 − u 2 dx                                 ∫ csc udu = − ctg u
                                                                                                                                                                                    2
                                                              dx                u dx                          dx
tgh 2 x =
          cosh 2 x + 1                                        d                log e du
                                                                 ( log a u ) = a ⋅ a > 0, a ≠ 1               d
                                                                                                                 sech −1 u =
                                                                                                                                 ∓1
                                                                                                                                        ⋅ ⎨
                                                                                                                                            ⎧−       −1
                                                                                                                                         du ⎪ si sech u > 0, u ∈ 0,1
                                                                                                                                                                        ∫ sec u tg udu = sec u                                           ∫
                                                                                                                                                                                                                                                   du
                                                                                                                                                                                                                                              u 2 ± a2
                                                                                                                                                                                                                                                                      (
                                                                                                                                                                                                                                                                = ln u + u 2 ± a 2              )
                                                              dx                 u      dx                                                           −1
tgh x =
          sinh 2 x                                                                                            dx             u 1 − u 2 dx ⎪ + si sech u < 0, u ∈ 0,1
                                                                                                                                            ⎩
        cosh 2 x + 1
                                                              d u
                                                                 ( e ) = eu ⋅ du                              d       −1             1      du                          ∫ csc u ctg udu = − csc u                                        ∫u
                                                                                                                                                                                                                                                       du 1
                                                                                                                                                                                                                                                            ln   =
                                                                                                                                                                                                                                                                    u
                                                              dx               dx                                csch u = −                ⋅ , u≠0                                                                                                 a2 ± u 2
                                                                                                                                                                                                                                                          a a + a2 ± 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                                                                                                                                                               du    1       a
                                                                                                                                                                                                                                          ∫ u u 2 − a 2 = a ∠ cos u
                                                              dx                    dx                        20. INTEGRALES DEFINIDAS, PROPIEDADES                     ∫ ctg udu = ln sin u
12. OTRAS
 ax 2 + bx + c = 0
                                                              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
                                                                                                                                                                                                                                                        = ∠ sec
                                                                                                                                                                                                                                                                 u
                                                                                                                                                                                                                                                          a      a
        ⇒ x=
                −b ± b 2 − 4ac
                                                             16. DERIVADA DE FUNCIONES TRIGO                  integración).
                                                                                                                                                                        ∫ csc udu = ln csc u − ctg u                                                       u 2        a2  u
                                                                                                              ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx                                                                     ∫ a − u du = 2 a − u + 2 ∠ sen a
                                                                                                               b                                         b       b                                                                            2     2               2
                                                              d                     du
                       2a                                        ( sin u ) = cos u                                                                                                  u 1
                                                                                                                                                                        ∫ 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
                                                                                                                                                                                    2 4                                                                    u 2        a
                                                                                                                                                                                                                                          ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a
                                                                                                                                                                                                                                              2     2               2       2   2
                                                              d                       du
exp (α ± i β ) = e    α
                          ( cos β ± i sin β )   si α , β ∈       ( cos u ) = − sin u                           a                                     a
                                                                                                                                                                                     u 1
                                                                                                                                                                        ∫ cos udu = 2 + 4 sin 2u
                                                                                                                                                                              2
                                                              dx                      dx
                                                                                                              ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
                                                                                                               b                         c                   b
13. LÍMITES                                                                                                                                                                                                                              28. MÁS INTEGRALES
                                                              d                     du
                                                                 ( tg u ) = sec2 u
                                                                                                               a                     a                       c

                                                                                                                                                                        ∫ tg udu = tg u − u
                                                                                                                                                                            2
                                                                                                                                                                                                                                                                       e au ( a sin bu − b cos bu )
                                                                                                              ∫ f ( x ) dx = − ∫ f ( x ) dx
              1                                                                                                b                                 a
lim (1 + x ) x = e = 2.71828...                               dx                    dx
                                                                                                                                                                                                                                         ∫e        sin bu du =
                                                                                                                                                                                                                                              au

                                                                                                                                                                                                                                                                                    a 2 + b2
                                                                                                                                                                        ∫ ctg udu = − ( ctg u + u )
 x →0                                                                                                          a                             b
                                                              d                        du
                                                                 ( ctg u ) = − csc2 u
                                                                                                                                                                             2


                                                                                                              ∫ f ( x ) dx = 0                                                                                                                                            e au ( a cos bu + b sin bu )
                  x                                                                                             a
      ⎛ 1⎞                                                    dx                        dx
 lim ⎜1 + ⎟ = e                                                                                                                                                                                                                          ∫e        cos bu du =
                                                                                                               a                                                                                                                              au
 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                                                                a2 + b2
       sen x                                                  dx                         dx                                                                                                                                                            1              1
                                                                                                                                                                        ∫ u cos udu = cos u + u sin u                                     ∫ sec u du = 2 sec u tg u + 2 ln sec u + tg u
                                                                                                                                     a
             =1
                                                                                                                                                                                                                                                   3
                                                                                                                ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈
 lim
 x →0                                                         d                              du
         x
                                                                 ( csc u ) = − csc u ctg u                                                                              24. INTEGRALES DE FUNCS TRIGO INV
      1 − cos x                                               dx                             dx                                                                                                                                          29. ALGUNAS SERIES
                                                                                                              ∫ f ( x ) dx ≤ ∫ g ( x ) dx
                                                                                                               b                         b
 lim            =0
                                                                                                                                                                        ∫ ∠ sin udu = u∠ sin u + 1 − u                                                                                                     f '' ( x0 )( x − x0 )
                                                                                                                                                                                                                  2                                                                                                                  2
 x →0                                                         d                       du
                                                                 ( vers u ) = sen u
                                                                                                               a                     a
           x
       ex −1                                                  dx                      dx                        ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a , b ]                                                                                        f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) +
              =1                                                                                                                                                        ∫ ∠ cos udu = u∠ cos u − 1 − u                                                                                                                2!
                                                                                                                                                                                                                   2
 lim
                                                                                                               ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b
                                                             17. DERIV DE FUNCS TRIGO INVER                     b                            b
                                                                                                                                                                                                                                                                     f ( n ) ( x0 )( x − x0 )
 x →0    x                                                                                                                                                                                                                                                                                          n

       x −1                                                                                                                                                             ∫ ∠ tg udu = u∠ tg u − ln 1 + u                                                     +    +
                                                                                                                a                        a                                                                         2
                                                              d                     1       du                                                                                                                                                                                                          : Taylor
 lim         =1                                                  ( ∠ sin u ) =            ⋅                   21. INTEGRALES                                                                                                                                                         n!
  x →1 ln x                                                   dx                  1 − u 2 dx                                                                                                                                                                                              f '' ( 0 ) x 2
                                                                                                                                                                        ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u
                                                                                                                                                                                                                        2
14. DERIVADAS                                                d                      1       du                ∫ adx =ax                                                                                                                   f ( x ) = f (0) + f ' ( 0) x +
                                                                ( ∠ cos u ) = −
                                                                                                                                                                        ∫ ∠ sec udu = u∠ sec u − ln ( u + u                          )
                                                                                          ⋅
                        f ( x + ∆x ) − f ( x )
                                                                                                                                                                                                                                                                                               2!
                                                        ∆y                                                                                                                                                                  2
                                                                                                                                                                                                                                −1
                                                                                                              ∫ af ( x ) dx = a ∫ f ( x ) dx
               df                                                                 1 − u 2 dx
 Dx f ( x ) =     = lim                        = lim         dx
                                                                                                                                                                                                                                                                     f ( n) ( 0 ) x n
               dx ∆x →0          ∆x              ∆x → 0 ∆x
                                                                                                                                                                                                                                                            +    +                         : Maclaurin
                                                             d
                                                                ( ∠ tg u ) =
                                                                                1     du                                                                                                        = u∠ sec u − ∠ cosh u
 d                                                                                  ⋅
                                                                                                              ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ±                                                                                                 n!
                                                                                                                                                                        ∫ ∠ csc udu = u∠ csc u + ln ( u +                            )
    (c) = 0                                                  dx              1 + u 2 dx
                                                                                                                                                                                                                        u2 − 1                       x 2 x3          xn
                                                                                                              ∫ udv = uv − ∫ vdu ( Integración por partes )                                                                              ex = 1 + x +   + + +           +
 dx                                                          d                     1     du
 d                                                              ( ∠ ctg u ) = −        ⋅                                                                                                                                                             2! 3!           n!
    ( cx ) = c                                               dx                 1 + u 2 dx
                                                                                                                            u n +1
                                                                                                                                                                                 = u∠ csc u + ∠ cosh u                                                3    5      7
                                                                                                                                                                                                                                                                                      x 2 n −1
                                                                                                              ∫ u du = n + 1
                                                                                                                                                                                                                                                    x    x      x
                                                                                                                                             n ≠ −1                                                                                                                 + + ( −1)
 dx                                                                                                                 n                                                                                                                                                         n −1
                                                             d                       1       du ⎧+ si u > 1                                                             25. INTEGRALES DE FUNCS HIP                                      sin x = x − +       −
                                                                ( ∠ sec u ) = ±             ⋅ ⎨                                                                                                                                                                                    ( 2n − 1)!
    ( cx n ) = ncx n−1
 d                                                                                                                                                                                                                                                  3! 5! 7!
                                                             dx                 u u 2 − 1 dx ⎩− si u < −1       du                                                      ∫ sinh udu = cosh u
 dx
                                                                                           du ⎧− si u > 1     ∫u        = ln u
                                                                                                                                                                                                                                         cos x = 1 −
                                                                                                                                                                                                                                                                x2 x4 x6
                                                                                                                                                                                                                                                                  +  −   +                    + ( −1)
                                                                                                                                                                                                                                                                                                           n −1     x 2 n− 2
 d                      du dv dw
    (u ± v ± w ± ) = ± ± ±
                                                             d
                                                                ( ∠ csc u ) = ∓
                                                                                    1
                                                                                          ⋅ ⎨                                                                           ∫ cosh udu = sinh u                                                                     2! 4! 6!                                          ( 2n − 2 ) !
 dx                     dx dx dx                             dx                 u u 2 − 1 dx ⎩+ si u < −1
                                                                                                                                                                        ∫ sech udu = tgh u
                                                                                                                                                                                        2
                                                                                                                                                                                                                                                       x2 x3 x 4              n −1 x
                                                                                                                                                                                                                                                                                       n
 d              du                                           d                     1       du                                                                                                                                            ln (1 + x ) = x −
                                                                                                                                                                                                                                                         + −      + + ( −1)
    ( cu ) = c                                                  ( ∠ vers u ) =           ⋅
                                                                                                                                                                        ∫ csch udu = − ctgh u
                                                                                                                                                                                        2                                                              2   3    4                     n
 dx             dx                                           dx                  2u − u 2 dx                                                                                                                                                                                      2 n −1
                                                                                                                                                                                                                                                     x3 x 5 x 7           n −1 x
                                                                                                                                                                                                                                         ∠ tg x = x − + −       + + ( −1)
                                                                                                                                                                        ∫ sech u tgh udu = − sech u                                                  3   5   7                 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.
30. 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

31. 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.

Formulas De Calculo

  • 1.
    Fórmulas de CálculoDiferencial e Integral (Página 1 de 3) http://www.geocities.com/calculusjrm/ Jesús Rubí M. Fórmulas de ( a + b ) ⋅ ( a − ab + b ) = a + b 2 2 3 3 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 2 2 Cálculo Diferencial HIP CO 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 ACTUALIZADO AGO-2007 θ 3 2 2 ( a + b ) ⋅ ( a5 − a 4 b + a 3b 2 − a 2 b3 + ab 4 − b5 ) = a 6 − b 6 1 1 CA cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β ) Jesús Rubí Miranda (jesusrubim@yahoo.com) 2 2 2 Móvil. Méx. DF. 044 55 13 78 51 94 1 1 ⎛ ⎞ cos α − cos β = −2 sin (α + β ) ⋅ sin (α − β ) n ( a + b ) ⋅ ⎜ ∑ ( −1) k +1 a n − k b k −1 ⎟ = a n + b n ∀ n ∈ impar θ sin cos tg ctg sec csc 1 2 2 ⎝ k =1 ⎠ 1. VALOR ABSOLUTO 0 0 1 0 ∞ 1 ∞ sin (α ± β ) 0 ⎛ n ⎞ tg α ± tg β = ( a + b ) ⋅ ⎜ ∑ ( −1) k +1 12 ⎧a si a ≥ 0 a n − k b k −1 ⎟ = a n − b n ∀ n ∈ par 30 3 2 1 3 3 2 3 2 a =⎨ ⎝ k =1 ⎠ -1 arc ctg x cos α ⋅ cos β 45 1 2 1 2 2 2 ⎩− a si a < 0 1 1 arc sec x arc csc x 1 ⎡sin (α − β ) + sin (α + β ) ⎤ 5. SUMAS Y PRODUCTOS 60 3 2 12 3 1 3 2 2 3 -2 sin α ⋅ cos β = a = −a 2⎣ ⎦ -5 0 5 ∞ ∞ n + a n = ∑ ak 90 1 0 0 1 a ≤ a y −a ≤ a a1 + a2 + 8. IDENTIDADES TRIGONOMÉTRICAS 1 k =1 ⎡ π π⎤ 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 y = ∠ sin x y ∈ ⎢− , ⎥ ∑ c = nc ⎣ 2 2⎦ 1 + ctg 2 θ = csc 2 θ cos α ⋅ cos β = 1 ⎡cos (α − β ) + cos (α + β ) ⎤ y = ∠ cos x y ∈ [ 0, π ] n n ∏a = ∏ ak 2⎣ ⎦ k =1 ab = a b ó k n n tg 2 θ + 1 = sec 2 θ k =1 k =1 ∑ ca = c ∑ ak y = ∠ tg x y∈ − π π tg α + tg β sin ( −θ ) = − sin θ k n n k =1 k =1 , tg α ⋅ tg β = a+b ≤ a + b ó ∑a ≤ ∑ ak n n n 2 2 ctg α + ctg β cos ( −θ ) = cos θ k k =1 k =1 ∑(a k + bk ) = ∑ ak + ∑ bk y = ∠ ctg x = ∠ tg 1 y ∈ 0, π 9. FUNCIONES HIPERBÓLICAS tg ( −θ ) = − tg θ k =1 k =1 k =1 2. EXPONENTES x ex − e− x sinh x = n a p ⋅ a q = a p+q ∑(a k =1 k − ak −1 ) = an − a0 y = ∠ sec x = ∠ cos 1 y ∈ [ 0, π ] sin (θ + 2π ) = sin θ 2 ap x = a p−q e x + e− x n n 1 ⎡ π π⎤ cos (θ + 2π ) = cos θ cosh x = aq ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤ ⎣ ⎦ ⎣ ⎦ y = ∠ csc x = ∠ sen y ∈ ⎢− , ⎥ ⎣ 2 2⎦ tg (θ + 2π ) = tg θ 2 (a ) p q =a pq k =1 x tgh x = sinh x e x − e − x = n (a + l ) Gráfica 1. Las funciones trigonométricas: sin x , sin (θ + π ) = − sin θ cosh x e x + e− x (a ⋅b) = a ⋅b = p p p 2 cos x , tg x : cos (θ + π ) = − cos θ 1 e x + e− x p n 1 − r n a − rl ctgh x = = ⎛a⎞ ap ⎜ ⎟ = p ∑ ar k −1 = a 1 − r = 1 − r 2 tg (θ + π ) = tg θ tgh x e x − e − x ⎝b⎠ b k =1 1 2 sin (θ + nπ ) = ( −1) sin θ sech x = = 1.5 n ∑ k = 2 ( n2 + n ) n a = a 1 cosh x e x + e − x p/q q p 1 cos (θ + nπ ) = ( −1) cos θ n 3. LOGARITMOS k =1 1 2 n ( n + 1)( 2n + 1) 0.5 csch x = = ∑ k 2 = 6 ( 2n3 + 3n2 + n ) = tg (θ + nπ ) = tg θ sinh x e x − e − x n log a N = x ⇒ a x = N 1 0 log a MN = log a M + log a N k =1 6 sinh : → -0.5 sin ( nπ ) = 0 ∑ k = 4 ( n + 2n + n ) → [1, ∞ n M 3 1 4 3 2 cosh : = log a M − log a N cos ( nπ ) = ( −1) -1 n log a k =1 N -1.5 sen x tgh : → −1,1 tg ( nπ ) = 0 cos x ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n ) n log a N = r log a N r 1 − {0} → −∞ , −1 ∪ 1, ∞ tg x -2 -8 -6 -4 -2 0 2 4 6 8 ctgh : k =1 ⎛ 2n + 1 ⎞ π ⎟ = ( −1) log b N ln N → 0 ,1] n log a N = = 1+ 3 + 5 + + ( 2n − 1) = n 2 Gráfica 2. Las funciones trigonométricas csc x , sin ⎜ sech : log b a ln a ⎝ 2 ⎠ sec x , ctg x : csch : − {0} → − {0} ⎛ 2n + 1 ⎞ n log10 N = log N y log e N = ln N n! = ∏ k cos ⎜ π⎟=0 4. ALGUNOS PRODUCTOS k =1 2.5 ⎝ 2 ⎠ Gráfica 5. Las funciones hiperbólicas sinh x , a ⋅ ( c + d ) = ac + ad ⎛n⎞ n! 2 ⎛ 2n + 1 ⎞ cosh x , tgh x : ⎜ ⎟= , k≤n tg ⎜ π⎟=∞ ⎝ k ⎠ ( n − k ) !k ! 1.5 ( a + b ) ⋅ ( a − b ) = a 2 − b2 1 ⎝ 2 ⎠ 5 4 n ⎛n⎞ ⎛ π⎞ ( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b 2 ( x + y ) = ∑ ⎜ ⎟ x n −k y k 2 n 0.5 sin θ = cos ⎜ θ − ⎟ 3 k =0 ⎝ k ⎠ ⎝ 2⎠ 0 ( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2 2 2 -0.5 ⎛ π⎞ 1 ( x1 + x2 + + xk ) = ∑ n! cos θ = sin ⎜ θ + ⎟ ( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd x1n1 ⋅ x2 2 n -1 n xknk ⎝ 2⎠ 0 n1 ! n2 ! nk ! -1.5 ( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd csc x -1 sin (α ± β ) = sin α cos β ± cos α sin β -2 sec x 6. CONSTANTES ctg x -2 ( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd -2.5 π = 3.14159265359… cos (α ± β ) = cos α cos β ∓ sin α sin β -8 -6 -4 -2 0 2 4 6 8 senh x -3 cosh x tgh x ( a + b ) = a3 + 3a 2b + 3ab 2 + b3 e = 2.71828182846… 3 -4 Gráfica 3. Las funciones trigonométricas inversas tg α ± tg β -5 0 5 tg (α ± β ) = 10. FUNCIONES HIPERBÓLICAS INV ( a − b ) = a 3 − 3a 2b + 3ab 2 − b3 3 7. TRIGONOMETRÍA arcsin x , arccos x , arctg x : 1 ∓ tg α tg β ( a + b + c ) = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc 2 sen θ = CO cscθ = 1 4 sin 2θ = 2sin θ cos θ ( sinh −1 x = ln x + x 2 + 1 , ∀x ∈ ) sen θ ( ) HIP cos 2θ = cos θ − sin θ 2 2 cosh −1 x = ln x ± x 2 − 1 , x ≥ 1 ( a − b ) ⋅ ( a 2 + ab + b 2 ) = a 3 − b3 3 CA 1 cosθ = secθ = 2 tg θ HIP cosθ tg 2θ = 1 ⎛1+ x ⎞ ( a − b ) ⋅ ( a 3 + a 2 b + ab 2 + b3 ) = a 4 − b 4 1 − tg 2 θ 2 sen θ CO 1 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 1 ⎛ x +1⎞ 0 2 ctgh −1 x = ln ⎜ ⎟, x >1 ⎛ n ⎞ 2 ⎝ x −1⎠ ( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n ∀n ∈ π radianes=180 1 cos 2 θ = (1 + cos 2θ ) ⎝ k =1 ⎠ -1 arc sen x 2 ⎛ 1 ± 1 − x2 ⎞ sech −1 x = ln ⎜ ⎟, 0 < x ≤ 1 arc cos x arc tg x 1 − cos 2θ ⎜ ⎟ -2 -3 -2 -1 0 1 2 3 tg 2 θ = ⎝ x ⎠ 1 + cos 2θ ⎛1 x2 + 1 ⎞ csch −1 x = ln ⎜ + ⎟, x ≠ 0 ⎜x x ⎟ ⎝ ⎠
  • 2.
    Fórmulas de CálculoDiferencial e Integral (Página 2 de 3) http://www.geocities.com/calculusjrm/ Jesús Rubí M. ∫ tgh udu = ln cosh u 11. IDENTIDADES DE FUNCS HIP d dv du 18. DERIVADA DE FUNCS HIPERBÓLICAS 22. 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 x = sech x d dw dv du dx dx ∫ ctgh udu = ln sinh u ( uvw ) = uv + uw + vw 2 2 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 ⎠ ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟ 2 tgh u = sech 2 u u v cosh ( − x ) = cosh x dx dx ⎝ ⎠ 1 = ln tgh u tgh ( − x ) = − tgh x d n dx ( u ) = nu n−1 du dx d ctgh u = − csch 2 u du ∫ ue du = e ( u − 1) u u 2 dx dx 26. 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 2 dx dx u ∫ u log a udu = 4 ⋅ ( 2log a u − 1) a a dF dF du 1 u−a sinh 2 x = 2sinh x cosh x 19. DERIVADA DE FUNCS HIP INV = ∫ u 2 − a 2 = 2a ln u + a ( u > a ) du 2 2 dx dx du d 1 du u 2 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 ⎪ 23. INTEGRALES DE FUNCS TRIGO cosh −1 u = ⋅ , u >1 ⎨ 1 15. DERIVADA DE FUNCS LOG & EXP dx u 2 − 1 dx ⎪− si cosh u < 0 ⎩ -1 ∫ sin udu = − cos u 27. INTEGRALES CON RAIZ sinh 2 x = ( cosh 2 x − 1) du ∫ a 2 − u 2 = ∠ sin a u 2 d ( ln u ) = du dx 1 du = ⋅ d tgh −1 u = 1 du ⋅ , u <1 ∫ cos udu = sin u 1 dx u u dx dx 1 − u 2 dx cosh x = ( cosh 2 x + 1) ∫ sec udu = tg u 2 2 u d log e du d 1 du = −∠ cos 2 ( log u ) = ⋅ ctgh −1 u = ⋅ , u >1 a cosh 2 x − 1 1 − u 2 dx ∫ csc udu = − ctg u 2 dx u dx dx tgh 2 x = cosh 2 x + 1 d log e du ( log a u ) = a ⋅ a > 0, a ≠ 1 d sech −1 u = ∓1 ⋅ ⎨ ⎧− −1 du ⎪ si sech u > 0, u ∈ 0,1 ∫ sec u tg udu = sec u ∫ du u 2 ± a2 ( = ln u + u 2 ± a 2 ) dx u dx −1 tgh x = sinh 2 x dx u 1 − u 2 dx ⎪ + si sech u < 0, u ∈ 0,1 ⎩ cosh 2 x + 1 d u ( e ) = eu ⋅ du d −1 1 du ∫ csc u ctg udu = − csc u ∫u du 1 ln = u dx dx csch u = − ⋅ , u≠0 a2 ± u 2 a a + a2 ± 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 du 1 a ∫ u u 2 − a 2 = a ∠ cos u dx dx 20. INTEGRALES DEFINIDAS, PROPIEDADES ∫ ctg udu = ln sin u 12. OTRAS ax 2 + bx + c = 0 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 = ∠ sec u a a ⇒ x= −b ± b 2 − 4ac 16. DERIVADA DE FUNCIONES TRIGO integración). ∫ csc udu = ln csc u − ctg u u 2 a2 u ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx ∫ a − u du = 2 a − u + 2 ∠ sen a b b b 2 2 2 d du 2a ( sin u ) = cos u u 1 ∫ 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 2 4 u 2 a ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a 2 2 2 2 2 d du exp (α ± i β ) = e α ( cos β ± i sin β ) si α , β ∈ ( cos u ) = − sin u a a u 1 ∫ cos udu = 2 + 4 sin 2u 2 dx dx ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx b c b 13. LÍMITES 28. MÁS INTEGRALES d du ( tg u ) = sec2 u a a c ∫ tg udu = tg u − u 2 e au ( a sin bu − b cos bu ) ∫ f ( x ) dx = − ∫ f ( x ) dx 1 b a lim (1 + x ) x = e = 2.71828... dx dx ∫e sin bu du = au a 2 + b2 ∫ ctg udu = − ( ctg u + u ) x →0 a b d du ( ctg u ) = − csc2 u 2 ∫ f ( x ) dx = 0 e au ( a cos bu + b sin bu ) x a ⎛ 1⎞ dx dx lim ⎜1 + ⎟ = e ∫e cos bu du = a au 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 a2 + b2 sen x dx dx 1 1 ∫ u cos udu = cos u + u sin u ∫ sec u du = 2 sec u tg u + 2 ln sec u + tg u a =1 3 ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈ lim x →0 d du x ( csc u ) = − csc u ctg u 24. INTEGRALES DE FUNCS TRIGO INV 1 − cos x dx dx 29. ALGUNAS SERIES ∫ f ( x ) dx ≤ ∫ g ( x ) dx b b lim =0 ∫ ∠ sin udu = u∠ sin u + 1 − u f '' ( x0 )( x − x0 ) 2 2 x →0 d du ( vers u ) = sen u a a x ex −1 dx dx ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a , b ] f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) + =1 ∫ ∠ cos udu = u∠ cos u − 1 − u 2! 2 lim ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b 17. DERIV DE FUNCS TRIGO INVER b b f ( n ) ( x0 )( x − x0 ) x →0 x n x −1 ∫ ∠ tg udu = u∠ tg u − ln 1 + u + + a a 2 d 1 du : Taylor lim =1 ( ∠ sin u ) = ⋅ 21. INTEGRALES n! x →1 ln x dx 1 − u 2 dx f '' ( 0 ) x 2 ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u 2 14. DERIVADAS d 1 du ∫ adx =ax f ( x ) = f (0) + f ' ( 0) x + ( ∠ cos u ) = − ∫ ∠ sec udu = u∠ sec u − ln ( u + u ) ⋅ f ( x + ∆x ) − f ( x ) 2! ∆y 2 −1 ∫ af ( x ) dx = a ∫ f ( x ) dx df 1 − u 2 dx Dx f ( x ) = = lim = lim dx f ( n) ( 0 ) x n dx ∆x →0 ∆x ∆x → 0 ∆x + + : Maclaurin d ( ∠ tg u ) = 1 du = u∠ sec u − ∠ cosh u d ⋅ ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ± n! ∫ ∠ csc udu = u∠ csc u + ln ( u + ) (c) = 0 dx 1 + u 2 dx u2 − 1 x 2 x3 xn ∫ udv = uv − ∫ vdu ( Integración por partes ) ex = 1 + x + + + + + dx d 1 du d ( ∠ ctg u ) = − ⋅ 2! 3! n! ( cx ) = c dx 1 + u 2 dx u n +1 = u∠ csc u + ∠ cosh u 3 5 7 x 2 n −1 ∫ u du = n + 1 x x x n ≠ −1 + + ( −1) dx n n −1 d 1 du ⎧+ si u > 1 25. INTEGRALES DE FUNCS HIP sin x = x − + − ( ∠ sec u ) = ± ⋅ ⎨ ( 2n − 1)! ( cx n ) = ncx n−1 d 3! 5! 7! dx u u 2 − 1 dx ⎩− si u < −1 du ∫ sinh udu = cosh u dx du ⎧− si u > 1 ∫u = ln u cos x = 1 − x2 x4 x6 + − + + ( −1) n −1 x 2 n− 2 d du dv dw (u ± v ± w ± ) = ± ± ± d ( ∠ csc u ) = ∓ 1 ⋅ ⎨ ∫ cosh udu = sinh u 2! 4! 6! ( 2n − 2 ) ! dx dx dx dx dx u u 2 − 1 dx ⎩+ si u < −1 ∫ sech udu = tgh u 2 x2 x3 x 4 n −1 x n d du d 1 du ln (1 + x ) = x − + − + + ( −1) ( cu ) = c ( ∠ vers u ) = ⋅ ∫ csch udu = − ctgh u 2 2 3 4 n dx dx dx 2u − u 2 dx 2 n −1 x3 x 5 x 7 n −1 x ∠ tg x = x − + − + + ( −1) ∫ sech u tgh udu = − sech u 3 5 7 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. 30. 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 31. 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.