Formulario Calculo1-2 (imprimir).docx
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Identidades fundamentales π πππ₯ πππ π₯ π πππ₯ππ ππ₯ = 1 β πππ π₯π πππ₯ = 1 β π‘πππ₯πππ‘π₯ = 1 β,π‘ππ = βπππ‘π₯ = βπ ππ2 π₯ + πππ 2 π₯ = 1 πππ π₯
π πππ₯
1 + π‘ππ2 π₯ = π ππ 2 π₯ β 1 + πππ‘ 2 π₯ = ππ π 2 π₯β π ππππ₯πππ ππ₯ = 1/2π ππ(π β π)π₯ + 1/2 πππ (π + π) π₯ Identidades para mΓΊltiplos y fracciones π ππ2π₯ = 2π πππ₯πππ π₯ β πππ 2π₯ = πππ 2 π₯ β sen2 π₯ β πππ 2π₯ = 1 β 2sen2 π₯ β πππ 2π₯ = 2cos2 π₯ β 1 π‘ππ2π₯ =
2π‘πππ₯ 1βtan2 π₯
β sen2 π₯ =
1βπππ 2π₯ 2
β cos2 π₯ =
1+πππ 2π₯ 2
β tan2 π₯ =
1βπππ 2π₯ 1+πππ 2π₯
-----------------------------------------------------------------------------------------------------------------------------------------------Identidades para sumas y diferencias π‘πππ’+π‘πππ£ π ππ(π’ + π£) = π πππ’πππ π£ + π πππ£πππ π’ βπ ππ(π’ β π£) = π πππ’πππ π£ β π πππ£πππ π’ β tan(π’ + π£) = πππ (π’ + π£) = πππ π’πππ π£ β π πππ£π πππ’ β πππ (π’ β π£) = πππ π’πππ π£ + π πππ£π πππ’β tan(u β v) =
1βπ‘πππ’π‘πππ£ tanuβtanv 1+tanutanv
βπππ π’πππ π£ = 1/2cos(π’ + π£) + 1/2cos(π’ β π£) β π πππ’π πππ£ = 1/2cos(π’ β π£) β 1/2cos(π’ + π£) β βπ πππ’πππ π£ = 1/2 sen(π’ + π£) + 1/2sen(π’ β π£) β πππ π’π πππ£ = 1/2sen(π’ + π£) β 1/2sen(π’ β π£) β -----------------------------------------------------------------------------------------------------------------------------------------------Integrales ππ’
1
β« ππ’ ππ’ = πππ + π β β« π π’ ππ’ = π π’ + πβ β« π πππ’ ππ’ = βπππ π’ + π β β« πππ π’ ππ’ = π πππ’ + πβ β« π’ ππ’ = ππ|π’| + π
β« π ππ 2 π’ ππ’ = π‘πππ’ + π β β« ππ π 2 π’ ππ’ = βπππ‘π’ + π ββ« π πππ’π‘πππ’ ππ’ = π πππ’ + π β β« ππ ππ’πππ‘π’ ππ’ = βππ ππ’ + π β« π‘πππ’ ππ’ = ππ|π πππ’| + π β β« πππ‘π’ ππ’ = ππ|π πππ’| + π β β« π πππ’ ππ’ = ππ|π πππ’ + π‘πππ’| + π β ππ₯
π₯
β« ππ ππ’ ππ’ = ππ|ππ ππ’ β πππ‘π’| + πβ β« πππ’ ππ’ = π’πππ’ β π’ + πβ β« π’ ππ£ = π’π£ β β« π£ππ’ββ« βπ2βπ₯2 = π ππβ1 π + πβ ππ₯
1
π₯
ππ₯
1
π₯
β« π2 +π₯2 = π π‘ππβ1 π + π ββ« π₯βπ₯2βπ2 = π π ππ β1 π + πβ -----------------------------------------------------------------------------------------------------------------------------------------------Derivadas 1 π·π₯ (π π’ ) = π π’ π·π₯ π’ β π·π₯ (ππ’ ) = ππ’ ππππ·π₯ π’ β π·π₯ (πππ’) = π·π₯ π’ββπ·π₯ (πππ π’) = βπ πππ’π·π₯ π’ ββπ·π₯ (π πππ’) = πππ π’π·π₯ π’ π’
π·π₯ (π‘πππ’) = π ππ 2 π’π·π₯ π’ βπ·π₯ (πππ‘π’) = βππ π 2 π’π·π₯ π’βπ·π₯ (π πππ’) = π πππ’π‘πππ’π·π₯ π’ βπ·π₯ (ππ ππ’) = βππ ππ’πππ‘π’π·π₯ π’ 1 β1 1 β1 (πππ‘ β1 π’) = π·π₯ (π ππβ1 π’) = π·π₯ π’βπ·π₯ (πππ β1 π’) = π·π₯ π’βπ·π₯ (π‘ππβ1 π’) = 2 π·π₯ π’βπ·π₯ 2 π·π₯ 2 2 β1βπ’ 1
βπ·π₯ (π ππ β1 π’) =
π’βπ’2 β1
1+π’
β1βπ’ β1
π·π₯ π’βπ·π₯ (ππ π β1 π’) =
π’βπ’2 β1
1+π’
π·π₯ π’βπ·π₯ (π ππβπ’) = πππ βπ·π₯ π’βπ·π₯ (πππ βπ’) = π ππβπ·π₯ π’
βπ·π₯ (π‘ππβπ’) = π ππβ2 π’π·π₯ π’βπ·π₯ (πππ‘βπ’) = βππ πβ2 π’π·π₯ π’βπ·π₯ (π ππβπ’) = βπ ππβπ’π‘ππβπ’π·π₯ π’ ------------------------------------------------------------------------------------------------------------------------------------------------Casos para integrales trigonomΓ©tricas sen & cos: β« π πππ’π π’ ππ’ & β« πππ π π’ ππ’ n es impar: se expresa en potencias: sen2 x + cos2 x = 1 β« π πππ’π π’ ππ’ & β« πππ π π’ ππ’ n es par: se expresa en potencias: sen2 π₯ =
1βπππ 2π₯
2 cos2 x
β cos2 π₯ =
=1 β« π πππ’π π’ πππ π π’ππ’ n o m es impar : se expresa en potencias: sen2 x + 1βπππ 2π₯ 1+πππ 2π₯ π π 2 β cos2 π₯ = β« π πππ’ π’ πππ π’ππ’ n es par: se expresa en potencias: sen π₯ = 2
1+πππ 2π₯ 2
2
β« π‘πππ’π π’ ππ’ n es par: π‘πππ = π‘ππ2 π‘πππβ2 = π‘πππβ2 (π ππ 2 β 1) lo mismo cot , sec ,csc , β« π πππ’π π’ππ’ n impar por partes u=secu β« π‘πππ π’ π ππ π π’ππ’ β« πππ‘ π π’ ππ π π π’ππ’ n: par positivo ,ej: β« π‘ππ5 π’ π ππ 4 π’ππ’ = β« π‘ππ5 π’ (π‘ππ2 π’ + 1)π ππ 2 π’ππ’ Dis y U β« π‘πππ π’ π ππ π π’ππ’ β« πππ‘ π π’ ππ π π π’ππ’ n: impar ,ej: β« π‘ππ5 π’ π ππ 7 π’ππ’ = β«(π ππ 2 + 1)2 π ππ 6 π’(π πππ’π‘πππ’) ππ’ Dis y mantener (π πππ’π‘πππ’)---β« π‘πππ π’ π ππ π π’ππ’; m=par y n=impar , β« π‘ππ2 π’ π ππ 3 π’ππ’ ππ: β«(π ππ 2 β 1) π ππ 3 π’ππ’ (partes) ----------------------------------------------------------------------------------------------------------------------------------------------SustituciΓ³n trigonomΓ©trica π π π βπ2 β π 2 π₯ 2 x = sen(t) , βπ2 + π 2 π₯ 2 x = tan(t) , βπ 2 π₯ 2 β π2 x = sec(t) π
π πππ = π/π
π
π
, πππ π = π/π , π‘πππ = π/π , πππ‘π = π/π, π πππ = π/π , ππ ππ = π/π
π πππ₯
1 + π‘ππ2 π₯ = π ππ 2 π₯ β 1 + πππ‘ 2 π₯ = ππ π 2 π₯β π ππππ₯πππ ππ₯ = 1/2π ππ(π β π)π₯ + 1/2 πππ (π + π) π₯ Identidades para mΓΊltiplos y fracciones π ππ2π₯ = 2π πππ₯πππ π₯ β πππ 2π₯ = πππ 2 π₯ β sen2 π₯ β πππ 2π₯ = 1 β 2sen2 π₯ β πππ 2π₯ = 2cos2 π₯ β 1 π‘ππ2π₯ =
2π‘πππ₯ 1βtan2 π₯
β sen2 π₯ =
1βπππ 2π₯ 2
β cos2 π₯ =
1+πππ 2π₯ 2
β tan2 π₯ =
1βπππ 2π₯ 1+πππ 2π₯
-----------------------------------------------------------------------------------------------------------------------------------------------Identidades para sumas y diferencias π‘πππ’+π‘πππ£ π ππ(π’ + π£) = π πππ’πππ π£ + π πππ£πππ π’ βπ ππ(π’ β π£) = π πππ’πππ π£ β π πππ£πππ π’ β tan(π’ + π£) = πππ (π’ + π£) = πππ π’πππ π£ β π πππ£π πππ’ β πππ (π’ β π£) = πππ π’πππ π£ + π πππ£π πππ’β tan(u β v) =
1βπ‘πππ’π‘πππ£ tanuβtanv 1+tanutanv
βπππ π’πππ π£ = 1/2cos(π’ + π£) + 1/2cos(π’ β π£) β π πππ’π πππ£ = 1/2cos(π’ β π£) β 1/2cos(π’ + π£) β βπ πππ’πππ π£ = 1/2 sen(π’ + π£) + 1/2sen(π’ β π£) β πππ π’π πππ£ = 1/2sen(π’ + π£) β 1/2sen(π’ β π£) β -----------------------------------------------------------------------------------------------------------------------------------------------Integrales ππ’
1
β« ππ’ ππ’ = πππ + π β β« π π’ ππ’ = π π’ + πβ β« π πππ’ ππ’ = βπππ π’ + π β β« πππ π’ ππ’ = π πππ’ + πβ β« π’ ππ’ = ππ|π’| + π
β« π ππ 2 π’ ππ’ = π‘πππ’ + π β β« ππ π 2 π’ ππ’ = βπππ‘π’ + π ββ« π πππ’π‘πππ’ ππ’ = π πππ’ + π β β« ππ ππ’πππ‘π’ ππ’ = βππ ππ’ + π β« π‘πππ’ ππ’ = ππ|π πππ’| + π β β« πππ‘π’ ππ’ = ππ|π πππ’| + π β β« π πππ’ ππ’ = ππ|π πππ’ + π‘πππ’| + π β ππ₯
π₯
β« ππ ππ’ ππ’ = ππ|ππ ππ’ β πππ‘π’| + πβ β« πππ’ ππ’ = π’πππ’ β π’ + πβ β« π’ ππ£ = π’π£ β β« π£ππ’ββ« βπ2βπ₯2 = π ππβ1 π + πβ ππ₯
1
π₯
ππ₯
1
π₯
β« π2 +π₯2 = π π‘ππβ1 π + π ββ« π₯βπ₯2βπ2 = π π ππ β1 π + πβ -----------------------------------------------------------------------------------------------------------------------------------------------Derivadas 1 π·π₯ (π π’ ) = π π’ π·π₯ π’ β π·π₯ (ππ’ ) = ππ’ ππππ·π₯ π’ β π·π₯ (πππ’) = π·π₯ π’ββπ·π₯ (πππ π’) = βπ πππ’π·π₯ π’ ββπ·π₯ (π πππ’) = πππ π’π·π₯ π’ π’
π·π₯ (π‘πππ’) = π ππ 2 π’π·π₯ π’ βπ·π₯ (πππ‘π’) = βππ π 2 π’π·π₯ π’βπ·π₯ (π πππ’) = π πππ’π‘πππ’π·π₯ π’ βπ·π₯ (ππ ππ’) = βππ ππ’πππ‘π’π·π₯ π’ 1 β1 1 β1 (πππ‘ β1 π’) = π·π₯ (π ππβ1 π’) = π·π₯ π’βπ·π₯ (πππ β1 π’) = π·π₯ π’βπ·π₯ (π‘ππβ1 π’) = 2 π·π₯ π’βπ·π₯ 2 π·π₯ 2 2 β1βπ’ 1
βπ·π₯ (π ππ β1 π’) =
π’βπ’2 β1
1+π’
β1βπ’ β1
π·π₯ π’βπ·π₯ (ππ π β1 π’) =
π’βπ’2 β1
1+π’
π·π₯ π’βπ·π₯ (π ππβπ’) = πππ βπ·π₯ π’βπ·π₯ (πππ βπ’) = π ππβπ·π₯ π’
βπ·π₯ (π‘ππβπ’) = π ππβ2 π’π·π₯ π’βπ·π₯ (πππ‘βπ’) = βππ πβ2 π’π·π₯ π’βπ·π₯ (π ππβπ’) = βπ ππβπ’π‘ππβπ’π·π₯ π’ ------------------------------------------------------------------------------------------------------------------------------------------------Casos para integrales trigonomΓ©tricas sen & cos: β« π πππ’π π’ ππ’ & β« πππ π π’ ππ’ n es impar: se expresa en potencias: sen2 x + cos2 x = 1 β« π πππ’π π’ ππ’ & β« πππ π π’ ππ’ n es par: se expresa en potencias: sen2 π₯ =
1βπππ 2π₯
2 cos2 x
β cos2 π₯ =
=1 β« π πππ’π π’ πππ π π’ππ’ n o m es impar : se expresa en potencias: sen2 x + 1βπππ 2π₯ 1+πππ 2π₯ π π 2 β cos2 π₯ = β« π πππ’ π’ πππ π’ππ’ n es par: se expresa en potencias: sen π₯ = 2
1+πππ 2π₯ 2
2
β« π‘πππ’π π’ ππ’ n es par: π‘πππ = π‘ππ2 π‘πππβ2 = π‘πππβ2 (π ππ 2 β 1) lo mismo cot , sec ,csc , β« π πππ’π π’ππ’ n impar por partes u=secu β« π‘πππ π’ π ππ π π’ππ’ β« πππ‘ π π’ ππ π π π’ππ’ n: par positivo ,ej: β« π‘ππ5 π’ π ππ 4 π’ππ’ = β« π‘ππ5 π’ (π‘ππ2 π’ + 1)π ππ 2 π’ππ’ Dis y U β« π‘πππ π’ π ππ π π’ππ’ β« πππ‘ π π’ ππ π π π’ππ’ n: impar ,ej: β« π‘ππ5 π’ π ππ 7 π’ππ’ = β«(π ππ 2 + 1)2 π ππ 6 π’(π πππ’π‘πππ’) ππ’ Dis y mantener (π πππ’π‘πππ’)---β« π‘πππ π’ π ππ π π’ππ’; m=par y n=impar , β« π‘ππ2 π’ π ππ 3 π’ππ’ ππ: β«(π ππ 2 β 1) π ππ 3 π’ππ’ (partes) ----------------------------------------------------------------------------------------------------------------------------------------------SustituciΓ³n trigonomΓ©trica π π π βπ2 β π 2 π₯ 2 x = sen(t) , βπ2 + π 2 π₯ 2 x = tan(t) , βπ 2 π₯ 2 β π2 x = sec(t) π
π πππ = π/π
π
π
, πππ π = π/π , π‘πππ = π/π , πππ‘π = π/π, π πππ = π/π , ππ ππ = π/π
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