Trabajo CáLculo Integral
Transcript of Trabajo CáLculo Integral
CÁLCULO CÁLCULO INTEGRALINTEGRAL
Sustitución trigonométricaSustitución trigonométricaRita DederléRita Dederlé
SUSTITUCIÓN SUSTITUCIÓN TRIGONOMÉTRICATRIGONOMÉTRICA
LA SUSTITUCIÓN TRIGONOMÉTRICA eS USAdA pARA ReSOLveR INTeGRALeS CUyOS INTeGRANdOS CONTeNGAN LOS RAdICALeS √A2 – U2 , √A2 + U2 , √U2 – A2
SUSTITUCION TRIGONOMÉTRICASUSTITUCION TRIGONOMÉTRICA
En integrales que contienen En integrales que contienen √a2 – u2, hacer : u= a senθ
En integrales que contienen √a2 + u2 , hacer :
u=a tg θ
En integrales que contienen √u2 – a2, hacer:
u= a sec θ
EJERCICIOSEJERCICIOS
∫∫dx⁄xdx⁄x2 2 √25 - xx22
x= 5 sen x= 5 sen θ dx= 5 cos θ dθ ∫ 5 cos θ dθ⁄(2sen2 θ) √25 -
25 sen2 θ ∫ 5 cos θ dθ⁄(25 sen2 θ)(5 cos
θ) 1 ⁄25 ∫ dθ ⁄ sen2 θ 1 ⁄25 ∫csc2 θ - 1 ⁄25 ctg ctg θ +c
∫ ∫ 3 dz__3 dz__ √ √9z9z22 –1 –13z= sec θ3z= sec θ3dz= sec θ tg θ dθ3dz= sec θ tg θ dθdz= dz= sec θ tg θ dθsec θ tg θ dθ 33∫ ∫ 3 * 1/3 sec θ tg θ dθ3 * 1/3 sec θ tg θ dθ √ √secsec22 θ –1 θ –1 ∫ ∫ sec θ tg θ dθ sec θ tg θ dθ tgθtgθ∫ ∫ sec θ dθsec θ dθLn= (sec Ln= (sec θθ + tg + tg θθ)) ____________Ln= (3z + √9zLn= (3z + √9z22 – 1 ) – 1 )
3. ∫ x√1 +x2 dx x= 1 tg dθ dx= sec2 θ ∫(tag θ sec θ) sec2 θ sec θ tag θ + c [√(1 +X2 ) x] + c
4. ∫√25 – x2 dx ∕xx=5 senӨ dӨdx= 5 cos Ө d Ө∫√25 –(25sen2 Ө) dӨ ∕ 5
sen Ө∫√25(1 – sen2 Ө)dӨ ∕ 5
sen Ө5 ∕ 5 ∫cosӨ ∕ senӨ dӨ∫tg Ө dӨLn sec Ө + c
n ∫√25 – x2 dx xx=5 sen θ dθdx= 5 cos θ d θ ____________∫√25 –(25sen 2 θ) dθ 5 sen θ∫√25(1 – sen 2 θ)dθ 5 sen θ5 /5 ∫cosθ dθ senθ∫tg θ dθLn sec θ + cLn = 5 + c x
∫ dy______ √25 + y2 y= 5 tgθdy= 5 sec2 θ dθ∫ 5 sec 2 θ dθ √ 25 + 25 tg2 θ∫ 5 sec 2 θ dθ √25 (1 + tg2 θ)∫ 5 sec 2 θ dθ 5 secθ ∫ secθ dθLn (sec θ + tg θ) + c _____________Ln= (√25 +y2) +(y) 5 5
∫ ∫ dx______dx______ x√xx√x22 – 16 – 16 x= 4 secθx= 4 secθdx= 4 sec θ tg θ dθdx= 4 sec θ tg θ dθ∫ ∫ 4 sec θ tg θ dθ_______ 4 sec θ tg θ dθ_______ 4 sec θ √ 16 sec4 sec θ √ 16 sec22 θ – 16 θ – 16 ∫ ∫ tg θ___________tg θ___________ √ √ 16 ( sec16 ( sec22 θ – 1 ) θ – 1 )∫ ∫ tg θ dθtg θ dθ 4 tg θ4 tg θ¼ ∫ dθ ¼ ∫ dθ ¼ θ+ c¼ θ+ c¼ Arcsec ¼ Arcsec x x + c+ c 4 4
∫ ∫ dx_______dx_______ 2x √ 4x2x √ 4x22 –1 –1 2x= sec θ2x= sec θdx= dx= sec θ tg θ dθ sec θ tg θ dθ 22∫ ∫ ½ secθ tg θ dθ½ secθ tg θ dθ sec θ √ secsec θ √ sec22 θ – 1 θ – 1 ∫ ∫ ½ sec θ tg θ dθ ½ sec θ tg θ dθ sec θ tg θsec θ tg θ½ ∫dθ ½ ∫dθ ½ θ + c½ θ + c½ Arcsec 2x + c½ Arcsec 2x + c
∫ dx______ √9 – 4x2 2x = 3 sen θdx= 3cos θ dθ 2∫3/2 cos θ dθ √(9 – 9 sen2 θ)∫3/2 cos θ dθ _ √9 (1 – sen2 θ)∫ 3/2 cos θ dθ 3 cos θ½ ∫dθ½ θ + c½ Arcsen 2x +c 3
∫ 3 dy____ √1 + 9y2 3y = tg θdy= sec 2 θ dθ 3 ∫ 3 * 1/3 sec2 θ dθ √ 1 + tg2 θ∫sec 2 θ dθ sec θ∫sec θ dθ Ln= (sec θ tg θ) ______Ln= (√ 1 +9y2 ) + (3y)
∫ dz_____ √ z2 - 4 z= 2 sen θdz = 2 cos θ dθ∫2 cos θ dθ___ √ 4 sen2 θ – 4∫2 cos θ dθ 2 cos θ∫dθ θ + c Arcsen z + c
2
∫dx________ 2x – x2 - 10∫ dx__________ - [ ( x – 1)2 + 9]∫ dx_______- (x – 1)2 + 9 x –1= 3 tg θdx= 3 sec2 θ dθ ∫3 sec 2 θ dθ_ (3 tg θ)2 + 9 ∫3 sec 2 θ dθ 9 tg2 θ + 9∫3 sec 2 θ dθ 9 ( tg2 θ + 1)∫3 sec 2 θ dθ 9 sec2 θ1/3 ∫dθ 1/3 θ+ c1/3 Arctg (x – 1/ 3) +c
∫ dx________ x2 √ 25 – x2 x= 5 senθ dx= 5 cos θ dθ∫5cos θ dθ_______________ (25 sen2 θ)√ 25– 25 sen2 θ∫5 cosθ dθ________ (25 sen2 θ) (5 cos θ)1/25 ∫dθ___ sen2 θ1/25 ∫csc2 θ dθ1/25 ctg θ +c- 1/25 (√25 – x 2 ) + c x
∫ dx ______ √ 16 – 9x2 3x= 4 sen θdx= 4 cos θ dθ 3∫4/3 cos θ dθ___ √16 – ( 4 sen θ)2 ∫4/3 cos θ dθ __ √ 16 – 16 sen2 θ∫4/3 cos θ dθ __√16 (1- sen2 θ)∫4/3 cos θ dθ √ 16 cos2 θ∫4/3 cos θ dθ 4 cos θ1/3∫dθ 1/3 θ + c1/3 Arcsen 3x / 4
∫ dx____ √4x2+ 12x= tgθdx= sec 2 θ dθ 2∫1/2 sec 2 θ dθ √ tg2 θ + 1½ ∫ sec 2 θ dθ secθ½ ∫secθ dθ½ Ln (sec θ + tgθ) + c½ Ln [(4x2 + 1)+ (2x)]+ c
∫ dx____ √x2 - 4x= 2 sec θ dx= 2 sec θ tg θ∫ 2 secθ tgθ dθ √ (4 sec2 θ) – 4∫ 2 secθ tgθ dθ__ √ a (sec2 θ – 1)∫ 2 secθ tgθ dθ 2 tgθ∫ secθ dθ = Ln (secθ + Tgθ) +CLn (x ∕ 2 + 2∕ √x2 + 4)+ c
∫ dx ____ (1 +4x2 )2x= tgθ2dx = sec2 θ dθdx = sec 2 θ dθ 2∫ ½ sec 2 θ dθ (1 +tg2)½ ∫ sec 2 θ dθ sec2 θ ½ ∫ dθ½ θ ½ Arctg 2x
∫ du _______ √4 –(u + 3)2 u + 3= 2 senθdu= 2 cosθ dθ∫ 2 cosθ dθ____ √(4 – 4sen2 θ)∫ 2 cosθ dθ__ √4 (1 –sen2 θ)∫ 2 cosθ dθ √(4 cos2 θ)∫ 2 cosθ dθ 2 cosθ∫ dθ θ +cArcsen ( u + 3 ∕ 2) + c