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