CIN_U1_A7_JURR
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Transcript of CIN_U1_A7_JURR
Materia:
Calculo Integral
Facilitador:
Fidel Esteban Flores Ocampo
Alumno:
Juan José Rodríguez Ramírez
Tema:
Actividad 7, Integración usando reglas de sustitución
Aguascalientes, Ags. Marzo 2013
INSTRUCCIONES:
Resolver las siguientes integrales usando sustitución:
1. ∫√2 x−1dx u=2x-1 du=2dx dx= 12
du
∫ √u 12
du = 12∫ √u du = 1
2∫ u
12 du = 1
2 [ 23U32 ] =
13
(2 x−1)32 +c = √2x−1dx
2. ∫ (x2−1 )2(2x )dx → u=(x2-1) du=2x
∫u2du=u3
3=13(x2−1)3+c = (x2-1)2(2x)dx
3. ∫5cos5 x dx=5∫cos (5 x )dx → u=5x du=5dx
∫cos (u )du=sen (u )=sen (5 x )+c =5cos5xdx
4. ∫ sen23 xcosxdx → u=3x du=3dx dx=du/3
∫ sen2(u)cos (u) du3 =13∫ sen
2 (u ) cos (u )du→ p=sen (u )dp=cos (u )du
13∫ p2dp 1
3∫ p2+1
2+1=13∫ p3
3=∫ p3
9=∫ sen
3u9
= sen33x9
+c ¿ sen23xcosxdx
5. ∫u3√u4+2du → p=u4+2 dp=4u3du du=dp
4u3
∫u3√ p dp4u3
=14∫ u3√ p dp
u3=14∫√ pdp=1
4∫ p
12=
14∗2 p
32
3=2 p
32
12
¿ 16p32=16(u4+2)
32+c = u3√u4+2du
6. ∫√9−t 2 (−2 t )dt=∫−2 t√9−t2dt=−2∫ t√9−t2dt → u=9-t2 du=-2tdt
∫ √udu=∫ u12 du=
2u32
3=23(9−t2)
32+c =∫√9−t 2 (−2 t )dt
7. ∫ sec 2 y tg2 y dy → u=2y du=2dy dy=du2
∫ sec (u ) tg (u ) du2
=12∫ sec (u ) tg (u )=1
2• sec (u )= sec 2 y
2 + c = sec2y tg2y dy
8. ∫ sec (1−x )tg (1−x )dx → u=1-x du=-1dx dx=du−1
∫ sec (u ) tg (u ) du−1
=−1∫ sec (u )tg (u )du=−sec (u )=−sec(1−x)+c
=sec(1-x)tg(1-x)dx
9. ∫ e5 xdx→u=5 xdu=5dx dx=du5∫ eu du
5=15∫ eudu=1
5• eu=1
5• e5x= e
5 x
5=e5 xdx