Actividad 7 Integracion Usando Reglas De Sustitucion

|

CALCULO INTEGRAL |
Actividad 7. |
|
Integración usando reglas de sustitución |
|
|
06/02/2013 |
|

u= 2x-1

du= 2 dx

du = dx
2
=ʃ √u du
2
= 1/2 ʃ u1/2 du = ½ u 3/2
3/2
=2u3/2
6
= (2x-1)3/2
3
= √(2x-1)3 + c
3u=x2-1

du= 2x dx

du = dx
2x

= ʃu2 2x du
2x
= ʃu2du

= u3
3
= (x2-1)3 + c
3

= 5ʃ cos 5x du

= 5 1 sen 5x
5
=sen 5x + c

u= sen 3x

du= cos 3x 3 dx

= ʃ u2 cos 3x du
3cos3x
=1/3 ʃ u2 du

= 1/3 u3
3
= (sen3x)3
9

h=u4+2

dh= 4u3 du

= ʃ u3 h1/2 dh
4u3
= ¼ ʃ h1/2 dh

= ¼ h3/2
3/2
= ¼ 2h3/2
3
= 1/6 (u4+2)3/2

= 1/6 √(u4+2)3u= 9-t2

du= -2t dt

= ʃ u1/2 (-2t) du
-2t
= 2u3/2
3
= 2(9-t2)3/2
3
= 2√ (9-t2)3 + c
3

= sec 2y + c2

u= 1-x
du= -1 dx

= ʃ sec u tng u (-du)

= - sec u

= -sec (1-x) + c

1/2x + ¼ - ¾ / 2x-1
2x - 1 x2 - 1-x2 + 1/2x
1/2x - 1
-1/2x + ¼
¾
u= 2x-1

du= 2 dx

du= dx
2

= ʃ(1/2x + ¼ - ¾ ) dx
2x-1
= ½ x2 + ¼ x – ¾ ʃ 1 dx
2 u
= x2 + 1/4x – ¾ ʃ 1 ½ du
4u
= x2 + x - 3/8 ln (2x-1) + c
4 4

u= 5x

du= 5 dx

du= dx
5

= ʃ e4 du
5
=1/5 ʃ e4 du

= 1/5 e4

=1/5 e5x + c