Aporte 2 Calculo Integral
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Transcript of Aporte 2 Calculo Integral
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 1/7
5.
∫ x3 ( x 4
+3 )2 dx
u= x4+3,dx=
1
4 x3 du
∫ x3 (u )2 1
4 x3 du
∫ (u )2 1
4 du=
1
4∫ (u )2 du
14 (
u
3
3 )=u
3
12
( x4+3 )
3
12 +c
6.
∫0
13
4+√ xdx
u=√ x , dx=2udu
3∫0
11
4+u 2udu
3∫0
1
2− 8
4+u du
3
(2u−8ln
(4+u
) )
3 (2√ x−8 ln (4+√ x ) )
x →0+¿3 (2√ x−8ln (4+√ x ) )=−24ln (4 )lim¿¿
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 2/7
x →1−¿3 (2√ x−8 ln (4+√ x ) )=6−24 ln (5 )lim¿¿
6−24ln (5 )+24 ln (4 )
7.
∫
1
x2
√ 4+ x2 dx
x=2 tan (u ) , dx=2 sec2
u du
∫ 1
(2tan ( x ) )2
√ 4+(2tan ( x ) )22 sec
2u du
1
2∫
csc2
u
√ 4+4 tan2
udu
1
2∫
csc2
u
2√ sec2
udu
1
2
1
2∫
csc2
u
secu du
1
4∫
cosu
sin2
udu
v=sin u,du= 1
cosu dv
1
4∫
cosu
v2
1
cosu dv
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 3/7
1
4∫
1
v2 dv
1
4 (−1
v )1
4 ( −1
sin(arctan 12 x))8.
∫ x
2
√ x2−4dx
x=2 secu,dx=2 tan usecudu
∫ 4 sec2
u
√ 4 sec2
u−42tanusecudu
∫ 8 sec
3u
2√ tan2utanu du
4
∫
sec3
u
tanu tanu du
4∫ sec3
u du
∫ sec3
u du=sec
2u sinu
2 +
1
2∫secudu
sec2
u sinu
2 +
1
2ln ( secu+ tan u )
Volvemos a nuestra variable
u=arcsec( 12 x )
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 4/7
sec2(arcsec( 12 x))sin(arcsec( 12 x ))
2+1
2ln(sec(arcsec( 12 x))+ tan(arcsec ( 12 x)))
9.
∫ x2sin x dx
∫uv' =uv−∫ u
' v
u= x2
,u' =2 x , v
' =sin x , v=−cos x
− x2cos x−∫−2 x cos x dx
− x2cos x+2∫ x cos xdx
∫u v' =uv−∫u
' v
u= x , u' =1, v
' =cos x , v=sin x
x sin x−∫sin x dx
−∫sin x dx=−(−cos x )
− x2cos x+2 ( x sin x+cos x )+c
10.
∫ (3 x+5 )
x3− x
2− x+1
dx
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 5/7
∫ 1
2 ( x+1)−
1
2( x−1 )+
4
( x−1 )2
1
2∫
1
( x+1 ) dx−
1
2∫
1
( x−1 ) dx+4∫
1
( x−1 )2 dx
1
2 ln ( x+1 )−
1
2 ln ( x−1 )−
4
x−1+c
11.
∫0
π
4
sin3 (2 x ) cos4 (2 x ) dx
u=2 x , dx=1
2 du
1
2∫0
π
4
sin3 (u )cos4 (u ) du
1
2∫0
π
4
(1−cos2u )sin ucos
4 (u ) du
v=cosu , d u= −1
sin u dv
−1
2 ∫
0
π
4
(1−v2 )sin u v
4 1
sinu dv
−1
2 ∫
0
π
4
(1−v2 )v4
dv
−1
2 ∫
0
π
4
v4−v
6dv
8/18/2019 Aporte 2 Calculo Integral
http://slidepdf.com/reader/full/aporte-2-calculo-integral 6/7
−1
2 ( v5
5 −
v7
7 )−1
2
(
cos5
u
5 −
cos7
u
7
)−1
2 ( cos52 x
5 −
cos72 x
7 )Evaluando en 0
−1
2 ( cos50
5 −
cos70
7 )=−1
2 ( 15−1
7 )=−1
35
Evaluando en pi/4
−1
2 ( cos
5 π
2
5 −
cos7 π
2
7 )=−1
2 (0−0 )=0
0−(−1
35 )= 1
35
12.
∫e xcosh x+ ln x dx
∫e
x ( e x+e
− x )2
+ ln x dx
∫e2 x
2 +
1
2+ ln x dx
u=2 x , dx=12
du
∫e2 x
2 dx