Trabajo final calculo
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Transcript of Trabajo final calculo
CΓ‘lculo ll
Contenido del curso
Martin Eduardo Gonzalez Miranda
Matricula: 131430
Profesor: Carlos LΓ³pez Ruvalcaba
2
Integrales de Monomios Algebraicos
2. β« βπ₯3 ππ₯ = β β« π₯3 ππ₯ = βπ₯4
4+ π
4. β« 5π₯ππ₯ = 5π₯2
2+ π
6. β« 7π₯2ππ₯ = 7π₯3
3+ π
8. β« β5π₯4ππ₯ = β5π₯5
5= βπ₯5 + π
10. β«3π2
2=
3
2β« π₯2ππ₯ =
3
2β
π₯3
3=
π₯3
2+ π
12. β«β ππ₯ ππ₯ = βππ₯2
2+ π
14. β«4ππ₯3
πππ₯ =
4π
πβ« π₯3ππ₯ =
4π
πβ
π₯4
4=
ππ₯4
π+ π
16. β« π₯β2ππ₯ = π₯β1
β1+ π =
β1
π₯+ π
18. β« β4π₯β2ππ₯ = β4 β« π₯β2ππ₯ = β4 β1
π₯+ π =
β4
π₯+ π
3
20. β«β4π₯β3
3ππ₯ =
β4
3β« π₯β3ππ₯ =
β4
3β
1
π₯2=
β4
3π₯2+ π
22. β« 2π₯3
2 ππ₯ = 2 β« π₯3
2 ππ₯ =4π₯
52
5+ π
24. β«1
2π‘
1
2ππ‘ =π‘
32
3+ π
26. β« 3 βπ₯ππ₯ = 3 β« π₯1
2 ππ₯ = 2π₯3
2 + π
28. β« ββπ₯23
2ππ₯ = β« β
π₯23
2ππ₯ = β
3π₯52
10
30. β«ππ₯
π₯2= β« π₯β2ππ₯ =
π₯β1
β1+ π = β
1
π₯+ π
32. β«2ππ₯
π₯2= β« 2π₯β2 = 2 β« π₯β2 = β
2
π₯+ π
34. β«3πππ‘
π‘4= β« 3ππ‘β4ππ‘ = 3π β« π‘β4ππ‘ = β
9π
π‘3+ π
36. β«ππ’
π’12
= β« π’β1
2ππ’ = 2π’1
2 + π
38. 2 β«3πππ¦
βπ¦= 2(3π) β« π¦β
1
2ππ¦ = 12ππ¦1
2 + π
4
40. β« βππ’
3βπ’= β
1
3β« π’β
1
2ππ’ =2
3π’
1
2 + π
Integrales que conducen a la funciΓ³n logaritmo natural
1. β«2
π₯ππ₯ = 2 β«
ππ₯
π₯= 2 ln|π₯| + π = ln|π₯2| + π
2. β« βππ₯
π₯= β ln|π₯| + π = ln|π₯β1| + π
3. β«2ππ₯
3π₯=
2
3β«
ππ₯
π₯ =
2
3ln|π₯| + π
4. 3 β«ππ₯
5π₯=
3
5β«
ππ₯
π₯=
3
5ln|π₯| + π
5. β«πππ₯
π₯= π ln|π₯| + π = ln|π₯π| + π
6. β2
3β«
6ππ₯
π₯= β4 β«
ππ₯
π₯= β4 ln|π₯| + π
7. β«π2ππ₯
π₯= π2 β«
ππ₯
π₯= π2 ln|π₯| + π
8. β«4ππ
π= 4 β«
ππ
π= 4 ln|π| + π
5
Emplear diferenciales y la grafica de f para aproximar a) f (1.9) y b)
f (2.04).
21.-
23.-
04.0
)04.0(1
04.104.01
)2()04.2(
)1.0(
)1.0(1
9.01.01
)2()9.1(
dy
dy
dyff
dy
dy
dyff
112
02
)0,1(),1,2(
)('
)04.2(
)9.1(
m
ndxdxxfdy
f
f
2
1
02
21
)1,2()2,0(
)04.2(
)9.1(
m
mdxdy
y
f
f
98.002.01
02.0)04.0(2
1
)2()04.2(
05.0
)1.0(2
1
05.15.01)9.1(
)2()9.1(
dy
dyff
dy
dy
f
dyff
6
27.-Area. Se encuentra que la mediciΓ³n del lado de un cuadrado es
igual a 12 pulgadas, con un posible error de 64
1 de pulgada. Usar
diferenciales para aproximar el posible error propagado en el calculo
del area del cuadrado.
29.- Area. Se mide el radio del extremo de un tronco y se encuentra que
es igual a 14 pulgadas, con un posible error de de pulgada. Utilizar
diferenciales para aproximar el posible error propagado en el calculo
del area del extremo del tronco.
errorindv
inindxxdv
indx
xv
3
22
3
75.6
))64
1)(12((33
64
1
errordv
ininxdxdv
indx
375.0
)64
1)(12(22
64
1
errorinininxdxda
indx
xa
2
2
99.21)4
1)(14(22
41
7
31.- Area. La mediciΓ³n del lado de un cuadrado produce un valor igual
a 15 cm, con un posible error de 0.05 cm.
a) Aproximar el error porcentual en el calculo del area del
cuadrado.
b) Estimar el mΓ‘ximo error porcentual permisible en la mediciΓ³n
del lado si el error en el calculo del area no fue mayor de 2.5%
A)
%66.0100.5.2
5.1_
_15)05.0)(15(22
05.0
2
porcentualError
areaerrorininxdxdx
indx
xa
B)
MΓ‘ximo error porcentual de lado= 1.25%
%25.1187.0
%10015
1875.030
625.5_
625.5)_)(15(2
625.5)100
%5.2(25.2
100.25.2
%5.2
ladoerror
ladoerror
error
error
8
Integral de la potencia de una suma
2) β« (7x2 β 1)3/2 x dx = 1/2 β« (7x2 β 1)3/2 x dx= (1/2)(1/14) β«(7x2 β
1)3 x dx=1/28 * (7x2 β 1)4/4 = 1/112 *(7x2 β 1)4 + c
4) β« x (2+x2) dx = Β½ * (2 + x2)2 /2 = ΒΌ (2 + x2)2 + c
6) β«(x3 + 2)3 x2 dx = 1/3 * β«(x3 + 2)3 x2 dx = 1/3 * (x3 + 2)4 /4 = 1/12
* (x3 + 2)4 +c
8) β«- (4-x)3 2 dx= 2 β« -(4-x)3 dx = 2 * (4-x)4/4 = Β½ (4-x)4 + c
10) β« u βπ β πππ du = β«u * (3- 2u2)1/2 du = -1/4β« u * (3- 2u2)1/2 du =
-1/4 * (3-2u2)3/2 / 3/2 = -1/6 β(3 β 2π’2)3 + c
12) β«3x dx/ (x2 + 3)2 = β« 3x dx * (x2 + 3)-2 = 3/2 β« x dx * (x2 + 3)-2 =
3/2 * (x2 + 3)-1/-1 =-3/2 * 1/(x2 + 3) + c
14) β« 2x2 dx / βπ + πππ = β« 2x2 dx * (a + bx3)-1/2 = 2/3b * (a + bx3)1/2/
Β½ = 4/3b * βπ + ππ₯3 + c
16) β« dv / βπ βπ
π = β« dv * (1-v/2)-1/2 = -2 β« dv * (1-v/2)-1/2 = -2 * (1-
v/2)1/2/ Β½ = -4 * β1 βπ£
2 + c
9
18) β«x2 + 4x β 10 dx / (x + 2)2 = β« x2 +4x -10 dx * (x+2)-2 = x +
14/(x+2) + c
20) β« βππ β ππ dx = β« (x4-x2)1/2 dx = β(π₯2 β 1)3 / 3 + c
Casos especiales
2) β« (4x2-1) 5x dx = 5/8 (4x2-1)2 / 2 = 5/16 (4x2-1)2 + c
10
Integrales de las funciones exponenciales
2) β« 8x/2 * Β½ dx = (8x/2 / ln8) + c
4) β« -3a5x * 5 dx = (-3a5x / lna) + c
6) β« bax^2 x dx = Β½ *(bax^2 / lna) +c
8) β« 10x^2 + 1 x dx = Β½ * (10x^2 + 1 / ln10) + c
10) β« dx / 74x = -ΒΌ * 7-4x / ln7 = (-1 / (4*74x * ln7)) + c
12) β«5e2t dt = Β½ * 5e2t / lne = (5e2t / 2) + c
14) β« 5eay dy = 1/a * 5eay / lne = (5eay / a) + c
16) β«πβπ / βπ dx = β«πβπ₯ * x-1/2 dx = 2 πβπ₯ + c
18) β«πβπ dx = β« ex^1/2 dx = 2 * βππ₯ + c
20) β« (πβπ *πβπ) dx / βπ = (2 *( 2π)βπ₯ ) / (ln2 + lne) + c
22) β« (e2x + 3)2 dx = β«e4x^2 + 6e2x + 9 dx = ΒΌ e4x + 3e2x + 9x + c
11
24) β« (e(x/2) + 4) dx / ex = β« e(x/2) dx / ex + β« 4 dx / ex = (-2 / βππ₯ ) β (4
/βππ₯) + c
12
Integrales en que intervienen la tangente, cotangente, secante y
cosecante
2) β«tg x3 x2dx = 1/3 ln |sec x3| +c
4) β«3 ctg 2x dx = 3/2 ln |sen 2x| + c
6) β« ctg βπ dx / βπ = 2 ln |sen βπ₯| + c
8) β«sec (x2 /3) x dx = 3/2 ln |sec(x2/3) + tg (x2/3)| + c
10) β« ax sec ax dx = 1/ lna * ln |sec(ax) + tg(ax)| +c
12) β«-2 csc (3-2x) dx = ln |csc(3-2x) - ctg(3-2x)| + c
14) β«sec (x/2) βtg (x/2) dx =β«sec (x/2) dx - β«tg(x/2) dx =
2 ln |sec(x/2) + tg(x/2)| - 2 ln |sec(x/2)| + c
2) β« sen 2x dx / (3 + cos 2x) = -1/2β« sen 2x dx / (3 + cos 2x) =
-1/2 ln |3 + cos 2x|+ c
4) β« csc2 u du / (3-ctg u) = ln |3-ctg u| + c
2) β« b dt / ctg(a βbt) = -β« tg(a- bt) b dt = -ln |sec(a β bt)| + c
13
4) β« a dx / (βπ πππ βπ) = 2a β« βπ₯ csc βπ₯ a dx = 2a ln |csc βπ₯ β ctg βπ₯|
+ c
6) β«xex^2 dx / ctg ex^2 = Β½ β« tg ex^2 xex^2 dx = Β½ ln |sec ex^2| + c
14
Integrales que conducen a las funciones trigonomΓ©tricas
2) β« cos (x/2) dx = 2β«cos(x/2) dx = 2 sen(x/2) + c
4) β« cos (1- x2) x dx = -1/2β«cos (1-x2) x dx = -1/2 sen (1-x2) +c
6β« β«2/3 sen (a βx/2) dx = (2/3)*(-2) β«sen (a βx/2) dx =4/3 cos (a-
x/2) +c
8) β« csc2 (1- βπ) dx / βπ = 2β« csc2 (1- βπ₯) dx / βπ₯ = 2 ctg (1 - βπ₯) + c
10) β« sec e-x tg e-x e-x dx = -Sec e-x + c
Caso especial
2) β«2 dx / 1 β cos 2x =2 β« (dx/ (1-cos 2x)) * ((1+cos 2x) /(1+cos2x))
dx = 2 (1/2) β«(1 + cos 2x)/(1-cosx)2 = β« (1+ cos 2x)/ sen22x =
β«csc22x + β« ctg 2x * csc 2x = -ctg 2x β csc 2x + c
4) β« 5 dx / (1 β sen 2x) = 5 β« dx / (1- sen2x) = 5(1/2) (β«
1+sen2x/(cos22x) ) =5/2 β« sec22x dx + 5/2 β«tg2x * sec 2x dx = 5/2 tg
2x + 5/2 sec2x + c
15
16
Integrales de las formas β«π π
βππβππ, β«
π π
βππ+ππ, β«
π π
πβππβππ
54
2
2
42x
dx
427 xa
xdxc
axarcsen
x
72
1 2
2)3(4
3
x
dx
22232
334
3x
dx
x
dxc
xarcsen
2
3
2
13
Casos Especiales:
Caso 1.-
222 12
3423
3x
dx
xx
dx c
xarcsen
2
13
54
42x
dx
221 ua
bdu
427 xa
xdx
223 v
dv
19 2yy
dy
x
x
e
dxe21
6.-
4.-
2.-
8.-
10.-
12.-
c
yarc
yy
dy
1
3sec
1
1
133
1
3
22 cyarc 3sec
23 2xx
dx2.-
222
2
3
4
1
4
1
2
324
9
4
93 x
dx
x
dx
xx
dx
cx
arcsen
x
dx
21
23
2
3
2
122 cxarcsen 32
223
3
xx
dx4.-
245
3
tt
dx6.-
cx
arctg 5
2
5
2
221 au
dub c
auarctg
a
b
1cauarctg
a
b)(
v
dv
23
2
2
1c
varcsen
3
2
2
1
carcsenex
14.-
22223
38845
345
3tt
dx
tt
dx
tt
dxc
tarcsen
3
23
17
Caso 2.-
Caso 3.-
8.-
10.-
12.-
2352 xx
dx
52xx
dx
544 2 xx
dx
dx
x
x29
23
dx
x
x2161
35
dx
x
x
254
22
2.-
4.-
6.-
4
534
95412
954121282
2
21
2
xx
dx
xx
dxdxxxx
2225623
1
11513
1
11513 x
dx
xx
dx
xx
dx
cx
arcsen
7
56
3
1
222
215112 x
dx
xx
dxc
xarctg
2
1
2
1
212
4
4
1
5224422
x
dx
xx
dxc
xarctg
2
12
4
1
2
21
2
22 9392
99
32
x
dxdxxx
x
dx
x
xdx
cx
arcsenx 3
293 2
2
21
2
22 1615161
4
3
161161
53
x
dxdxxx
x
dx
x
xdx
cxarcsenx
44
3161
16
5 2
cx
arctgx 5
2
5
1254ln
8
1 2
2548
1
525
1
254254 22222 x
xdx
x
dx
x
xdx
x
dx
18
723
32
51212129
32
5129
32222
x
dxx
xx
dxx
xx
dxx
dx
xx
x
5412
382
2
2
22
2
31
2
9
12
32
9
4
5
4
9
4
93
2
9
x
dx
x
dx
xx
dx
cxarcsen
xx
2
3
2
9
21
5412 21
2
2.-
dx
xx
x
5129
322
dx
xx
x23
54
4.-
6.-
ca
varctg
aav
aav
dv
av
vdv
av
dvv
3)ln(
13
3 22
222222
cxarctgxx 235129ln9
1 2
ca
varcsenva
va
dvdvvva
va
dv
va
vdv
va
dvv
x
dxx
xx
dxx
4
422
14
4
323
54
333
54
22
22
21
22
222222
22
cx
arcsenxx
3
3234 2
19
Integrales de las formas β«π π
ππβππ, β«
π π
ππβππ
2) Κ x dx / 4x4 β 1 = 1/2 * Κ x dx / 4x4 β 1=resultado 1/4β3 ln |(π₯2 β
β3)/(π₯2 + β3)| + c
4) Κ2x dx / (25-36x2) = 1/6 * Κ2x dx / (25-36x2) = (1/6)*(1/10)
ln|(5+6x2)/(5-6x2) = resultado
1/60 * ln|5+6x2 / 5-6x2 | +c
6) Κdx/3-2x2 = 1/β2 * Κdx/3-2x2 =resultado β6/ 12 ln |(β3 - β2 π₯)/
(β3 + β2 π₯)| + c
8)
2 2 2 2
1 1 6ln
2 5 (x 2 1) 5 1 ( 1) ( 6) 2(
6 1 6ln
12
6
1 66) 1 6
1
dx dx dx x
x x x x x
a
v x
dv d
xC
x
x
10) Κ du / (9-6u-3u2) = -Κ du/(3u2 + 6u-9) = Κdu/-(3u2-6u+9) -9 +9 =-
Κdu/(3u-3)*(u+3) =
resultado= 1/12 ln| (3+u)/(1-u)| + c
12 Κ(2-3z) dz/ 9-16z2 = ΒΌ *Κ2 dz/ (9-16z2) β ΒΌ *Κ3z dz /(9-16z2) =
resultado =
1/12 ln| (3+4z)/(3-4z)| + c
20
14)
2 2 2 2 2 2 2 2 2
2
2 2 2
2
2 2 2 2
( 3) ( 3) (x 3) 3
4 5 (x 4 4) 5 4 ( 2) 9
2
2
( 2) 1int
1#1 ln 4 5
4 5 2
4 5
2 4
1int#2 3 3 (3)(
2
2
2
( 2) (3
5
) 2(3
x x dx vdv dvdx dx dv
x x x x v a v a v a
v x
x v
vdv x dxx x
v a x x
v x x
dv x
dv dx
v a x
v
2
2 3 1 1) ln ln
1 1 1ln 4 5 ln
2 2 5
) 2 3 2 5
3
2
x xC
x x
a
v x
x
dv dx
resu xl o Ct xdx
a
21
Integrales de la forma β«π π
βππ+ππ π ππππ β«
π π
βππβππ
2)β«ππ₯
βπ₯2+2π₯+5= β«
ππ₯
β(π₯+1)2+4= ππ|π₯ + 1 + β(π₯ + 1)2 + 22| + π =
ππ|π₯ + 1 + β(π₯2 + 2π₯ + 5)| + π
V=x+1 a=2 ππ£ = ππ₯
4) β«(2yβ1)dy
β2y2+4y+10= β«
(2yβ1)dy
β2(y2+2y+5)=
1
β2β«
(2yβ1)dy
β(y2+2y+1+4)=
1
β2β«
(2yβ1)dy
β(y+1)2+22)=
1
β2β«
(2yβ1)dy
β(y+1)2+22)=
1
β2β«
((tanΞΈβ1)β1)2secΞΈ2dΞΈ
2secΞΈ=
1
β2β«(4tanΞΈ β 3)secΞΈdΞΈ =
4
β2β«(tanΞΈ)secΞΈdΞΈ β
3
β2β« secΞΈdΞΈ =
4
β2secΞΈ β
3
β2ln|secΞΈ + tanΞΈ| + c =
2β2β(y2+2y+5)
2β
3β2
2ln |
β(y2+2y+5)
2+
y+1
2| + c
π πππ =π¦+1
β(π¦+1)2+22 πππ π =
2
β(π¦+1)2+22 π‘πππ =
π¦+1
2 β(π¦ + 1)2 + 22 = 2π πππ 2π πππ2ππ = ππ¦
6) β«(2π₯+1)ππ₯
β3π₯2β5= β«
(2π₯+1)ππ₯
ββ3π₯2ββ52
= β«2π₯ππ₯
ββ3π₯2ββ52
+ β«ππ₯
ββ3π₯2ββ52
=1
3β3π₯2 β 5 +
1
β3ππ|β3π₯ + β3π₯2 β 5| + π
22
Integrales de la formaβ« βππ + ππ π π Γ β« βππ Β± πππ π
2) β« βπ + πππ π π = π₯
2β
5
3β π₯2 +
5
6ππππ ππ [
π₯
β5
3
] + π
De la forma β« βππ β πππ π ; a= β5
3 , v=x, dv=dx
4) β« βπ β ππ β ππ π π = (π₯+1)
2β4 β (π₯ + 1)2 + 2ππππ ππ [
(π₯+1)
2] + π
De la forma β« βππ Β± πππ π ; a= 2, v= (x+1), dv=dx
6) β« βπππ β ππ + ππ π π = (π₯β
1
2)
2β(π₯ β
1
2)
2+
9
4+
9
8ln |(π₯ β
1
2) +
β(π₯ +1
2)
2+
9
4| + π
De la forma β« βππ Β± πππ π ; a=3
2, v=(π₯ β
1
2), dv=dx
23
Integral de las potencias del seno y/o coseno.
Primer caso.
2) β«1
2π ππ34π₯ πππ 4π₯ ππ₯ =
1
2β« π ππ34π₯ πππ 4π₯ ππ₯ =
1
2
1
4β« π ππ34π₯ cos 4π₯ 4ππ₯=
1
32π ππ44π₯ + π ππ£ = πππ 4π₯ 4ππ₯
4)β« π ππ5 5π₯
3 πππ
5π₯
34ππ₯ = 4 β« π ππ5 5π₯
3 πππ
5π₯
3 ππ₯ =
43
5β« π ππ5 5π₯
3 πππ
5π₯
3 5
3ππ₯ = (
12
5) (
1
6) π ππ6 5π₯
3 =
2
5π ππ6 5π₯
3+ π
ππ£ = πππ 5π₯
3 5
3ππ₯
6)β«1
3πππ 2 π₯
2 π ππ
π₯
2 ππ₯ =
1
3β« πππ 2 π₯
2 π ππ
π₯
2 ππ₯ =
β21
3β« πππ 2 π₯
2 (βπ ππ
π₯
2)
1
2 ππ₯ = (β
2
3) (
1
3) πππ 3 π₯
2 =
β2
9πππ 3 π₯
2+ π ππ£ = βπ ππ
π₯
2β
1
2
8)β« πππ (2 β π₯) π ππ(2 β π₯) ππ₯ = 1
2πππ 2(2 β π₯) + π
ππ£ = βπ ππ(2 β π₯) β 1ππ₯
= π ππ(2 β π₯) ππ₯
10)β«(β2 tg 3π₯ + π ππ25π₯ β π ππ2π₯πππ 2π₯)ππ₯ = β2 β« π‘π3π₯ +
β« π ππ25π₯ β β« π ππ2π₯πππ 2π₯ =
β2
3β« π‘π3π₯ β 3ππ₯ +
1
5β« π ππ25π₯ β 5ππ₯ β
1
2β« π ππ2π₯πππ 2π₯ β 2ππ₯ =
2
3ln|πππ 3π₯| +
1
5π‘π5π₯ β
1
4π ππ22π₯ + π
ππ£ = 3ππ₯ ππ£ = 5ππ₯ ππ£ = πππ 2π₯ β 2ππ₯
Segundo caso
2)β« π ππ3 π₯
2ππ₯ = β« π ππ2 π₯
2π ππ
π₯
2ππ₯ = β« (1 β πππ 2 π₯
2) π ππ
π₯
2ππ₯
=β« π πππ₯
2ππ₯ β β« πππ 2 π₯
2π ππ
π₯
2ππ₯ = 2 β« π ππ
π₯
2β
1
2ππ₯ β
2 β« πππ 2 π₯
2(βπ ππ
π₯
2)
1
2ππ₯ = β2πππ
π₯
2+
2
3πππ 3 π₯
2+ π
ππ£ =1
2ππ₯ ππ£ = βπ ππ
π₯
2β
1
2ππ₯
24
4)β« πππ 35π₯ ππ₯ = β« πππ 25π₯ πππ 5π₯ ππ₯ = β«(1 β π ππ25π₯)πππ 5π₯ ππ₯ =
β« πππ 5π₯ ππ₯ β β« π ππ25π₯πππ 5π₯ ππ₯ = 1
5β« πππ 5π₯ β 5ππ₯ β
1
5β« π ππ25π₯ πππ 5π₯ β
5ππ₯ = 1
5π ππ5π₯ β
1
15π ππ35π₯ + π
ππ£ = 5ππ₯ ππ£ = πππ 5π₯ β 5ππ₯
6)β« π ππ2π₯πππ 3π₯ ππ₯ = β« π ππ2π₯ (1 β π ππ2π₯)ππ₯ = β« π ππ2π₯ ππ₯ β
β« π ππ4π₯ ππ₯ = 1
3π ππ3π₯ β
1
5π ππ5π₯ + π
8)β« π ππ33π₯ πππ 53π₯ ππ₯ = β« π ππ23π₯ π ππ3π₯πππ 53π₯ ππ₯ = β«(1 β
πππ 23π₯) πππ 53π₯ π ππ3π₯ ππ₯ = 1
3β« πππ 53π₯ (βπ ππ3π₯)3 ππ₯ β
1
3β« πππ 73π₯ (βπ ππ3π₯)3 ππ₯ = β
1
18πππ 63π₯ +
1
24πππ 83π₯ + π
ππ£ = βπ ππ3π₯ β 3ππ₯ ππ£ = βπ ππ3π₯ β 3ππ₯
Tercer caso
2)β« πππ 2π₯ ππ₯ = β« 1
2+
1
2πππ 2π₯ ππ₯ =
1
2β« ππ₯ +
1
2
1
2β« πππ 2π₯ 2ππ₯ =
1
2π₯ +
1
4π ππ2π₯ + π
ππ£ = 2π₯ β 2ππ₯
4)β« π ππ4π₯ ππ₯ = β« (1
2β
1
2πππ 2π₯)
2ππ₯=β« (
1
4β 2
1
2β
1
2πππ 2π₯ +
1
4πππ 22π₯) ππ₯=
1
4β« ππ₯ β
1
2β« πππ 2π₯ ππ₯ +
1
4β« πππ 22π₯ ππ₯ =
1
4β« ππ₯ β (
1
2) (
1
2) β« πππ 2π₯ 2ππ₯ +
β« (1
8+
1
8πππ 4π₯) ππ₯ =
1
4β« ππ₯ β (
1
2) (
1
2) β« πππ 2π₯ 2ππ₯ +
1
8β« ππ₯ +
(1
8)
1
4β« πππ 4π₯ 4ππ₯ =
3
8π₯ β
1
4π ππ2π₯ +
1
32π ππ4π₯ + π
ππ£ = 2π₯ β 2ππ₯ ππ£ = 4π₯ β 4ππ₯
Cuarto caso
25
2)β« π ππ3π₯ πππ 3π₯ ππ₯ = β« (1
2π ππ2π₯)
3ππ₯ = β«
1
2π ππ2π₯ (
1
4π ππ22π₯) ππ₯
=1
4
1
2β« π ππ2π₯ (1 β πππ 2π₯)ππ₯ =
1
8
1
2β« π ππ2π₯ 2ππ₯ β
1
8
1
2β« πππ 2π₯ (βπ ππ2π₯) 2ππ₯ = β
1
16πππ 2π₯ +
1
32πππ 22π₯ + π
ππ£ = 2π₯ β 2ππ₯ ππ£ = (βπ ππ2π₯)2ππ₯
Quinto caso
2)β« π ππ3π₯ πππ π₯
2ππ₯ =β«
1
2[π ππ (3π₯ β
π₯
2) + π ππ (3π₯ +
π₯
2)] ππ₯ =
1
2β« π ππ
5π₯
2ππ₯ +
1
2β« π ππ
7π₯
2ππ₯ =
1
2
2
5β« π ππ
5π₯
2
5
2ππ₯ +
1
2
2
7β« π ππ
7π₯
2
7
2ππ₯ =
1
5πππ
5π₯
2β
1
7πππ
7π₯
2+ π
ππ£ = 5π₯
2
5
2ππ₯ ππ£ =
7π₯
2
7
2ππ₯
4)β« πππ π₯ πππ 4π₯ ππ₯ = β«1
2[πππ (π₯ β 4π₯) + πππ (π₯ + 4π₯)] ππ₯ =
1
2β« πππ (β3π₯) ππ₯ +
1
2β« πππ 5π₯ ππ₯ =
1
2(β
1
3) β« πππ (β3π₯) β 3ππ₯ +
1
2
1
5β« πππ 5π₯ 5ππ₯ =
1
6π ππ3π₯ +
1
10π ππ5π₯ + π
ππ£ = (β3π₯) β 3ππ₯ ππ£ = 5π₯ β 5 aplicando cos(-A)=cosA
26
Integrales de las potencias de la tangente y cotangente
β« 4π‘πππππ₯π ππ2πππ₯ππ₯ = 4
ππ
π‘ππ2πππ₯
2 =
π
ππππππππππ + π
π£ππ£ = π ππ2πππ₯
β«ππ‘πβπ₯ππ π2βπ₯
βπ₯ππ₯ = 2 β« ππ‘πβπ₯ππ π2βπ₯ Β· (π₯)β
1
2 = β2ππ π2βπ₯
2 = βππππβπ + π
π£ππ£ = 1
2(π₯)β
12
β« ππ‘ππ2π₯ππ π2π2π₯π2π₯ = 1
2
ππ‘π2π2π₯
2 =
π
ππππππππ + π
π£ππ£ = ππ π2π2π₯2
β« π‘ππ32π₯ππ₯ = β« π‘ππ2π₯(π‘ππ22π₯)ππ₯ = β« π‘ππ2π₯(π ππ22π₯ β 1)ππ₯ =
β« π‘ππ2π₯π ππ22π₯ππ₯-β« π‘ππ2π₯ππ₯ = ππππππ
πβ
π
ππ₯π§|πππππ| + π
β« π‘ππ53π₯ππ₯ = β« π‘ππ33π₯(π‘ππ23π₯)ππ₯ = β« π‘ππ33π₯(π ππ23π₯ β 1)ππ₯ =
β« π‘ππ33π₯π ππ23π₯ππ₯ β β« π‘ππ33π₯ππ₯
β β« π‘ππ33π₯ππ₯ = β β« π‘ππ3π₯(π ππ23π₯ β 1)ππ₯ = β β« π‘ππ3π₯π ππ23π₯ +
β« π‘ππ3π₯= ππππππ
ππβ
ππππππ
π+ π₯π§|πππππ| + π
27
β« ππ‘π6π₯ππ₯ = β« ππ‘π4π₯(ππ‘π2π₯)ππ₯ = β« ππ‘ππ₯4π₯(ππ π2π₯ β 1)ππ₯ =
β« ππ‘π4π₯ ππ π2π₯ππ₯ β β« ππ‘π2π₯(ππ π2π₯ β 1)ππ₯ = β β« ππ‘π2π₯ππ π2π₯ππ₯ +
β« ππ‘π2π₯ππ₯ = βππππ
π+
ππππ
πβ ππππ β π + π
β«(π‘πππ₯ + 3)π = β« π‘πππ₯π + 2π‘πππ₯3 + 9 = ππππ β π + πππ|ππππ| +
ππ + π
28
SustituciΓ³n trigonometrica
1.-
249 xx
dx
8.-
10.-
dxx
x6
23
216
cctg
dctgsen
d
x
x
5
5
16
1csc
16
1
4
cos16 24
62
4
6
32
dd
sendctg
xx
dxcsc
3
1cos
cos
1
3
1sec
3
1
)2(3 22
cctg |csc|ln3
1
dxdxtgx
tg
xx
x
xsen
2
2
2
2
sec2
3
2
3
3
2
cos
349
49
3cos
49
2
cx
x
|
2
349|ln
3
1 2
722 xx
dx
csend
xx
dx
7
1cos
7
1
)7( 222
777
)7(
sec7cos
77cos
)7(
2
222
222
xtgxx
tg
xx
x
xxsen
c
x
x
7
7
1 2
cx
x
5224
18
1
22
2222
4
4cos44
4cos
44
x
xtg
xx
xsenx
sen
29
IntegraciΓ³n por partes
1.- dxx
xcoc2
dxdxx
senx
xsendxx
senx
xsen2
1
2)2)(2(
22
22
22
4.- xdxln dxx
xxx1
ln
6.- xdxx ln2 dxx
x
x
x
dx
xx
x 2ln1ln
1.- xdxx cos2 xsenxdxsenxx 22
xdxxxsenxx coscos22
cxx
xsen 2
cos42
2
22
2cos
xsenv
dxx
dv
xu
dxdu
cxx
x
1ln
1
2
ln
1
xv
dxxdv
xu
dxx
du
cxxx ln
xv
dxdv
xu
dxx
du
ln
1
senxv
xdxdv
xu
xdxdu
cos
2
2xv
senxdxdv
xu
dxdu
cos
csenxxxsenxx 2cos22
30
4.- dxex x22 dxxeex xx 222
2
1
dxexeex xxx 2222
2
1
2
1
2
1
1.- arctgxdx
dxx
xxarctgx
21
2
2
1
5.- dxxArcSenx2
dxxxarcsenxx
dxx
x
xarcsenx
x4)1(
221
2
22
1432
22
4
22
6.- xdxSenxSen3
x
x
ev
dxedv
xu
xdxdu
2
2
2
2
1
2
x
x
ev
dxedv
xu
dxdu
2
2
2
1
cexeex xxx 2222
4
1
2
1
2
1
cxarcsenxx
21
422
12
1
2
2
41
2
2
2
2
xv
xdxdv
arcsenxu
dxx
du
xdxxxxsen 3coscos3cos3
dxxxxxsen 4cos)2cos(2
13cos3
cxsenxsenxxsen 48
32
4
3cos3
xv
senxdxdv
xsenu
xdxdu
cos
3
3cos3
xv
dxdv
arctgxu
dxx
du
21
1
cxxarctgx |1|ln2
1 2
31
IntegraciΓ³n por sustituciΓ³n algebraica
2.- xdxx 9 cmm
dmmmmdmmm3
185
2922935
242
3.-
dxx
x
1
dss
sssds
s
s)
1(22
1 22css arctan2)(2
4.- 1xe
dx
cpp
dpdp
pp
p
p
dpp
p
arctan21
21
2
2
22
2
7.-
x
dx
9 cp
pdpp
p
dppp
36
3
494
)9(4 32
2
cex 1arctan2
1
2
|1|ln
1
1
2
2
2
p
pdx
px
ep
ep
x
x
mdmdx
mx
mx
xm
2
9
9
9
2
2
cxx 35
9695
2
cx 936393
4 3
dpppdx
px
xp
xp
xp
)9(4
)9(
9
9
9
2
22
2
2
cxx arctan22
sdsdx
sx
xs
2
2
32
IntegraciΓ³n por fracciones parciales
CASO 1
1.- 42x
dx
5.-
dz
zzz
z
2
6323
2
dz
zzz
z
)2(
632
2
)2()1()1)(2(Β΄63
12)1)(2(
63
)1)(2(
2
2
zczzbzzzaz
z
c
z
b
z
a
zzz
z
zzz
si z=-2 si z=1 si z=0
221
)2)(2(22
1
2222
1
xbxa
xxx
b
x
a
x
b
x
a
xx
cxxdxx
bdx
x
a
xx
dx|2|ln
4
1|2|ln
4
1
22)2)(2(
cx
x
|
2
2|ln
4
1
12)1)(2(
63 2
z
dzc
z
dzb
z
dzadz
zzz
z
czzz
czzza
|1|ln3|2|ln3||ln3
|1|ln3|2|ln3||ln
cz
zz
|
)1)(2(|ln3
3
618
b
b
3
39
c
c
3
26
a
a
33
CASO 2.-
dx
x
cdx
x
bdx
x
adx
xx
xxdx
xxx
xx22
2
2
2
)1(1)1(
18
12
18
)()1()1(18
)1(1)1(
18
22
22
2
xcxbxxaxx
x
c
x
b
x
a
xx
xx
Si x=0 si x=-1 si x=1
6
6
c
c
4.-
du
uu
u23
4
2
8
2)2(
822
4
u
c
u
b
u
adu
uu
u
22 )2()2(8 cuubuauau
Si x=-2 si x=0 si x=1
cx
x
1
6||ln
cx
cxbxa
|1
)1(|ln|1|ln||ln
1
a1
0
62410
2410
b
b
cba
2
84
42
82
2
82
2
23
3
34
423
u
u
uu
u
au
uuu
duu
cdu
u
bdu
u
adu
duuu
uudu
uu
u
22
2
842
2
8
2
23
2
23
4
2
48
c
c2
21234
a
a
4
28
b
b
cuu
auu
u |2|ln2||ln22
2
2
34
IntegraciΓ³n por fracciones parciales
1.-
dx
xx
x
41 22
2
Si x=0 si x= i si x2= -4
6.-
dx
x
xxx22
23
)1(
222
)()1)((222 223 dcxxbaxxxx
Si x= 0 x= i si= 1
22222
22
)2(411
2
2
41
x
dxd
x
xdxc
x
dxb
x
xdxa
dxx
dxdx
x
bax
cx
arctgd
xc
arctgxbxa
22
|4|ln21
1||ln
2
22
)1)(()4)((
41)4)(1(
222
2222
2
xdcxxbaxx
x
dcx
x
bax
xx
x
34
0
364
)3(24
2
14)1(44
d
c
dci
dci
ix
x
0
03
31
0
3310
a
a
b
a
baiidb 40
cx
arctgxarctgxx 23
4|4|ln
3
1||ln 22
22222
222
)1()1(11
2
2
)1(1
x
dxddx
x
xc
x
dxb
x
xdxa
x
dcxdx
x
bax
2
2
b
db
0
1
222
d
c
dcii
dciii
1
01)4(227
a
a
cx
arctgxx
1
1
2
12|1|ln
2
12
2
cx
arctgxx
)1(2
12|1|ln
2
12
2