2. Integrimi me metodën e zëvendësimit Duke zbatuar metodën e zëvendësimit të njehsohen integralet: 1.
(1 x ) dx. 7
2.
(3 2x ) dx. 6
3.
1 xdx .
4.
4
1 x dx .
Zgjidhja. 1.
7 (1 x ) dx
1x t t8 (1 x )8 t7 dt C C. 8 8 dx dt
3 2x t 1 1 2. (3 2x ) dx 2dx dt t 6 dt t 6 dt 2 2 1 dx dt 2 6
1 t7 t7 (3 2x )7 C C C 2 7 14 14
3. Mënyra e parë: 1x t
1 x dx dx dt
dx dt
1
1 2
1
t2 t ( dt ) t dt C 1 1 2
2
2 2 t3 C (1 x )3 C. 3 3
Mënyra e dytë: 1 x t2 1 x dx
2 t( 2t )dt 2 t 2dt t 3 C 3 dx 2tdt dx 2tdt
t 1x
4.
4
1 xdx
2 (1 x )3 C. 3
1 x t4 t 4 1 x dx 4t dt 3
3
t 4t dt 4 t 4 dt
INTEGRALI I PACAKTUAR
2
4
t5 4 C 4 (1 x )5 C. 5 5
Detyra për ushtrime Të njehsohen integralet: 1.
(1 2x ) dx.
4.
7.
((1 x )
4
5
3
1 3x dx . 3
2 2. x dx . 3
3.
5x 1 dx .
6.
(
5.
5
1
x dx . 2
x 1 x 1)dx .
3 1 x )dx .
Të njehsohen integralet: 5.
2x 1 x 2 x 3 dx.
8.
x
2
1 x 3 dx .
x3 x 4 1 dx. x2 3 dx . 9. 3 x 2
6.
7.
x
1 2x 2 dx .
Zgjidhja. 5.
6.
x2 x 3 t 2x 1 dt dx ln|t | ln( x 2 x 3) C. x2 x 3 t (2x 1)dx dt
x
x
4
3
1
x4 1 t dx 4 x 3 dx dt 1 x 3 dx dt 4
1 dt ln( x 4 1) 1 dt 1 4 C. ln|t | 4 t 4 t 4
1 2x 2 t 2 t 1 2x 2 1 7. x 1 2x 2 dx 4 xdx 2tdt t t dt 2 1 xdx tdt 2 3 1 t 1 (1 2x 2 )3 C. 2 3 6
ANALIZA MATEMATIKE II
3
1 x 3 t2 t 1 x 3
8.
1 x 3 x 2 dx 3x 2 dx 2tdt
x 2 dx
9.
x2 3 3
x 2
dx
2 tdt 3
2 2 t3 t tdt C 3 3 3
2 (1 x 3 )3 C 9
x 2 t 3 x t 3 2; t 3 x 2 dx 3t 2 dt
(t 3 2)2 3 3t 2 dt 3 (t 6 4t 3 4 3) t dt t t8 t5 t2 3 (t7 4t 4 t )dt 3 12 3 8 5 2 33 12 3 3 ( x 2)8 ( x 2)5 3 ( x 2)2 C. 8 5 2
Detyra plotësuese Të njehsohen integralet 8.
2x 1 x 2 x 4 dx.
9.
2x 3 x 2 3x 1 dx.
10.
x5 x 6 1 dx.
11.
2x a 2x 3 , a-const.12. dx x 2 ax 1 4x 4 1 dx.
13.
14.
2 x 1 x dx.
15.
2 x 2 3x dx.
16.
17.
2 3 x 2 5x dx.
18.
x 3 x 5
dx . x 2 x2 2
Të njehsohen integralet: 10.
12.
dx x 1 x 1 1x dx . 1x
.
1x
11.
13.
x
1 x2
dx .
dx x2 1
.
x 2 x 3
dx .
x ( x 2 1) 3
x2 1
dx .
INTEGRALI I PACAKTUAR
4
Zgjidhja. 10. Pas racionalizimit të emëruesit kemi: dx 1 x 1 x 1 I dx x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 dx dx x 1 ( x 1) 2 1 1 x 1dx x 1dx ( I1 I 2 ) , 2 2 ku I1 x 1 dx ; I 2 x 1 dx .
Zgjidhim veçmas integralet I1 , I 2 . Merret: x 1 t2
I1
t3 2 x 1dx t x 1 t 2tdt 2 ( x 1)3 C. 3 3 dx 2tdt x 1 t2
I2
t3 2 x 1dx t x 1 t 2tdt 2 ( x 1)3 C. 3 3 dx 2tdt
Pra
12 2 1 ( x 1)3 ( x 1)3 C ( x 1)3 ( x 1)3 C. 23 3 3 1x dx xdx 11. I dx arcsin x I1 . 2 2 1x 1x 1 x2 I
I1
1 x 2 t2 2xdx 2tdt 1 x2 xdx tdt xdx
tdt dt t 1 x 2 t
Pra, I arcsin x 1 x 2 C 12. Pasi të kryejmë racionalizimin merret 1x 1x 1x 1x 1 x dx 1 x 1 x dx 1 x 2 dx.
Integrali që morëm është i ngjashëm me integralin e detyrës paraprake. 1x Pas zgjidhjes merret dx arcsin x 1 x 2 C. 1x
ANALIZA MATEMATIKE II
13.
x
dx x2 1
5
dx 1 x x 2 1 2 x
dx 1 x x sgn x 1 x
2
dx 1 x |x | 1 x dx
1 sgn x
1 x2 1 x
2
2
1 t dx dt x sgn x sgn x 2 1 1 t2 1 2 dt x 1 x2 x 1 sgn x arcsin t C sgn x arcsin C. x
Detyra për ushtrime Të njehsohen integralet: 19.
22.
25.
dx
.
dx
20.
x 3 dx . 1 (2x )2
23.
1x dx . 1x
26.
15.
x
18.
1 x
x 1 x 1
1
21.
dx .
24.
.
27.
x
16.
x ln x ln(ln x ) .
2x 1 2x 1
x 1 1 9x
2
xdx 4
x x
2
.
x 2 dx . 1 x2
3
1x 1 4x 2 dx x2 1
dx . .
Të njehsohen integralet: dx
14.
x (1 2 ln x ).
17.
x cos
19.
dx
2
ln x
.
ln( x 1 x 2 ) dx . 1 x2
dx 1 ln x 2
1
2
ln
.
1x dx. 1x
dx
INTEGRALI I PACAKTUAR
6
Zgjidhja.
(1 2 ln x ) t dx 1 dt 1 14. dx 1 ln|t | x (1 2 ln x ) 2 t 2 dt x 2 1 ln|1 2ln x |C. 2 ln x t dx dt 15. dx arcsin t C arcsin(ln x ) C. 2 dt x 1 - ln x 1 t2 x ln(ln x ) u dx du 16. dx ln|u | C ln|ln(ln x )| C. x ln x ln(ln x ) u du x ln x ln x t dx dt 17. dx tan t C tan(ln x ) C. 2 x cos ln x cos2 t dt x 1 1x 18. I ln dx 2 1x 1x 1x Zëvendësojmë ln u 1x Pas diferencimit merret: '
1 1 x dx du 1 x 1 x 1x 1 x (1 x )' (1 x ) (1 x )(1 x )' dx du 1x (1 x )2 1 1 x (1 x )( 1) dx du 1x 1x 1 x 1 x dx du 1 x2 1 1 dx du 2 1x 2 Pra, kemi: 2
1x ln 1 x 2 du u 1 ln 2 1 x C. I u 2 4 4 4 1x
ANALIZA MATEMATIKE II
19. I
7
ln( x 1 x 2 ) ln( x 1 x 2 ) dx dx 1 x2 1 x2
Zëvendësojmë ln( x 1 x 2 ) u Pas diferencimit merret: ( x 1 x 2 )'
dx du x 1 x2 x 1 1 x 2 dx du x 1 x2
x 1 x2 1 x2
dx du x 1 x2 dx Pra du. 1 x2 Merret: I
1 2
3
u2 2 udu u du ln 3 ( x 1 x 2 ) C. 3 3 2
Detyra për ushtrime Të njehsohen integralet: dx
28.
2x (ln x 1).
31. 34.
dx
29.
x
ln(sin x ) dx.
32.
cos
1 1x 1 x 2 ln 1 x dx.
35.
cos x
1 ln x 2
.
dx . x ln(ctgx )
sin x
30.
ln(cos x ) dx.
33.
x sin
dx 2
ln x
.
ln( x 1 x 2 ) dx . 1 x2
Të njehsohen integralet: 20.
cos x e sin x dx.
21.
ex 1 e x x dx. 22.
ex
1 e 2 x
dx . 23.
x
2 x3
e
dx .
INTEGRALI I PACAKTUAR
8 1
ex 24. 2 dx . x
25.
Zgjidhja. sin xdx
dx ex ex .
cos x t
26.
ex ex 1
et dt et e cos x C.
20.
e
21.
ex x t ex 1 dt dx ln|t | ln| e x x | C. ex x x t ( e 1)dx dt
22.
cos x
ex ex
1 2
dx
sin xdx dt
1 t dt 1 ln|t | ln e x C. 2 t 2 e x dx dt ex
x 3 t
23.
x
2 x3
e
dx .
dx 3x 2 dx dt 1 x 2 dx dt 3
1 t 1 1 3 e dt et e x C. 3 3 3
1 1 t 1 ex x et dt et e x C. 24. 2 dx 1 x 2 dx dt x dx dx dx e x dx 25. x e2x 1 e2x 1 e ex x 1 e x e ex
26.
ex e 1 x
ex t e x dx dt 2 arctan t arctan e x C. x 2 x ( e ) 1 e dx dt t 1
dx
ex 1 t2 t ex 1 e dx 2tdt x
2tdt 2 e x 1 C. t
Detyra për ushtrime Të njehsohen integralet: 36.
sin x e cos xdx.
37.
e tgx cos2 x dx.
38.
e
dx . 1
x
ANALIZA MATEMATIKE II
39. 42.
9
2e 2 x 1 e2x x dx.
e x dx 3
ex 3
40.
x
2 2 x 3
e
dx .
41.
xe
x 2 a2
dx .
.
Të njehsohen integralet: dx 8x 3 x 2 dx 27. 2 28. 29. 2 . dx . . 2 6 x a x 7 9x Zgjidhja. dx x 2 t dx 1 dx 1 a 2 2 2 27. 2 a 2 2 2 x a x a a x a 1 dx adt 2 a a 1 dt 1 1 x 2 arctan t arctan C. a t 1 a a a
t
adt 2 1
dt x dx x dx 1 dt 2 3x 2 dx dt 2 3 2 2 2 . 28. I 6 3 2 3 3 t 9x 3 (x ) 3 t dt 2 x dx 3 Në bazë të detyrës paraprake dt 1 t 1 x3 arctan C arctan C1 . 1 32 t 2 3 3 3 3 1 x3 Pra I arctan C. 9 3 8x 3 8x dx 29. 2 dx 8 2 dx 3 2 8 I1 3 I 2 , ku x 7 x 7 x ( 7 )2 xdx dx I1 2 ; I2 2 x 7 x ( 7 )2 2
2
x2 7 t
I1
Pra
x3 t
xdx 1 dt 1 1 2xdx dt ln|t | ln( x 2 7). 2 2 t 2 2 7 dt xdx 2
x
INTEGRALI I PACAKTUAR
10
1 1 x ln( x 2 7) 3 arctan C 2 7 7 3 x 4 ln( x 2 7) arctan C. 7 7
I 8
Detyra për ushtrime Të njehsohen integralet 43.
dx 2x 2 a 2 .
46.
2x
x2 dx . 6 5
44.
dx 2x 2 3 .
47.
x
9x 1 dx . 2 15
45.
x3 7 x 8 dx.
48.
(ex )
dx . 2 2
Të njehsohen integralet: 30.
a arctan x 1 x 2 dx , a R \ {1}, a 0.
31.
dx
arcsin x
1 x2 e arctg2 x x 33. dx . 1 4x 2
.
arctanx x ln(1 x 2 ) dx . 1+x 2 Zgjidhja. arctan x u a arctan x au a arctan x u 30. dx a du C. dx ln| a | ln a 1 x2 du 2 1x arcsin x u dx du 31. ln|u | ln|arcsin x | C. dx 2 du u arcsin x 1 x 1 x2
32.
arctan x x ln(1 x 2 ) arctan x x ln(1 x 2 ) dx dx 1 x2 1 x 2 dx 1 x2 I1 I 2 ,
32. I
arctan x u u2 (arctan x )2 u du . dx 2 2 du 2 1x ln(1 x 2 ) u ln(1 x 2 ) 1 u2 ln 2 (1 x 2 ) I2 dx udu . 1 du 2 4 4 1 x2 xdx 2 1 x2 Pra
ku I1
arctan x 1 x 2 dx
ANALIZA MATEMATIKE II
11
(arctan x )2 ln 2 (1 x 2 ) C. 2 4 e arctan 2 x x e arctan 2 x xdx 33. I dx dx I1 I 2 , 2 2 1 4x 1 4x 1 4x 2 I
arctan 2x u
I1
I2
e 1 1 u eu e arctan 2 x dx 2 dx du e du . 2 2 1 4x 2 2 2 1 4x dx 1 du 2 2 1 4x arctan 2 x
1 4 x 2 t 1 dt 1 xdx 1 2 1 4x 2 8xdx dt 8 t 8 ln|t | 8 ln(1 4x ).
Pra, I
e arctan 2 x 1 ln(1 4 x 2 ) C. 2 8
Detyra për ushtrime Të njehsohen integralet: 49.
3arctan x 1 x 2 dx.
50.
52.
x ln(1 3x 2 ) 1 3x 2 dx.
53.
Të njehsohen integralet: 34. sin ax dx. 35. 37. 40. 43.
cot ax dx. tan x dx. sin x sin 2x dx. 4
51.
3
arcsin 2 x 1 x2
dx .
e arctan 4 x e arctan 4 x dx . 1 x2 16
cos ax dx. cos 2x
36. tan ax dx .
sin x cos x dx. 39. sin x dx. 41. sin x dx . 42. cot x dx . 44. cos x cos 2x cos 3x dx. 38.
dx 2 sin2 x cos2 x . 46. sin x sin 3 x 48. dx . 49. 2 cos2 x sin 2 x
45.
arctan 3 x 1 9x 2 dx.
4
5
sin 2x tan4 x dx. dx sin x .
5
47.
sin 3 x cos2 x dx.
50.
sin
4
x cos4 x dx .
INTEGRALI I PACAKTUAR
12
Zgjidhja. ax t
34.
sin ax dx adx dt dx
1 1 1 sin t dt cos t cos ax C. a a a
1 1 1 cos t dt sin t sin ax C. a a a
1 dt a
ax t
35.
cos ax dx adx dt dx
1 dt a
cos ax u sin ax 36. tan ax dx dx 1 cos ax sin ax dx du a 1 du 1 1 ln|u| ln|cos ax |C. a u a a sin ax u cos ax 37. cot ax dx dx 1 sin ax cos ax dx du a 1 du 1 1 ln|u| ln|sin ax |C. a u a a 38. Mënyra e parë: cos 2x cos2 x sin 2 x cos x sin x I dx dx dx dx sin x cos x sin x cos x sin x cos x cot x dx tan x dx ln|sin x | ln|cos x | ln|sin x cos x |C.
Mënyra e dytë: cos 2x 2 cos 2x cos 2x I dx dx 2 dx 2 cot 2x dx sin x cos x 2 sin x cos x sin 2x ln|sin 2x | C1 . Shënim.
ANALIZA MATEMATIKE II
13
Lexuesi mund të ketë përshtypjen se rezultatet që morrëm nga zgjidhja në dy mënyrat janë të ndryshme por lehtë vërejmë se: ln|sin 2x | C1 ln|2 sin x cos x | C1 ln 2 ln|sin x cos x | C1 ln|sin x cos x | C. 2
1 cos 2x 4 2 2 sin xdx (sin x ) dx 2 dx 1 (1 2 cos 2x cos2 2x ) dx 4 1 1 1 1 cos4 x dx cos2x dx dx 4 2 4 2 1 1 1 x I1 I 2 . 4 2 4 2x u 1 1 1 I1 cos 2x dx cos u du sin u sin 2x C1 . 1 2 2 dx du 2 2 1 cos 4 x 1 1 1 1 I2 dx dx cos 4 x dx x sin 4 x C2 . 2 2 2 2 8 Përfundimisht merret: 1 1 1 1 1 1 3x sin 2x sin 4 x I x sin 2x x sin 4 x C. 4 22 4 2 8 8 4 32 1 1 dx 40. tan 4 x dx tan 2 tan 2 x dx tan 2 x 2 cos x 1 tan 2 x dx tan 2 x dx I1 I 2 . cos2 x tan x u 1 u3 tan 3 x 2 2 I1 tan x dx u du . dx 3 3 cos2 x du 2 cos x 2 sin x 1 cos2 x dx I 2 tan 2 x dx dx dx dx 2 2 cos x cos x cos2 x tan x x C. Përfundojmë se tan3 x 4 tan x dx tan x x C. 3 cos x t 41. sin 5 xdx sin x sin 4 x dx sin x (1 cos2 x )2 dx sin xdx dt
39.
INTEGRALI I PACAKTUAR
14
(1 t 2 )2 dt dt 2 t 2 dt t 4 dt t 2
t3 t5 3 5
2 1 cos3 x cos5 x C. 3 5 5 cos x cos x cos4 x cos x (1 sin 2 x )2 42. cot 5 xdx dx dx dx sin5 x sin5 x sin 5 x sin x u (1 u2 )2 1 2u2 u4 du du cos x dx du u5 u5 du 1 1 u 5 du 2 u 3 du 4 2 ln|u | u 4u u 1 1 ln|sin x | C. 4 sin 4 x sin 2 x Në detyrat vijuese zbatohen formulat: 1 sin x sin y (cos( x y) cos( x y)) 2 1 cos x cos y (cos( x y ) cos( x y )) 2 1 sin x cos y (sin( x y) sin( x y)). 2 1 43. sin x sin 2x dx (cos( x 2x ) cos( x 2x )) dx 2 1 1 1 cos x dx cos 3x dx sin x sin 3x C. 2 2 6 1 44. cos x cos 2x cos 3x dx cos 2x (cos( x 3x ) cos( x 3x )) dx 2 1 cos 2x (cos( 2x ) cos 4 x ))dx 2 1 cos 2x cos 2x dx cos 2x cos 4 x dx 2 1 dx cos 4 x dx cos 2x dx cos 6x dx 4 1 1 1 1 x sin 2x sin 4 x sin 6 x C. 4 2 4 6 dx dx tan x u 2 2 dx cos x cos x 45. dx 2 sin 2 x cos2 x sin 2 x 2 tan 2 x 1 du 2 1 cos2 x 2 cos x cos x
ANALIZA MATEMATIKE II
15
du du 2 arctan 2u 2 1 ( 2u )2 1
2u
2 arctan( 2 tan x ) C. sin 2x 2 sin x cos x sin x cos x cos 4 x 46. dx dx 2 dx 4 4 tan x sin x sin 4 x cos4 x sin x u sin x (1 sin 2 x )2 cos x 2 dx 4 cos x dx du sin x u(1 u2 )2 du du du du 2 3 4 2 2 4 u u u u 1 4 1 4 2 2 ln|u | 2 ln|sin x | C. u u sin 2 x sin x sin3 x sin x sin 2 x sin x (1 cos2 x ) 47. dx dx dx cos2 x cos2 x cos2 x 2
cos x u
sin x dx du
1 u2 1 1 du u cos x C. u cos x u2
sin x sin 3 x sin x (1 sin 2 x ) sin x cos2 x dx dx dx 2 2 2 x sin x cos x 1 cos2 x 1 cos x t t2 dt 2 dt dt 2 sin x dx dt t 1 t 1 t arctan t cos x arctan(cos x ) C. x x sin 2 cos2 dx dx 2 2 dx 49. x x x x sin x 2 sin cos 2 sin cos 2 2 2 2 x x sin cos 1 1 2 2 dx dx ( I 2 I 2 ). x x 2 2 sin cos 2 2
48.
2 cos
INTEGRALI I PACAKTUAR
16
x u 2 x sin 2 dx 1 sin x dx du du 2 ln|u | 2 ln|cos x |. I1 u x 2 2 2 cos x 2 sin dx 2 du 2 cos
x x sin u du x 2 2 dx I2 2 2 ln|u | 2 ln sin . x u 2 x sin cos dx 2du 2 2 x sin 1 x x 2 ln tan x C. Pra I 2 ln cos 2 ln sin ln x 2 2 2 2 cos 2 cos
4
2 sin x cos x 50. sin x cos x dx (sin x cos x ) dx dx 2 4
4
4
2
(sin 2 2x )2 1 1 cos 4 x 16 dx 16 2 dx 1 (1 2 cos 4 x cos2 4 x ) dx 64 1 dx 2 cos4 x dx cos2 4 x dx 64 1 1 1 cos 8x x 2 sin 4 x dx 64 4 2
1 1 1 1 x sin 4 x x sin 8 x 64 2 2 16
1 3 1 1 x sin 4 x sin 8 x C. 64 2 2 16
Detyra për ushtrime Të njehsohen integralet: cos x dx . x x sin cos 2 2
54.
57.
cos
5
x dx .
55.
cos
58.
tan xdx.
4
5
x dx .
56.
cot
59.
sin 3x sin x dx.
4
x dx .
ANALIZA MATEMATIKE II
17
dx . x 5 cos2 x
60.
sin x sin 2x cos 3xdx.
61.
sin
62.
cos 2x cot4 x dx.
63.
sin 2x cot4 x dx.
65.
cos2 x sin 2 x sin x sin3 x dx.
66.
cos x .
2
dx
Të njehsohen integralet dx 51. . 2 x a2
52.
x
a 2 x 2 dx .
54.
53.
64.
Zgjidhja.
67.
dx a2 x 2 dx (1 x 2 )3
sin5 x dx . cos2 x
dx . x sin 2
. .
x 2 a2 t x , t x x 2 a2 x 2 a 2 t 2 2tx x 2
51.
dx
x 2 a2
t2 a2 t2 a2 , dx dt 2t 2t 2 t2 a2 x 2 a2 t 2t 2 t a2 x 2 a2 2t
2tx t 2 a 2 x
t2 a2 2t 2 dt dt ln|t | ln| x x 2 a 2 | C. t t2 a2 2t
x a tan u dx du 52. dx a 2 2 cos2 u x a x a2 x 2 a
du 2 cos u 2 a tan u a a 2 tan 2 u a
1 cos u
INTEGRALI I PACAKTUAR
18
du du 2 1 du cos u cos2 u 2 1 a sin u a tan u 1 tan u a sin u cos u 1 u ln tan . a 2 a x Meqë cos u atëherë sin u dhe 2 2 2 a x a x2 x tan
Pra
53.
u sin u 2 1 cos u
x
dx 2
a x
2
1
a2 x 2 a
a2 x 2
x a a2 x 2
.
1 x ln C. a a a2 x 2
x a sin t dx a cos t dt a 2 x 2 dx sin t x a x t arcsin a
a 2 a 2 sin 2 t a cos t dt
a 2 1 sin 2 t cos t dt a 2 cos2 t dt a 2
dt cos 2tdt a2 t 12 sin 2t 2
1 x x x2 a2 t 2 sin t cos t arcsin 1 2 2 a a a 2 a2 x x arcsin a 2 x 2 C. 2 a 2 dt x tan t dx dt cos2 t dt 3 dx (1 x 2 )3 (1 tan 2 t )3 sin 2 t 2 cos2 t cos t 1 cos2 t
54.
a2 2
1 cos 2t dt 2
a2 2
ANALIZA MATEMATIKE II
Shënim.
19
dt cos2 t x
1 cos3 t
1 x2
cos t dt sin t C
tan t 1 tan 2 t
C.
sin 2 t sin t tan 2 t tan t cos2 t sin 2 t sin t . 2 2 2 2 2 2 sin t cos t sin t cos t 1 tan t 1 tan t cos2 t cos2 t 2
Detyra për ushtrime Të njehsohen integralet: 68.
71.
dx 2
x a
69.
2
dx ( a 2 x 2 )3
.
x
dx x 1 2
.
70.
dx 9 x2
.