Analiza-matematike-2-Integrali-i-pacaktuar-A.Sh_.Shabani.pdf
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1 Integrali i pacaktuar 1. Integralet tabelare Duke zbatuar tabelën e integrale të zgjidhen integralet: 1.
(x
2.
3.
4.
5.
1 x
6.
4
7.
8x 2 2 2x 1 dx.
2
3x 4)dx .
x 3 dx . x4 dx 3
x2
.
x 3 x dx .
1
16 x x
x x dx .
dx .
Zgjidhja. 1.
2 2 ( x 3x 4)dx x dx 3xdx 4dx
2.
x 3 dx x4
x
x
4
x3 3 x dx 4 dx 3
x3 x2 3 4 x C. 3 2
dx
3 dx dx dx 3 3 4 x 3 dx 3 x 4 dx 4 x x x
INTEGRALI I PACAKTUAR
2
x 31 x 4 1 1 1 3 2 3 C. 3 1 4 1 2x x
3.
dx 3
x2
dx x
2 3
2 1
2 3
1
x 3 x3 x dx 3 3 x C. 2 1 1 3 3
1 3
2
2 3
5.
1 1 1 1 x x x 1 x x 2 x 4 dx 1 x x 4 dx
x x dx
x x dx
1
3 4
x dx x 7
1
x3 3 x dx x dx 3 x 5 C. 2 5 1 3
4.
3
4 3
1
1
3
3
3 4
1
3
1 x4 x dx x 4 dx 3 1 4
3
4
7
x4 x4 4 4 4 3 x x C. 7 3 7 3 4 4
6. Mënyra e parë. 16 x
4
x
dx
4 2 ( x )2 4 x
dx
(4 x )(4 x ) 4 x
dx (4 x ) dx
3
1 2
x2 2 4 dx x dx 4 x 4x x 3 C. 3 3 2
Mënyra e dytë. Kryjemë racionalizimin e emëruesit: 16 x
4
x
dx
16 x
4 x
4 x 4 x
4x
2 x 3 C. 3
dx
(16 x )(4 x ) dx (4 x )dx 16 x
ANALIZA MATEMATIKE II
7.
3
8x 2 2 2(4 x 2 1) (2x 1)(2x 1) dx 2x 1 2x 1 dx 2 (2x 1) dx 2 (2x 1)dx
2 2 x dx dx 4
x2 2x 2x 2 2x C. 2
Detyra plotësuese Të njehsohen integralet: 1. ( x 3 x 2) dx. dx
4.
x
7.
( x 1) dx.
.
4
3
10.
x
12.
x4 1 x 2 dx.
14.
2
x
dx .
Të njehsohen integralet: 8.
ex x e 1 x dx.
9.
e
2 x
10.
(e
11.
e
12.
(2
13.
2
x
2x dx .
x e ) dx .
e3 x 8 dx . x 2
x
x
x
.
6.
x
8. ( x 2 1)2 dx .
9.
x
5.
x x x2 3 x
9 x
3
3.
2.
3x ) dx . 3 x dx.
x ( x 1) dx. dx 4
x
1 dx . x
13. 15.
11.
dx . 3x 1
3
x
x3 dx.
dx . x
x 2x 3
( x 1)( x 2 1) dx . x3 9x 2 3
4
3
(1 x 2 ) dx .
dx .
INTEGRALI I PACAKTUAR
4
14.
(3
15.
3x 1 5x 1 15x dx.
x
4 x )2 dx .
Zgjidhja. 8.
ex x e 1 x
9.
dx x e x ln| x | C. dx e dx x
e 2 x 2x dx ( e 2 x 2x ) dx e 2 x dx 2x dx e 2 e x dx e 2 e x dx
2x ln 2
2x 2x 2x e2 e x e 2 x C. ln 2 ln 2 ln 2 x e 1 C. e 1
10.
x e x e x (e x ) dx e dx x dx e
11.
e3 x 8 ( e x )3 23 ( e x 2)( e 2 x 2e x 4) dx dx ex 2 ex 2 ex 2 e 2 x dx 2 e x dx 4 dx ( e 2 )x dx 2e x 4 x
( e 2 )x e2x e2x x x 2 e 4 x 2 e 4 x 2e x 4 x C. 2 ln e 2 ln e 2
12. (2x 3x ) dx 2x dx 3x dx
2x 3x C. ln 2 ln 3 x
2 x 3 2 13. 2x 3 x dx 2x (31 )x dx dx C. 2 3 ln 3
14.
(3
x
4 x )2 dx (32 x 2 3x 4 x 4 2 x )dx 9 x dx 2 12x dx 16 x dx
9x 12x 16 x 2 C. ln 9 ln12 ln16 x
15.
x
3x 1 5x 1 3 3x 51 5x 1 1 1 dx dx 3 dx dx 15x 5 5 3 15x
ANALIZA MATEMATIKE II
5 x
x
1 1 5 1 3 3 x 1 3 5 3 x C. 1 5 1 ln 5 5 ln 5 ln ln 5 3
Detyra plotësuese Të njehsohen integralet: e x 16. e 1 2 dx. x x
x 1 1 x 2 17. e 2 dx . 2
18.
2x 2e (e x ) dx.
19.
e4 x 1 e2x 1dx.
20.
e3 x 27 e x 3 dx.
21.
3
22.
(2
23.
(3
24.
2x 1 3x 1 6x dx.
25.
(e
x
2 x ) dx .
Të njehsohen integralet: 16.
x2 x 2 1 dx.
17.
x4 1 x 2 dx.
18.
(1 x )2 x (1 x 2 ) dx.
19.
x2 4 x 2 1 dx.
20.
x3 x 2 x 2 1 dx.
21.
4e
x
dx . 4 4x 2 5
5 x dx .
x
x
3 x )2 dx .
x e
x e 1 ) dx .
INTEGRALI I PACAKTUAR
6
Zgjidhja. 16. Mënyra e parë. Pjesëtojmë polinomit x 2 me polinomin x 2 1. Merret: x2 1 1 2 . x2 1 x 1
Pra, x2 1 dx x 2 1 dx 1 x 2 1 dx dx x 2 1 x arctan x C.
Mënyra e dytë. x2 x 2 1 dx
x2 1 x2 1 1 1 dx x2 1 x 2 1 x 2 1 dx
1 1 2 dx x arctan x C. x 1
17.
18.
x4 1 x 2 dx
x4 1 1 x4 1 1 dx 1 x2 1 x 2 dx 1 x 2 dx
( x 2 1)( x 2 1) dx arctan x ( x 2 1)dx arctan x 1 x2
x3 x arctan x C. 3
(1 x )2 1 2x x 2 dx x (1 x 2 ) x (1 x 2 ) dx
19.
x2 4 x 2 1 dx
1 x2 2x x (1 x 2 ) dx x (1 x 2 ) dx
dx dx 2 ln| x | 2 arctan x C. x 1 x2
x2 1 3 3 dx x 2 1 dx 1 x 2 1 dx dx 3 x 2 1
x 3 arctan x C.
20. Pas pjesëtimit të polinomeve x 3 x 2 me x 2 1 merret: x3 x 2 2 x 2 . 2 x 1 x 1
ANALIZA MATEMATIKE II
7
Kemi: x3 x 2 2 x2 dx x dx 2 arctan x C. x2 1 x 2 1 2
21.
4e
x
dx 5 dx 4 e x 5 4e x arcsin x C. 2 4 4x 2 2 1 x2 5
Detyra plotësuese Të njehsohen integralet: 26.
3x 2 1 x 2 1 dx.
27.
x4 1 x 2 1 dx.
28.
(1 x )3 x 2 x 3 dx . x (1 x )
29.
x3 x 3 x 2 1 dx.
x8 1 x 2 1 dx.
31. 3x
30.
1 dx . 2 1 x 9 3
Të njehsohen integralet: 22.
( x sin x 2 cos x ) dx.
23.
sin
2
24.
sin
2
25.
tan
26.
cos
27.
cos 2x dx . x cos2 x
2
2
dx . x cos2 x x dx .
x dx . 2
1 sin 2x dx .
Zgjidhja. 22.
( x sin x 2 cos x ) dx x dx sin x dx 2 cos x dx
INTEGRALI I PACAKTUAR
8
x2 ( cos x ) 2 sin x C 2
x2 cos x 2 sin x C. 2
23. Zbatojmë formulën cos 2x cos2 x sin 2 x.
Merret: cos 2x cos2 x sin 2 x dx sin2 x cos2 x sin2 x cos2 x dx
24.
cos2 x sin 2 x dx sin2 x cos2 x sin2 x cos2 x dx
sin
dx 2
x
dx cot x tan x C. cos2 x
dx sin 2 x cos2 x dx dx sin2 x cos2 x sin2 x cos2 x dx cos2 x sin 2 x tan x cot x C.
25.
2 tan x dx
sin 2 x 1 cos2 x dx dx cos2 x cos2 x dx cos2 x dx
tan x x C.
26. Meqë cos2 x
1 cos 2x x 1 cos x atëherë cos2 . 2 2 2
D.m.th.
cos 27.
2
x 1 cos x 1 1 1 1 dx dx dx cos x dx x sin x C. 2 2 2 2 2 2
1 sin 2x dx
sin 2 x cos2 x 2 sin x cos x dx
(sin x cos x ) |sin x cos x |dx (sin x cos x ) sgn(sin x cos x ) dx
2
( cos x sin x ) sgn(sin x cos x )
ANALIZA MATEMATIKE II
9
(sin x cos x ) sgn(sin x cos x ) (sin x cos x ) sgn(cos x sin x ).
Detyra plotësuese Të njehsohen integralet: x 32. 2sin x 3cos x dx. 2
4
34.
sin
36.
2 sin
38.
1 sin3 x sin3 x dx.
2
2x
dx .
x dx . 2
cos 2x dx . 2 2x
33.
sin
35.
cot
37.
x x sin 2 cos 2 dx.
2
x dx . 2
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 3
dx 4t dt
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: 4
1.
(1 2x ) dx.
2 2. x dx . 3
4.
5.
7.
((1 x )
5
3
1 3x dx . 3
5
5x 1 dx .
3.
6.
(
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
3
4
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: dx
10.
x 1 x 1
12.
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 2
1 ln x
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
3
1 2
u2 2 udu u du ln3 ( 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.
dx ex ex .
26.
ex ex 1
dx .
Zgjidhja. cos x t
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
sin xdx
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
et dt et e cos x C.
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 x
e 1
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 x
e dx 2tdt
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
x3 t
x2 7 t I1
Pra
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
arctan 2 x
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 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.
36. tan ax dx .
cos 2x
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 tan3 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 cos4 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 .
Të njehsohen integralet dx 51. . 2 x a2 53.
a 2 x 2 dx .
2
64.
dx
52.
x
54.
67.
dx a2 x 2 dx (1 x 2 )3
sin5 x dx . cos2 x
dx . x sin 2
. .
Zgjidhja. 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
x a a2 x 2
.
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
19
dt cos2 t
1 cos3 t
x 1 x2
cos t dt sin t C
tan t 1 tan 2 t
C.
Shënim. 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 2
x 1
.
70.
dx 9 x2
.
3. Integrimi me metodën parciale Të njehsohen integralet: 1.
ln x dx.
2.
x ln x dx.
3.
x
n
ln x dx .
4.
x
4
ln3 x dx .
Zgjidhja. ln x u dx du 1. ln x dx x
x ln x x
dx x ln x dx x
dx v v x x ln x x C.
ln x u
2.
x ln x dx
dx du x v xdx ; v
x2 x 2 dx ln x 2 2 x
x2 2
x2 1 x2 x2 ln x xdx ln x C. 2 2 2 4
ln x u dx du 3. x n ln x dx x v x n dx ; v
x n1 x n1 dx ln x n 1 n 1 x
x n1 n 1
x n1 1 x n1 1 x n1 n ln x x dx ln x n 1 n 1 n 1 n 1 (n 1)
x n1 x n1 ln x C. n 1 (n 1)2
INTEGRALI I PACAKTUAR
2
ln3 x u
4.
4 3 x ln x dx
3 ln 2 x x5 3 3x 5 ln 2 x dx du ln x dx x 5 5 x x5 v x 4 dx 5
x5 3 3 x5 3 3 ln x x 4 ln 2 xdx ln x I1 , 5 5 5 5
ku ln 2 x u I1 x 4 ln 2 dx
2 ln x x5 2 x 5 2 ln x dx du ln x dx x 5 5 x x5 v 5
x5 2 2 x5 2 2 ln x x 4 ln x ln x I 2 . 5 5 5 5
Në bazë të detyrës 3. I 2 x 4 ln xdx
Pra I1
x5 x5 ln x 2 5 5
x5 2 2 x5 x5 x5 2 2 5 2 5 ln x ln x 2 ln x x ln x x . 5 5 5 25 125 5 5
Përfundimisht, I
x5 3 3 x5 2 5 2 5 ln x ln 2 x x ln x x C. 5 5 5 25 125
Detyra për ushtrime Të njehsohen integralet: 1.
x
2
ln x dx .
2.
x ln
2
x dx .
3.
x
3
ln 2 x dx .
ANALIZA MATEMATIKE II
3
Të njehsohen integralet: 5.
ln 2 x x 2 dx.
6.
(x
2
7.
ln( x
8.
x
ln
1 x 2 )dx .
2
x ) ln( x 1) dx .
1x dx. 1x
Zgjidhja. ln 2 x u ln 2 x 1 1 1 1 5. 2 dx 2 ln x dx du ln 2 x 2 ln x dx x x x x x 2 1 v x dx x
1 2 ln x 1 ln x 2 2 dx ln 2 x 2 I1 , x x x
ku ln x u ln x dx 1 dx 1 I1 2 dx du x 1 ln x ln x x 2 dx x x x x x v x 1
1 1 ln x . x x
Pra, I
1 2 1 1 1 ln x 2 ln x (ln 2 x 2 ln x 2) C. x x x x
ln( x 1) u x3 x2 dx du 6. ( x 2 x ) ln( x 1)dx ln( x 1) x 1 2 3 x3 x2 v 3 2
x 3 x 2 dx x3 x2 2x 3 3x 2 dx ln( x 1) 2 x 1 3 2 6( x 1) 3
INTEGRALI I PACAKTUAR
4
x3 x2 ln( x 1) I1 , 2 3
ku I1
2x 3 3x 2 2x 3 2x 2 x 2 2 x 2 ( x 1) dx dx dx 6( x 1) 6( x 1) 6 x 1
1 x2 1 1 1 dx x 2 dx x 1 dx 6 x 1 3 6 x 1
x3 1 x2 x ln( x 1) . 9 6 2
Pra x3 x2 x3 x2 x 1 I ln( x 1) ln( x 1) 2 9 12 6 6 3 x3 x2 1 x3 x2 x ln( x 1) C. 2 6 9 12 6 3 ln( x 1 x 2 ) u ( x 1 x 2 )'
dx du x 1 x2 1 1 2 x 1 dx du 7. ln( x 1 x 2 )dx x 1 x2 dx du x2 1 v dx v x
x ln( x 1 x 2 )
xdx 2
x 1
x ln( x 1 x 2 ) 1 x 2 C.
ANALIZA MATEMATIKE II
5
1x u 1x 1x 2 x3 1 x x3 2 dx 2 dx du ln 2 dx 8. x 2 ln 1x 3 1x 3 x 1 x 1 x3 v 3 ln
x3 1 x 2 x x3 1 x 2 x2 ln x 2 dx ln 3 1x 3 3 1x 3 2 x 1
2 1 x3 1 x x2 1 ln| x 2 1| ln ln| x 2 1| C. 3 2 3 1x 3 3
Detyra për ushtrime Të njehsohen integralet:
4.
2
x ln 2 x dx .
5.
ln x x dx.
6.
ln x dx . x3
Të njehsohen integralet: 9.
xe
x
dx .
10.
x e
3 x2
dx .
11.
e
x
dx .
12.
e
x ln x
dx .
Zgjidhja. x u
9.
x xe dx
dx du v e x dx
xe x e x dx xe x e x C.
v e x x2 u 2xdx du
10.
x e
3 x2
dx x 2 x e x dx v xe x 2 dx 2
v
2 1 2 x2 1 x e 2xe x dx 2 2
1 x2 e 2
2 2 1 2 x2 1 1 2 1 2 x e xe x dx x 2 e x e x e x ( x 2 1) C. 2 2 2 2
INTEGRALI I PACAKTUAR
6
1
x u
11.
x e dx
2 xe
x
x
x
e
x
2e
dx
x t e dx v dx dt 2 et dt 2e x 2 x dx 2dt x
dx
x
dx du
2 x
2 x
x
x
2 xe
e
x
x
dx 2 xe
x
2e
x
x
2( x 1)e x .
12.
e
x ln x
dx e x e ln x dx xe x dx
x u, dx du ve
x
xe x e x dx
xe x e x e x ( x 1) C.
Detyra për ushtrime Të njehsohen integralet: 7.
x
10.
2 x
e dx .
x 3 ln x e dx.
8. 11.
x
n x n 1
e
9.
dx .
x2 e x dx.
e
12.
2 x 1
x
2
dx .
3x dx .
Të njesohen integralet 13.
e
15.
x sin 2x dx.
2x
( x 2 3x 4) dx .
14.
(x
16.
x
2
2
x 1) sin x dx.
cos 5x dx .
Zgjidhja. x 2 3x 4 u 13. e 2 x ( x 2 3x 4)dx (2x 3)dx du v e 2 x dx v
1 2x e 2
1 2x 2 e ( x 3x 4) 2
ANALIZA MATEMATIKE II
7
1 2x 1 1 e (2x 3)dx e 2 x ( x 2 3x 4) I1 . 2 2 2 2x 3 u
1 1 I1 e (2x 3)dx 2dx du e 2 x (2x 3) 2 e 2 x dx 2 2 1 v e2x 2 2x
1 2x 1 1 e (2x 3) e 2 x dx e 2 x (2x 3) e 2 x . 2 2 2
Përfundimisht, I
1 2x 2 1 1 e ( x 3x 4) e 2 x (2x 3) e 2 x 2 4 4 1 1 2x e (2x 2 6x 8 2x 3 1) e2 x ( x 2 2x 6) C. 4 2 x2 x 1 u
14.
2 ( x x 1) sin xdx
(2x 1)dx du v sin xdx
cos x ( x 2 x 1)
v cos x cos x (2x 1)dx ( x 2 x 1) cos x I1 ,
ku
I1 (2x 1) cos xdx
2x 1 u 2dx du v cos xdx
(2x 1) sin x 2 sin xdx
v sin x (2x 1) sin x 2 cos x .
Pra, I ( x 2 x 1) cos x (2x 1) sin x 2 cos x (1 x 2 x ) cos x (2x 1) sin x C.
INTEGRALI I PACAKTUAR
8
x u dx du
15.
x sin 2xdx v sin 2xdx v
1 cos 2x 2
x cos 2x 1 sin 2x C. 2 4 x2 u
16.
2 x cos 5x dx 2xdx du 1 v sin 5x 5
x cos 2x 1 cos 2xdx 2 2
x 2 sin 5x 2 x sin 5x dx 5 5
x 2 sin 5x 2 I1 , ku 5 5
x u
1 1 I1 x sin 5xdx dx du x cos 5x cos 5xdx 5 5 1 v cos 5x 5
1 1 x cos 5x sin 5x. 5 25
Pra, I
x 2 sin 5x 2 1 1 x cos 5x sin 5x 5 5 5 25
x 2 sin 5x 2 2 x cos 5x sin 5x 5 25 125
x2 2 2 x cos5x C. sin 5x 25 5 125
ANALIZA MATEMATIKE II
9
Detyra për ushtrime Të njehsohen integralet: 13.
x
15.
(x
2
sin 3x dx. 2
x 1) cos 2x dx .
14.
( x 1) cos( x 2) dx.
16.
x
x
( x 2 x 3)dx .
Të njehsohen integralet 17.
e
x
20.
e
x
sin x dx.
23.
e
x
cos2 x dx .
cos x dx .
18.
sin(ln x ) dx.
19.
cos(ln x ) dx.
21.
e
22.
e
x
cos x dx .
x
sin 2 x dx .
Zgjidhja. cos x u
17.
e
x
cos x dx sin xdx du e x cos x e x sin x dx v e x
e x cos x I1 ,
ku sin x u I1 e
x
sin x dx cos xdx du e x sin x e x cos x dx v e x
e x sin x I .
Pra, I e x cos x ( e x sin x I ) I e x cos x e x sin x I 2 I e x (sin x cos x ) ex I (sin x cos x ) C. 2 18. sin(ln x )dx. 19. cos(ln x )dx . Në vijim, njëkohësisht do të zgjidhim integralet në detyrat 18, 19. Le të shënojmë: I1 sin(ln x )dx ; I 2 cos(ln x )dx. Zgjidhim I1 . Merret:
INTEGRALI I PACAKTUAR
10
sin(ln x ) u cos(ln x ) cos(ln x ) I1 sin(ln x )dx dx du x sin(ln x ) x dx x x vx x sin(ln x ) cos(ln x )dx x sin(ln x ) I 2 .
Pra, kemi: I1 x sin(ln x ) I 2 gjegjësisht I1 I 2 x sin(ln x )
(1)
Zgjidhim I 2 . Merret: cos(ln x ) u I 2 cos(ln x )dx
sin(ln x ) sin(ln x ) dx du x cos ln x x dx x x vx
x cos(ln x ) sin(ln x ) dx x cos(ln x ) I1 .
Pra, I 2 x cos(ln x ) I1 , gjegjësisht, I1 I 2 x cos(ln x ).
(2)
Nga (1), (2) merret sistemi: I1 I 2 x sin(ln x ) I1 I 2 x cos(ln x )
me zgjidhjen e të cilit merret: x x I1 (sin(ln x ) cos(ln x )), I 2 (sin(ln x ) cos(ln x )). 2 2 x x 20. e sin xdx. 21. e cos xdx . Veprojmë ngjashëm si në detyrat paraprake. Shënojmë: I1 e x sin xdx, Zgjidhim integralin I1 .
I 2 e x cos xdx .
ANALIZA MATEMATIKE II
11
sin x u I1 e sin xdx cos xdx du e x sin x e x cos xdx e x sin x I 2 . x
v ex
Pra, I1 I 2 e x sin x .
(1)
Zgjidhim integralin I 2 . cos x u I 2 e cos xdx sin xdx du e x cos x e x sin xdx x
v ex
e x cos x I1 .
Pra I1 I 2 e x cos x
(2)
Nga (1) dhe (2) merret sistemi: I1 I 2 e x sin x x I1 I 2 e cos x
me zgjidhjen e të cilit merret: I1
ex ex (sin x cos x ), I 2 (sin x cos x ). 2 2
Duke vepruar si më sipër tregoni se:
e
ax
sin bxdx
a sin bx b cos bx ax e a 2 b2
a cos bx b sin bx ax e . a 2 b2 1 cos 2x 1 1 22. e x sin 2 xdx e x dx e x dx e x cos 2xdx 2 2 2 1 x 1 cos 2x 2 sin 2x x e e C. 2 2 5 Shënim. Shprehja e fundit mund të transformohet në ex (sin 2 x 2 sin x cos x 2) C. 5 1 cos 2x 1 1 23. e x cos2 xdx e x dx e x dx e x cos 2xdx 2 2 2
e
ax
cos bxdx
INTEGRALI I PACAKTUAR
12
1 x 1 cos 2x 2 sin 2x x e e 2 2 5
1 x cos2 x sin 2 x 4 sin x cos x e 1 2 5
1 x e (cos2 x 2 sin x cos x 2) C. 5
Detyra për ushtrime Të njehsohen integralet: 17.
e
20.
sin
23.
e
2x
x
cos 3x dx .
18.
e
2x
sin 3x dx.
19.
cos (ln x ) dx.
2
21.
e
ax
sin 2 x dx .
22.
e
24.
e
x
(ln x ) dx.
sin 4 x dx.
2
ax
cos2 x dx .
cos4 x dx .
Të njehsohen integralet 25.
27.
x a dx. arctan x dx.
30.
x arc tgx dx.
33.
(1 x
24.
2
2
dx 2 2
)
.
x a dx. x arccos x dx.
26.
31.
2 x (arc tgx ) dx.
32.
34.
x
28.
2
2
2
2
29.
1 x 2 dx . 35.
arcsin x dx. (arcsin x ) dx. 2
x2 (1 x 2 )2 dx. x5 dx . 1 x4
Zgjidhja.
24. I
x 2 a2 u xdx xdx x 2 a 2 dx du x x 2 a 2 x 2 2 x a x 2 a2 vx
x x 2 a2
x 2 a2 a2 x 2 a2
dx
x 2 a2 dx x x 2 a2 dx a 2 2 2 2 x a x a2
x x 2 a 2 x 2 a 2 dx
ANALIZA MATEMATIKE II
dx
a 2
2
x a
13
x x 2 a 2 I a 2 ln|a x 2 a 2 |.
2
Pra 2 I a x 2 a 2 a 2 ln| x x 2 a 2 |
x x 2 a2 a2 ln| x x 2 a 2 | C. 2 2 25. Duke vepruar ngjashëm si në detyrën paraprake merret: I
x 2 a 2 dx
x x 2 a2 a2 ln| x x 2 a 2 | C. 2 2 arcsin x u
26.
1
arcsin xdx
1 x2 vx
dx du x arcsin x
xdx 1 x2
x arcsin x 1 x 2 C. arctan x u
27.
dx
arctan xdx 1 x
2
du x arctan x
xdx 1 x2
vx
x arctan x
1 ln(1 x 2 ) C. 2 arccos x u,
28. I x arccos xdx 2
v
1 1 x2
dx du
x3 3
x3 1 x3 x3 1 arccos x dx arccos x I1 , ku 2 3 3 3 3 1x
I1 x 2
x2 u 2xdx du
x 1x
2
dx
v
xdx 1 x2
v 1 x2
x 2 1 x 2 1 x 2 2dx
INTEGRALI I PACAKTUAR
14
x 2 1 x 2
Pra I
2 (1 x 2 )3 . 3
x3 1 2 arccos x x 2 1 x 2 (1 x 2 )3 3 3 3
C.
(arcsin x )2 u
29.
(arcsin x ) dx 2
2 arcsin x 1 x2 vx
x (arcsin x )2 2 x
dx du
arcsin x 1x
2
dx x (arcsin x )2 2 I1 ,
ku I1 arcsin x
x 1 x2
arcsin x u, dx v
1 x 2 arcsin x 1 x 2
x 1x dx
2
1x
dx 1 x2
du
dx v 1 x 2
2
1 x 2 arcsin x x.
Pra, I x (arcsin x )2 2 1 x 2 arcsin x 2x C.
arctan x u dx x2 x2 dx du arctan x 30. x arctan x dx 2 2 2 1 x2 1x x2 v 2 x2 1 1 arctan x 1 dx 2 2 1 x 2 x2 1 1 x2 1 1 arctan x x arctan x arctan x x C. 2 2 2 2 2
ANALIZA MATEMATIKE II
15
(arctan x )2 u 2 arctan x dx du 31. I x (arctan x )2 dx 1 x2 x2 v 2 2 2 x x arctan x x2 (arctan x )2 dx (arctan x )2 I1 , 2 2 1 x2 1 arctan x u, dx du 2 x 1 x2 ku I1 arctan x dx 1 x2 x2 v dx v x arctan x 1 x2 ( x arctan x ) ( x arctan x ) arctan x dx 1 x2 1 x arctan x x arctan x (arctan x )2 dx dx 2 2 1x 1 x2 1 (arctan x )2 x arctan x (arctan x )2 ln(1 x 2 ) 2 2 1 1 x arctan x (arctan x )2 ln(1 x 2 ). 2 2 Pra x2 1 1 I (arctan x )2 x arctan x (arctan x )2 ln(1 x 2 ) 2 2 2 2 x 1 1 (arctan x )2 x arctan x ln(1 x 2 ) C. 2 2 x u dx du 2 x xdx 1 x 1 dx xdx 32. dx x v 2 2 2 2 2 2 2 21x 2 1 x2 (1 x ) (1 x ) (1 x ) v
1 1 2 1 x2
1 x 1 arctan x C. 2 21x 2 dx 1 x2 x2 1 x2 x2 dx dx 33. I (1 x 2 )2 (1 x 2 )2 dx (1 x 2 )2 (1 x 2 )2
INTEGRALI I PACAKTUAR
16
dx x2 1 x 2 (1 x 2 )2 dx. x2 dx e zgjidhëm në detyrën paraprake, prandaj Integrali (1 x 2 )2 1 x 1 arctan x 1 x I arctan x arctan x C. 2 21x 2 2 2 1 x2 x u
dx du
34. I x 2 1 x 2 dx x x 1 x 2 dx v x 1 x 2 dx v
(1 x 2 ) 1 x 2 3
x 1 (1 x 2 ) 1 x 2 (1 x 2 ) 1 x 2 dx 3 3 x 1 1 (1 x 2 ) 1 x 2 (1 x 2 )dx x 2 1 x 2 dx 3 3 3 x 1 1 (1 x 2 ) 1 x 2 I1 I , 3 3 3
ku
I1
Pra
1 x 2 dx
x 1 1 x 2 ln| x 1 x 2 |. 2 2
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2dt ; sin x 1 t2
dx
1 t2 .# 1 t2
2t ; cos x 1 t2
=6:6&-6#
#
#
2dt 1 t2 ³ 2t 1 t 2 3 1 t2 1 t2
dx
³ sin x cos x 3 2³
dt 4t 2t 2 2
³ 2t
=##O'#!!
³ 3sin
!
3³
#
³
sin x 3cos x dx 2 x 5 cos 2 x
1 t2 2 dt 1 t 1 t2 4 t 1 1 t2 (1 t 2 ) 2 (1 t 2 ) 2 sin xdx ! 2 x) 5 cos 2 x
³ 3(1 cos
cos xdx 3sin x 5(1 sin 2 x)
³
2
t2 ³ (t 2 1)(t 4 1)dt. !
sin xdx
³ 3 2 cos
2
x
3³
cos xdx # 5 2 sin 2 x
d (cos x) d (sin x) 3³ .# 2 3 2 cos x 5 2 sin 2 x
2?##P'##!
2?#!
³ cos x cos a . # #
#
#
#
#
#
22#!
23##
dx sin 2 xdx . ## # # # # 24#! ³ 3sin x cos x 1 ³ sin 2 x cos4 x . #
dx
!
dx
³ sin x 2 cos x 5. #
sin 2 x cos 4 x sin x 2 cos x 25#! ³ dx. # dx. # # # # # 2;## ³ 4 4 3sin 2 x 7 cos 2 x cos x sin x 2#!
³
cos n1 sin n1
xa 2 dx, # # # xa 2
n
xa · § ¨ sin 2 ¸ 3?## ³ ¨ ¸ dx, n N . ! x a ¨ sin ¸ 2 ¹ © 32#!T'2&%&1/0#,%#
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