Uzdevumi 1 2 sem


R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
1. nodarb%2łba
6
3x - 2 x5 + 5 " x
1. piemrs. Apr7int dx .
+"
x3
5 1
6 6
# ś#
# -3 -3 ś#
3x - 2 x5 + 5 " x 3x 2 x5 5 " x
1-3
2 6
dx = - +
+" +"ś# x3 x3 x3 ź#dx = +"ś#3x - 2x + 5x ź#dx =
ś# ź#
ś# ź#
x3
# #
# #
1 11
-
1 17
2 6
- -
x-1 x x 3 30
2 6
= 3 x-2dx - 2 x dx + 5 x dx = 3" - 2 " + 5 " + C = - - 4 x - + C
+" +" +"
1
-1 x
-116
116 x11
2
2. piemrs. Apr7int (5x + 2- x sin x)" 2x dx .
+"
(5x + 2- x sin x)" 2x dx = (5x " 2x + 2- x " 2x sin x)dx = (10x + sin x)dx =
+" +" +"
10x
x
= dx + xdx = - cos x + C .
+"10 +"sin ln10
cos 2x
3. piemrs. Apr7int dx .
+"
sin2 x " cos2 x
2 2
# ś#
cos 2x cos2 x - sin x cos2 x sin x
ź#
dx = dx = -
+" 2 +" 2 +"ś# sin 2 2 ź#dx =
ś#
sin x " cos2 x sin x " cos2 x x " cos2 x sin x " cos2 x
# #
dx dx
= - = -ctg x - tg x + C .
+" +"
sin2 x cos2 x
2 - 9x2 + 4
4. piemrs. Apr7int dx .
+"
9x2 + 4
2 - 9x2 + 4 2dx 9x2 + 4 dx dx
dx = - dx = 2 - =
+" +" +" +" +"
9x2 + 4 9x2 + 4 9x2 + 4 9x2 + 4
9x2 + 4
1 d(3x) 1 d(3x) 2 1 3x 1
2
= 2 " - = " arctg - ln 3x + (3x) + 22 + C =
+" 2 +"
2
3 3
(3x) + 22
(3x) + 22 3 2 2 3
1. nodarb%2łba. 1. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
1 3x 1
= arctg - ln 3x + 9x2 + 4 + C .
3 2 3
dx
5. piemrs. Atrast .
+"
2x -1
dx d(2x -1) 1 d(2x -1) 1
= = = ln 2x -1 + C .
+" +" +"
2x -1 (2x -1)" 2 2 2x -1 2
dx
6. piemrs. Atrast .
+" 5
(2x -1)
-4
dx d(2x -1) 1 1 (2x -1)
-5 -5
= + C =
+" 5 +"(2x -1) 2 = 2 +"(2x -1) d(2x -1)= 2 "
- 4
(2x -1)
1
= - + C .
4
8(2x -1)
7. piemrs. Atrast
+"sin(5x + 4)dx .
+ 4) 1 1
+"sin(5x + 4)dx =+"sin(5x + 4)d(5x = 5 +"sin(5x + 4)d(5x + 4) = - cos(5x + 4)+ C .
5 5
8. piemrs. Atrast
+"e3x + 7dx .
d(3x + 7)= 1
e3x + 7 + C .
+"e3x + 7dx =+"e3x + 7
3 3
dx
9. piemrs. Apr7int .
+"
x ln x
dx 1 dx d(ln x)=[izmantojot integraanas pamatformulu
= " =
+" +" +"
x ln x ln x x ln x
du
= ln u + C ] = ln ln x + C .
+"
u
1. nodarb%2łba. 2. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
cos xdx
10. piemrs. Apr7int .
+"1+ sin 2
x
cos xdx d(sin x)
= = [izmantojot integraanas pamatformulu
+"1 + sin 2 +"1 + sin 2
x x
du
+"1+ u2 = arctgu + C ] = arctg(sin x)+ C .
11. piemrs. Apr7int x dx .
+"tg
sin xdx d(cos x)
= - = [izmantojot integraanas pamatformulu
+"tg x dx = +" +"
cos x cos x
du
= ln u + C ]= - ln cos x + C .
+"
u
dx
12. piemrs. Apr7int .
+" 2
4sin x + 9cos2 x
dx
= [skait%2łtju un saucju izdal%2łsim ar cos2 x ] =
+" 2
4sin x + 9cos2 x
dx
d(tg x) 1 d(tg x)
cos2 x
= = = = [izmantojot integraanas
+" 2 +" +" 2
4tg x + 9 4tg2 x + 9 4
3
# ś#
tg2 x +
ś# ź#
2
# #
du 1 1 tg x 1 2tg x
pamatformulu = arctg u + C ] = " " arctg + C = " arctg# ś# + C .
ś# ź#
+"1+ u 2
3 3
4 6 3
# #
2 2
dx
13. piemrs. Apr7int .
+"
sin x cos x
dx
= [skait%2łtju un saucju izdal%2łsim ar sin2 x ] =
+"
sin x cos x
dx
2
d(ctg x) = [izmantojot integraanas pamatformulu
sin x
= = -
+" +"
sin x cos x
ctg x
2
sin x
1. nodarb%2łba. 3. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
du
= ln u + C ] = - ln ctg x + C .
+"
u
2
arcsin xdx
14. piemrs. Apr7int .
+"
1 - x2
2
arcsin xdx
2
=
+" +"arcsin xd(arcsin x)=[izmantojot integraanas pamatformulu
1 - x2
ua+1 arcsin3 x
a
+"u du = a +1 + C, a `" -1] = 3 + C .
arctg xdx
15. piemrs. Apr7int .
+"
1 + x2
arctg xdx
= arctg xd(arctg x) = [izmantojot integraanas pamatformulu
+" +"
1 + x2
3
2
ua+1 (arctg x) 2
a
+"u du = a +1 + C, a `" -1] = 3 + C = 3 arctg3x + C .
2
1 1
16. piemrs. Apr7int tg dx .
+"
x2 x
1 1 1 1 1 1
tg dx = - tg " d# ś# = [aeit ievrojam, ka d# ś# = - dx ] =
ś# ź# ś# ź#
+" +"
x2 x x x x x2
# # # #
1 1
sin d cos
1
x x
= - d# ś# = = [izmantojot integraanas pamatformulu
ś# ź#
+" +"
1 1
x
# #
cos cos
x x
du 1
= ln u + C ] = ln cos + C .
+"
u x
xdx
17. piemrs. Apr7int .
+"
3
3x +1
1. nodarb%2łba. 4. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
`aj gad%2łjum ir jintegr iracionla funkcija, un nav reizintja, kuru panesot zem
diferenci<a z%2łmes, ms nonktu pie kdas no integraanas pamatformulm.
Atbr%2łvosimies no iracionalittes ar substitkcijas pal%2łdz%2łbu, apz%2łmjot 3x +1 = t3 . T
varam rakst%2łt:
3
t -1
2
2 2
" t dt
3x +1 = t3, 3dx = 3t dt, dx = t dt,
xdx 1
3 2
= = = (t -1)" tdt =
t3 -1
+" 3 +" +"
3
x = , t = 3x +1 t 3
3x +1
3
1 xa+1
= (t3 - t)dt = [izmantojot integraanas pamatformulu xadx = + C ] =
+" +"
3 a +1
4 2
# ś#
1 t t 1 1
4 2
3 3
ś# + C = [prejot uz skotnjo main%2łgo] = (3x +1) - (3x +1) + C .
= - ź#
ś# ź#
3 4 2 12 6
# #
dx
18. piemrs. Apr7int .
+"
ex -1
R%2łkosimies l%2łdz%2łgi k ieprieka. Lai atbr%2łvotos no kvadrtsaknes zemintegr<a funkcijas
2
saucj, izvlsimies substitkciju ex -1 = t . T iegksim
2tdt
2 2
t = ex -1, ex -1 = t , ex = t +1
2
dx 2tdt dt
t +1
= = = =
2tdt
+" 2 +" +" 2 2
x = ln(t +1), dx = t t(t +1)= 2+" t +1
ex -1
2
t +1
= 2arctgt + C = 2arctg ex -1 + C ,
du
jo atbilstoa integraanas pamatformula ir
+"1+ u2 = arctg u + C un t = ex -1 .
ln xdx
19. piemrs. Apr7int .
+"
x 1- ln2 x
R7inot ao uzdevumu, izmantosim gan main%2łga reizintja neaanu zem diferenci<a
z%2łmes, gan pieraksta rt%2łbas d< zemintegr<a funkcijas da<as apz%2łmaanu ar jaunu
main%2łgo. Redzam, ka zemintegr<a funkcija viegli prvraama par funkciju no funkcijas
1 1
ln x , ja reizintju nesam zem diferenci<a z%2łmes: dx = d(ln x). Varam rakst%2łt
x x
ln xdx ln xd(ln x) = [ ln x = t - ao aizvietoaanu veicam rt%2łbas d< ] =
=
+" +"
x 1- ln2 x 1- ln2 x
1. nodarb%2łba. 5. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
2 2
tdt d(- t +1) ] = - 1 d(- t +1) =
2
= = [izmantojam d(1- t )= -2tdt jeb dt =
+" +"
2 2
2t 2
1- t 1- t
1
1 -
2 2
2
= - (1- t ) d(1- t ) = [izmantojot integraanas pamatformulu
+"
2
2
ua+1 1 1- t
a 2
+ C = - 1- t + C = [prejam uz skotnjo
+"u du = a +1 + C, a `" -1] = -
1
2
2
main%2łgo x ] = - 1- ln2 x + C .
7cos3 xdx
20. piemrs. Apr7int .
+"
3
sin2 x
`aj uzdevum izmantosim to, ka cos xdx = d(sin x) un to, ka cos2 x = 1- sin2 x . L%2łdz
ar to ms zemintegr<a izteiksmi, l%2łdz%2łgi k ieprieka, varam prvrst par funkcijas sin x
funkciju. Tlk sin x apz%2łmsim ar t pieraksta %2łsumam.
2 2
7 cos3 xdx cos2 x " cos xdx (1- sin x)d(sin x) 1- t
= 7 = 7 = [sin x = t] = 7 dt =
+" +" +" +" 2
3 2 3 3 2
sin x sin2 x sin x
3
t
2 4
-
ua+1
a
3 3
= 7 dt - 7 dt =[izmantojot integraanas pamatformulu du = + C,a `" -1]
+"t +"t +"u a +1
1 7
3 3
t t
7
= 7 " - 7 " + C = 213 t - 33 t + C = 213 sin x - 33 sin7 x + C .
1 7
3 3
3dx
21. piemrs. Apr7int .
+"
x x2 + 9
1
`eit izmantosim substitkciju = t .
x
dt
-
2
3dx 1 1 1
t
= = t, x = , dx = - dt = 3 = [ zem saknes esoao
+" 2 +"
x t t
1 1
x x2 + 9
+ 9
2
t t
1. nodarb%2łba. 6. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
dt
dt
2 t
izteiksmi uzrakstm k da<u, kopsaucjs ir t ] = - 3 = - 3 =
+" +"
2 2
1+ 9t 1+ 9t
2
t
=[zem diferenci<a z%2łmes panesam konstanto reizintju 3, ievrojot, ka d(3t) = 3dt ] =
d(3t)
= - = [izmantojot integraanas pamatformulu
+"
2
1+ (3t)
du
2
2
= ln u + 1+ u + C ] = - ln 3t + 1+ (3t) + C = [prejam uz skotnjo
+"
1+ u2
2
1 3 3
# ś#
main%2łgo x, izmantojot t = ] = - ln + 1+ + C .
ś# ź#
x x x
# #
x5dx
22. piemrs. Apr7int .
+"
x3 + 2
Zem diferenci<a z%2łmes tagad nes%2łsim reizintju x2 , t no dx prejot uz dx3, jo
dx3 = 3x2dx .
x5dx 1 x3 "3x2dx 1 x3dx3 1 tdt
= = = [rt%2łbas labad apz%2łmsim x3 = t ] = =
+" +" +" +"
x3 + 2 3 x3 + 2 3 x3 + 2 3 t + 2
= [ pieskait%2łsim un atFemsim zemintegr<a funkcijas skait%2łtjam 2 ] =
1 t + 2 - 2 1 t + 2 2 1 2 dt
# ś#dt
= dt = = [ievrosim, ka
+" +"ś# - ź# = +"dt - +"
3 t + 2 3 t + 2 t + 2 3 3 t + 2
# #
1 2 d(t + 2)
d(t + 2) = dt ] = - = [izmantojot integraanas pamatformulas
+"dt +"
3 3 t + 2
du 1 2
= ln u + C ] = t - ln t + 2 + C = [prejam uz skotnjo
+"du = u + C un +"
u 3 3
1 2
main%2łgo x , izmantojot t = x3 ] = x3 - ln x3 + 2 + C .
3 3
x +1
23. piemrs. Apr7int dx .
+"
x x - 2
1. nodarb%2łba. 7. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
Vispirms novrs%2łsim iracionalitti zemintegr<a funkcijas saucj, izmantojot
2
substitkciju x - 2 = t .
2 2 2
x +1 x - 2 = t , x = t + 2, dx = 2tdt (t + 2 +1)" 2tdt
dx = = = [ ievrojam,
+" +" 2
x - 2 = t (t + 2)t
x x - 2
ka pc main%2łg t sa%2łsinaanas zemintegr<a funkcijas skait%2łtj k saskaitmais ieiet viss
2 2
# ś#
t + 2 +1 t + 2 1
ź#
da<as saucjs, tpc neveicam saskait%2łaanu ] = 2 dt = 2 + =
+" 2 +"ś# t 2 2 ź#dt
ś#
t + 2 + 2 t + 2
# #
dt dt
= 2 + 2 = 2 + 2 = [izmantojot integraanas pamatformulas
+"dt +" 2 +"dt +" 2
2
t + 2
t +( 2)
du 1 u 2 t
= arctg + C ] = 2t + arctg + C = [prejam uz
+"du = u + C un +"
a2 + u2 a a
2 2
x - 2
skotnjo main%2łgo x, ievietojot t = x - 2 ] = 2 x - 2 + 2arctg + C .
2
24. piemrs. Apr7int ln(x - 3)dx .
+"2x
`da veida integrlis jintegr parcili. Savukrt, ja integrlis, kura zemintegr<a
funkcija satur logaritmu, ir jintegr parcili (substitkcijas metode neved pie
atrisinjuma), tad par funkciju u ir jizvlas logaritms.
dx
u = ln(x - 3), du =
x2
dx = g
+"2x ln(x - 3)dx = dv = 2xdx, v = x - 3x2 = x2 ln(x - 3)- +"
x - 3
+"2xdx =
x2
Lai nointegrtu da<u var vispirms izdal%2łt polinomus: to, kas ir skait%2łtj izdal%2łt ar
x - 3
to polinomu, kas ir saucj, proti,
x2 : (x - 3) = x + 3
x2 - 3x
3x
3x - 9
9
x2 9
Esam ieguvuai dal%2łjumu x+3 un atlikumu 9, ttad = x + 3 + .
x - 3 x - 3
1. nodarb%2łba. 8. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
To paau varja iegkt ar%2ł savdk, pieskaitot un atFemot da<as skait%2łtjam 9, t.i.,
x2 x2 - 9 + 9 x2 - 9 9 (x - 3)(x + 3) 9 9
= = + = + = x + 3 + .
x - 3 x - 3 x - 3 x - 3 x - 3 x - 3 x - 3
Varam turpint integraanu.
9
# ś#dx = x2 ln(x - 3)- x2
g= x2 ln(x - 3)- x + 3 + - 3x - 9ln x - 3 + C .
ź#
+"ś#
x - 3 2
# #
2x cos x
25. piemrs. Apr7int dx .
+"
sin3 x
Ja ir parcili jintegr funkcija, kas ir trigonometriskas funkcijas un polinoma
reizinjums, tad par u izvlas polinomu, t k atvasinot polinoma krta samazins.
u = 2x, du = 2dx,
2x cos x
dx = cos xdx cos xdx sin-2 x 1 =
-3
+"
dv = , v = = = -
sin3 x
+"+"sin xd(sin x) =
sin3 x sin3 x - 2 2sin2 x
- 2x 2dx x dx x
= + = - + = - - ctg x + C .
+" 2 +" 2 2
2sin2 x 2sin2 x sin x sin x sin x
26. piemrs. Apr7int (4 - 9x2)cos3xdx .
+"
Tpat k iepriekaj uzdevum par u izvlsimies polinomu, tikai aoreiz parcili bks
jintegr divas reizes.
u = 4 - 9x2 , du = -18xdx
(4 - 9x2)cos3xdx = d(3x) 1 =
+"
dv = cos3xdx, v =
+"cos3xdx = +"cos3x 3 = sin 3x
3
18
= (4 - 9x2)sin 3x + xsin 3xdx =
+"
3 3
u = x, du = dx
= d(3x) 1 =
dv = sin 3xdx, v = 3xdx = 3x = - cos3x
+"sin +"sin 3 3
1 1
ś#
= (4 - 9x2)sin 3x + 6# - x cos3x +
ś#
+"cos3xdxź# =
3 3 3
# #
d(3x)=
= (4 - 9x2)sin 3x - 2x cos3x + 2
+"cos3x 3
3
1. nodarb%2łba. 9. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
2
= (4 - 9x2)sin 3x - 2x cos3x + sin 3x + C =
3 3
4 2
= sin 3x - 3x2 sin 3x - 2x cos3x + sin 3x + C =
3 3
= 2sin 3x - 3x2 sin 3x - 2x cos 3x + C .
arcsin x
27. piemrs. Apr7int dx .
+"
1+ x
Inverso trigonometrisko funkciju gad%2łjum pie parcils integraanas ar u tiek apz%2łmta
invers trigonometrisk funkcija.
dx
u = arcsin x, du =
1- x2
arcsin x
1
dx = =
1
2
+" dx dx (1+ x)
-
1+ x
2
dv = , v = = x) d(1+ x) = = 2 1+ x
+"+"(1+
1
1+ x 1+ x
2
1+ x 1+ x
= 2 1+ x arcsin x - 2 dx = 2 1+ x arcsin x - 2 dx =
+" +"
(1- x)(1+ x)
1- x2
1
dx
-
2
= 2 1+ x arcsin x - 2 = 2 1+ x arcsin x + 2 x) d(1- x) =
+" +"(1-
1- x
1
2
(1- x)
= 2 1+ x arcsin x + 2 " + C = = 2 1+ x arcsin x + 4 1- x + C .
1
2
28. piemrs. Apr7int (x2 + 4)dx .
+"ln
`aj uzdevum ar u apz%2łmsim visu zemintegr<a funkciju.
2x
x2
u = ln(x2 + 4) du = dx
(x2 = x ln(x2 + 4)- 2 dx =
x2 + 4
+"ln + 4)dx = dv = dx +"
x2 + 4
v = x
x2 + 4 - 4 1
= x ln(x2 + 4)- 2 dx = x ln(x2 + 4)- 2 + 8 dx =
+" +"dx +"
x2 + 4 x2 + 4
1. nodarb%2łba. 10. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
1 x x
= x ln(x2 + 4)- 2x + 8" arctg + C = x ln(x2 + 4)- 2x + 4 " arctg + C .
2 2 2
29. piemrs. Apr7int
+"sin(ln x)dx .
Ar%2ł aeit ar u apz%2łmsim visu zemintegr<a funkciju.
1
1
u = sin(ln x) du = cos(ln x)" dx
= xsin(ln x)- x cos(ln x)" dx =
x
+"sin(ln x)dx = dv = dx +"
x
v = x
1
u = cos(ln x) du = -sin(ln x)" dx
= x sin(ln x)- =
x
+"cos(ln x)dx = dv = dx
v = x
# 1 ś#
ś#dxź#
= x sin(ln x)- ś# x cos(ln x)- x# - sin(ln x)" =
ś# ź#
ś# +" ź#
x
# #
# #
= x sin(ln x)- x cos(ln x)-
+"sin(ln x)dx .
Prrakst%2łsim vlreiz, ko esam ieguvuai.
+"sin(ln x)dx = x sin(ln x)- x cos(ln x)- +"sin(ln x)dx .
Redzam, ka iegktaj viend%2łb mekljamais integrlis
+"sin(ln x)dx ieiet gan ts labaj
pus, gan kreisaj. Prnes%2łsim
+"sin(ln x)dx uz viend%2łbas kreiso pusi, t iegkstot:
2
+"sin(ln x)dx = x sin(ln x)- x cos(ln x).
L%2łdz ar to varam rakst%2łt
1
+"sin(ln x)dx = 2 (x sin(ln x)- x cos(ln x))+ C .
Uzdevums 26. Apr7int x2 - 9dx .
+"
3
`o integrli var r7int, izmantojot trigonometrisku substitkciju x = vai
cost
1. nodarb%2łba. 11. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
3 9 1
x = , t iegkstot zemsaknes izteiksmi - 9 =9# -1ś# = 9tg2t pirmaj
ś# ź#
sin t cos2 t cos2 t
# #
9 1
gad%2łjum un - 9 =9# -1ś# =9ctg2t otraj gad%2łjum. `da pieeja <auj mums
ś# ź#
2
sin2 t sin t
# #
izvilkt sakni un zemintegr<a izteiksme prvraas par racionlu da<u. Racionlu da<u
integraana tiks apskat%2łta nkoaaj nodarb%2łb. Bet ir iespjams r%2łkoties ar%2ł citdi, proti,
integrt parcili. T aeit ar%2ł dar%2łsim.
2x xdx
u = x2 - 9 du = dx =
x2 - 9dx = =
+" 2 x2 - 9 x2 - 9
dv = dx v = x
xdx x2dx x2 - 9 + 9
= x x2 - 9 - x " = x x2 - 9 - = x x2 - 9 - dx =
+" +" +"
x2 - 9 x2 - 9 x2 - 9
x2 - 9 9 1
= x x2 - 9 - dx - dx = x x2 - 9 - x2 - 9dx -9 dx =
+" +" +" +"
x2 - 9 x2 - 9 x2 - 9
= x x2 - 9 - x2 - 9dx -9ln x + x2 - 9 .
+"
Izveidojusies ir tieai tda pati situcija k iepriekaj uzdevum. Mekljamais integrlis
x2 - 9dx atrodas viend%2łbas
+"
x2 - 9dx = x x2 - 9 - x2 - 9d - +9ln x + x2 - 9
+" +"
abs puss. Prnesot to uz kreiso pusi, iegkstam
2 x2 - 9dx = x x2 - 9 - 9ln x + x2 - 9 .
+"
Ttad uzdevuma atrisinjums ir
1
# ś#
x2 - 9dx = x x2 - 9 - 9ln x + x2 - 9 + C .
ś# ź#
+"
# #
2
1. nodarb%2łba. 12. lpp. Augstk matemtika.
I. Volodko


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