university of Žilinaobsahzoznam01112131415161718191 zoznamintegrálov–príklady001–100 001. z...
TRANSCRIPT
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Matematická analýza 1
2018/2019
12. Neurčitý integrálRiešené príklady
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Obsah – príklady 001–100
1 Riešené príklady 001–0102 Riešené príklady 011–0203 Riešené príklady 021–0304 Riešené príklady 031–0405 Riešené príklady 041–0506 Riešené príklady 051–0607 Riešené príklady 061–0708 Riešené príklady 071–0809 Riešené príklady 081–09010 Riešené príklady 091–100
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Zoznam integrálov – príklady 001–100
001.
∫dxsin x 002.
∫dx
cos x 003.
∫dx
1+sin x 004.
∫dx
1+cos x 005.
∫dx
sinh x 006.
∫dx
cosh x 007.
∫dx
1+sinh x 008.
∫dx
1+cosh x 009.
∫sin2 x dx 010.
∫cos2 x dx 011.
∫sin3 x dx
012.
∫sin2n+1 x dx 013.
∫sinn x dx 014.
∫cos3 x dx 015.
∫cos2n+1 x dx 016.
∫cosn x dx 017.
∫sinh2 x dx 018.
∫cosh2 x dx 019.
∫sinh3 x dx
020.
∫sinh2n+1 x dx 021.
∫sinhn x dx 022.
∫cosh3 x dx 023.
∫cosh2n+1 x dx 024.
∫coshn x dx 025.
∫tg2 x dx 026.
∫tg3 x dx 027.
∫cotg2 x dx
028.
∫cotg3 x dx 029.
∫tgh2 x dx 030.
∫cotgh2 x dx 031.
∫cos x
4+3 sin x dx 032.∫
1+3 sin x+2 cos xsin 2x dx 033.
∫dx
sin2 x cos2 x 034.
∫[tg x + cotg x ] dx
035.
∫dx
a2 cos2 x+b2 sin2 x 036.
∫dx
a2 cos2 x−b2 sin2 x 037.
∫dx
4 cos2 x+sin2 x 038.
∫dx
4 cos2 x−sin2 x 039.
∫x2 dxsin x3 040.
∫x2 sin x3 dx 041.
∫cos x dx3√sin2 x
042.
∫sin x dx√cos5 x
043.
∫cos x−sin xcos x+sin x dx 044.
∫ln cos xsin2 x dx 045.
∫dx
sin x cos x 046.
∫dx
cos x+sin x 047.
∫(sin x−cos x) dx
4√sin x+cos x048.
∫1−tg x1+tg x dx 049.
∫cosn ax ·sin ax dx 050.
∫sin ax dxcosn ax
051.
∫sinn ax ·cos ax dx 052.
∫cos ax dxsinn ax 053.
∫sin ax ·cos bx dx 054.
∫cos ax ·cos bx dx 055.
∫sin ax ·sin bx dx 056.
∫sin ax ·cos ax dx
057.
∫cos2 ax dx 058.
∫sin2 ax dx 059.
∫x tg2 x dx 060.
∫x cotg2 x dx 061.
∫x sin ax dx 062.
∫x cos ax dx 063.
∫x2 sinh ax dx 064.
∫x2 sin ax dx
065.
∫x2 cosh ax dx 066.
∫x2 cos ax dx 067.
∫xn sin ax dx 068.
∫xn cos ax dx 069.
∫x3 sin ax dx 070.
∫x3 cos ax dx 071.
∫ √1+ 1sin x dx
072.
∫arctg x dx 073.
∫x arctg x dx 074.
∫ln x dx 075.
∫ln arctg x
x2+1 dx 076.∫
ln (x−1)5 dx 077.∫
ln xx dx 078.
∫dx
x ln x 079.
∫x2 ln
√1−x 080.
∫ln x√
x dx
081.
∫x ln x dx 082.
∫x2 ln x dx 083.
∫xn ln x dx 084.
∫(x +1)2 ln (x−1)5 dx 085.
∫xx (ln x +1) dx 086.
∫x ln2 x dx 087.
∫ln (x2+1) dx
088.
∫ln(√
1 + x +√1− x
)dx 089.
∫ln(
x +√
x2+1)
dx 090.∫
x(x−a)(x−b) dx 091.∫|x | dx 092.
∫min
x∈(0;∞)
{1, 1x}
dx 093.∫
dx5+4 ex
094.
∫dx√5+4 ex 095.
∫dx√
e2x + ex +1096.
∫ √1−ex1+ex dx 097.
∫(x + 1) ex dx 098.
∫x2 eax dx 099.
∫x8 eax dx 100.
∫xn ex dx
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 001∫dxsin x = ln
∣∣tg x2 ∣∣+ c1 = 12 ln 1−cos x1+cos x + c2=
[UGS: t =tg x2 x ∈(−π + 2kπ; 0 + 2kπ), t∈(−∞; 0) sin x 6=0dx = 2 dtt2+1 sin x =
2t2+1 x ∈(0 + 2kπ;π + 2kπ), t∈(0;∞) x 6=kπ, k∈Z
]=∫
2 dtt2+12t
t2+1=∫
dtt
= ln |t|+ c1 = ln∣∣tg x2 ∣∣+ c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
dx2 sin x2 cos
x2
=[Subst. t = x2 x ∈(0+kπ;π+kπ) sin x 6=0, k∈Z
dt = dx2 t∈(0+ kπ2 ;
π2 +
kπ2)
x 6=kπ, t 6= kπ2
]=∫
dtsin t cos t
=∫
(cos2t+sin2t) dtsin t cos t =
∫cos t dtsin t −
∫− sin t dtcos t = ln |sin t|−ln |cos t|+c1
= ln |tg t|+c1 = ln∣∣tg x2 ∣∣+c1, x ∈R−{kπ, k∈Z}, c1∈R.
=∫
sin x dxsin2 x =
∫sin x dx1−cos2 x =
[Subst. t =cos x x ∈(0+kπ;π+kπ) sin x 6=0dt =− sin x dx t∈(−1; 1) x 6=kπ, k∈Z
]=∫− dt1−t2 =
∫dt
t2−1
= 12 ln∣∣ t−1
t+1∣∣+ c2 = 12 ln 1−t1+t + c2 = 12 ln 1−cos x1+cos x + c2,
x ∈R−{kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 002∫dxcos x = ln
∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1 = 12 ln 1+sin x1−sin x + c2=
[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞; 1) x ∈
(π2 +2kπ;π+2kπ
), t∈(1;∞)
dx = 2 dtt2+1 cos x =1−t2t2+1 x ∈
(−π2 +2kπ;
π2 +2kπ
), t∈(−1; 1) cos x 6=0, x 6= π2 +kπ, k∈Z , t 6=±1
]
=∫
2 dtt2+11−t2t2+1
= −∫
2 dtt2−1 = −
22 ln∣∣ t−1
t+1∣∣+c1 = ln ∣∣ t+1t−1 ∣∣+c1 = ln ∣∣∣ tg x2 +1tg x2−1 ∣∣∣+c1,x ∈R −
{π2 +kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
cos x dxcos2 x =
∫cos x dx1−sin2 x =
[Subst. t =sin x x ∈(−π2 +kπ; π2 +kπ) cos x 6=0dt =cos x dx t∈(−1; 1) x 6= π2 +kπ, k∈Z
]
=∫
dt1−t2 = −
∫dt
t2−1 = −12 ln∣∣ t−1
t+1∣∣+ c2 = − 12 ln 1−t1+t + c2
= 12 ln1+t1−t + c2 =
12 ln
1+sin x1−sin x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 003∫dx
1+sin x = c1 −2
tg x2 +1= sin x−1cos x + c2
=[UGS: t =tg x2 x ∈
(−π+2kπ;−π2 +2kπ
), t∈(−∞;−1) sin x 6=−1, k∈Z
dx = 2 dtt2+1 sin x =2t
t2+1 x ∈(−π2 +2kπ;π+2kπ
), t∈(−1;∞) x 6=−π2 +2kπ
]=∫
2 dtt2+1
1+ 2tt2+1
=∫
2 dtt2+1
t2+1+2tt2+1
=∫
2 dt(t+1)2 =
[Subst. u = t+1 t∈(−∞;−1), u∈(−∞; 0)du =dt x ∈(−1;∞), t∈(0;∞)
]=2∫
duu2 =2
∫u−2 du
=2 u−1
−1 + c1=c1 −2u =c1 −
2t+1 =c1 −
2tg x2 +1
,x ∈R−
{−π2 +2kπ, π+2kπ, k∈Z
}, c1∈R.
=∫
(1−sin x) dx(1−sin x)(1+sin x) =
∫(1−sin x) dx1−sin2 x =
∫(1−sin x) dx
cos2 x =∫
dxcos2 x +
∫− sin x dxcos2 x
=[Subst. t =cos x x ∈
〈0+2kπ; π2 +2kπ
), t∈(0; 1〉 x ∈
(π2 +2kπ;π+2kπ
〉, t∈〈−1; 0) cos x 6=0
dt =− sin x dx x ∈〈π+2kπ; π2 +2kπ
), t∈〈−1; 0) x ∈
( 3π2 +2kπ;π+2kπ
〉, t∈(0; 1〉 x 6= π2 +kπ, k∈Z
]
=∫
dxcos2 x +
∫dtt2 =
∫dx
cos2 x +∫
t−2 dt =tg x + t−1
−1 + c2=tg x −1t + c2
= sin xcos x −1
cos x + c2 =sin x−1cos x + c2, x ∈R −
{π2 +kπ, k∈Z
}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 004∫dx
1+cos x = tgx2 + c1 =
1−cos xsin x + c2
=[UGS: t =tg x2 x ∈(−π+2kπ;π+2kπ), t∈Rdx = 2 dtt2+1 cos x =
1−t2t2+1 cos x 6=−1, x 6=π+2kπ, k∈Z
]=∫
2 dtt2+1
1+ 1−t2t2+1
=∫
2 dtt2+12
t2+1=∫
dt
= t + c1 = tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
dx2 cos2 x2
=[Subst. t = x2 x ∈(−π+2kπ;π+2kπ) cos x 6=−1, k∈Z
dt = dx2 t∈(−π2 +kπ;
π2 +kπ
)x 6=π+2kπ
]=∫
dtcos2 t = tg t + c1
= tg x2 + c1, x ∈R − {π+2kπ, k∈Z}, c1∈R.
=∫
(1−cos x) dx(1−cos x)(1+cos x) =
∫(1−cos x) dx1−cos2 x =
∫(1−cos x) dx
sin2 x =∫
dxsin2 x −
∫cos x dxsin2 x
=[Subst. t =sin x x ∈
(−π+2kπ;−π2 +2kπ
〉, t∈〈−1; 0) x ∈
〈−π2 +2kπ; 0+2kπ
), t∈〈−1; 0) sin x 6=0
dt =cos x dx x ∈(0+2kπ; π2 +2kπ
〉, t∈(0; 1〉 x ∈
〈π2 +2kπ;π+2kπ
), t∈(0; 1〉 x 6=kπ, k∈Z
]
=∫
dxsin2 x−
∫dtt2 =
∫dx
sin2 x−∫
t−2 dt = − cotg x − t−1
−1 + c2 =1t − cotg x + c2
= 1sin x −cos xsin x + c2 =
1−cos xsin x + c2, x ∈R − {π+kπ, k∈Z}, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 005∫dx
sinh x = ln∣∣tgh x2 ∣∣+c1 = 12 ln cosh x−1cosh x+1 +c2 = 12 ln ∣∣ ex −1ex +1 ∣∣+c3
=[UHS: t =tgh x2 x ∈(−∞; 0), t∈(−1; 0) sinh x 6=0dx = 2 dt1−t2 sinh x =
2t1−t2 x ∈(0;∞), t∈(0; 1) x 6=0
]=∫ 2 dt
1−t22t
1−t2=∫
dtt = ln t+c1
= ln∣∣tgh x2 ∣∣+c1, x ∈R−{0}, c1∈R.
=∫
sinh x dxsinh2x =
∫sinh x dxcosh2x−1 =
[ Subst. t =cosh x x ∈(−∞; 0), t∈(1;∞) sinh x 6=0dt =sinh x dx x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
dtt2−1
= 12 ln∣∣ t−1
t+1∣∣+c2= 12 ln t−1t+1 +c2= 12 ln cosh x−1cosh x+1 +c2, x ∈R−{0}, c2∈R.
=∫
dxex − e−x
2=∫
2 dxex − e−x =
[Subst. t =ex x = ln t x ∈(−∞; 0), t∈(0; 1) sinh x 6=0e−x = t−1= 1t dx =
dtt x ∈(0;∞), t∈(1;∞) x 6=0
]=∫
2 dtt(t− 1t )
=∫
2 dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+c3= 12 ln ∣∣ ex −1ex +1 ∣∣+c3, x ∈R−{0}, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 006∫dx
cosh x = 2 arctg tghx2 +c1 = arctg sinh x +c2 = 2 arctg e
x +c3
=[UHS: t =tgh x2 x ∈(−∞; 0〉, t∈(−1; 0〉dx = 2 dt1−t2 cos x =
1+t21−t2 x ∈〈0;∞), t∈〈0; 1)
]=∫ 2 dt
1−t21+t21−t2
=∫
2 dtt2+1 =2 arctg t+c1
=2 arctg tgh x2 +c1, x ∈R, c1∈R.
=∫
cosh x dxcosh2x =
∫cosh x dxsinh2x+1 =
[Subst. t =sinh x x ∈Rdt =cosh x dx t∈R
]=∫
dtt2+1 =arctg t+c2
= arctg sinh x +c2, x ∈R, c2∈R.
=∫
dxex + e−x
2=∫
2 dxex + e−x =
[Subst. t =ex x = ln t x ∈Re−x = t−1= 1t dx =
dtt t∈(0;∞)
]=∫
2 dtt(t+ 1t )
=∫
2 dtt2+1
=2 arctg t+c3=2 arctg ex +c3, x ∈R, c3∈R.
[email protected] http://frcatel.fri.uniza.sk/ beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01 02 03 04 05 06 07 08 09 10
Riešené príklady – 007∫dx
1+sinh x =√22 ln
∣∣∣ tgh x2−1+√2tgh x2−1−√2 ∣∣∣+ c1 = √22 ln ∣∣∣ ex +1−√2ex +1+√2 ∣∣∣+c2=[UHS: t =tgh x2 x ∈
(−∞; ln (
√2−1)
), t∈
(−1;√2−1
)sinh x 6=−1
dx = 2 dt1−t2 sinh x =2t
1−t2 x ∈(ln (√2−1);∞
), t∈
(√2−1; 1
)x 6= ln (
√2−1)
]=∫ 2 dt
1−t2
1+ 2t1−t2
=∫ 2 dt
1−t21−t2+2t1−t2
=∫−2 dt
t2−2t−1 =∫−2 dt
(t−1)2−2 =−∫
2 dt(t−1)2−(
√2)2 =−
22√2 ln∣∣∣ t−1−√2t−1+√2 ∣∣∣+c1
= 1√2 ln∣∣∣t−1+√2t−1−√2 ∣∣∣+c1= √22 ln ∣∣∣tgh x2−1+√2tgh x2−1−√2 ∣∣∣+c1, x ∈R−{ln(√2−1)}, c1∈R.
=∫
dx1+ ex−e−x2
=∫
2 dx2+ex−e−x =
[Subst