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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Matematická analýza 1
2018/2019
12. Neurčitý integrálRiešené príklady
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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.
beerb@frcatel.fri.uniza.sk http://frcatel.fri.uniza.sk/ beerb
mailto:beerb@frcatel.fri.uniza.skhttp://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
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