804
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 Matematická analýza 1 2018/2019 12. Neurčitý integrál Riešené príklady [email protected] http://frcatel.fri.uniza.sk/ beerb

University of ŽilinaObsahZoznam01112131415161718191 Zoznamintegrálov–príklady001–100 001. Z dx sinx 002. Z dx cosx 003. Z dx 1+sinx 004. Z dx 1+cosx 005. Z dx sinhx 006. Z dx

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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

    [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