Skripta - Neurčitý integrál

tgx, vprime(x) = x3
uprime(x) = 11+x2, v(x) = 14x4
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
= x
4
4 arctgx−
1
4
integraldisplay (x4 −1) + 1
x2 + 1 dx
= x
4
4 arctgx−
1
4
integraldisplay parenleftbigg
x2 −1 + 1x2 + 1
parenrightbigg
dx
= x
4
4 arctgx−
1
4
parenleftBiggx3
3 −x + arctgx
parenrightBigg
= x
4 −1
4 arctgx−
x3
12 +
x
4 + c.
13
Pˇr´ıklad 3.4. Vypoˇctˇete integr´al
I =
integraldisplay
ex sinxdx na R.
ˇReˇsen´ı.
integraldisplay
ex sinxdx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
u(x) = ex vprime (x) = sinx
uprime (x) = ex v(x) = −cosx
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle = −ex cosx +
integraldisplay
ex cosxdx
=
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
u(x) = ex vprime (x) = cosx
uprime (x) = ex v(x) = sinx
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle = ex(−cosx + sinx)−
integraldisplay
ex sinxdx.
Tedy
I = ex(sinx−cosx)−I,
a odtud pro x ∈ R
integraldisplay
ex sinxdx = 12ex(sinx−cosx) + c, c ∈ R.
Pozn´amka 3.1. Analogick´ym zp˚usobem lze poˇc´ıtat integr´aly
integraldisplay
eax sinbxdx,
integraldisplay
eax cosbxdx na R,
kde a, b jsou libovoln´e re´aln´e konstanty.
Pˇr´ıklad 3.5. Kombinac´ı prvn´ı substituˇcn´ı metody a metody per partes vypoˇc´ıtejte
integr´aly
I =
integraldisplay
x5 ex2 dx, J =
integraldisplay
arctgxdx na R.
ˇReˇsen´ı.
I =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
x2 = t
2xdx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle =
1
2
integraldisplay
t2 et dt
=
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
u(t) = t2, vprime(t) = et
uprime(t) = 2t, v(t) = et
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle =
t2 et
2 −
integraldisplay
t et dt
=
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
u(t) = t, vprime(t) = et
uprime(t) = 1, v(t) = et
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle =
t2 et
2 −t e
t +
integraldisplay
et dt
= e
x2 (x4 −2x2 + 2)
2 + c;
J =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
u(x) = arctgx, vprime(x) = 1
uprime(x) = 11 + x2, v(x) = x
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle = xarctgx−J1,
14
kde
J1 =
integraldisplay x
1 + x2 dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
1 + x2 = t
x dx = 12 dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle =
1
2 ln|t| =
1
2 ln(1 + x
2) + c.
Celkem tedy integraldisplay
arctgxdx = xarctgx− 12 ln(1 + x2) + c.
Cviˇcen´ı 3.1. Uˇzit´ım substituˇcn´ıch metod spoˇctˇete dan´e integr´aly na dan´ych oborech:
a)
integraldisplay 1
3−4x dx na
parenleftbigg3
4,∞
parenrightbigg
;
b)
integraldisplay sinx
cos4 x dx na
parenleftbigg
−pi2, pi2
parenrightbigg
;
c)
integraldisplay x3
√5 + x2 dx na R;
d)
integraldisplay arcsin3 x
√1−x2 dx na (−1,1);
e)
integraldisplay 1
x2 + 4x + 29 dx na R;
f)
integraldisplay 1
√5−4x−x2 dx na (−5,1).
Cviˇcen´ı 3.2. Uˇzit´ım metody per partes spoˇctˇete dan´e integr´aly na dan´ych oborech:
a)
integraldisplay
xcos(4x + 3) dx na R;
b)
integraldisplay
xsin2 x dx na R;
c)
integraldisplay
logx dx na (0,∞);
d)
integraldisplay
arctg3x dx na R;
e)
integraldisplay
e2x cos5x dx na R;
f)
integraldisplay ln2 x
x3 dx na (0,∞).
4 Integrace racion´aln´ıch funkc´ı.
Jak jiˇz v´ıme z teorie racion´aln´ıch funkc´ı, m˚uˇzeme zadanou ryz´ı racion´aln´ı funkci
15
rozloˇzit na parci´aln´ı zlomky tvaru
(I) A(ax + b)l (II) Bx + C(px2 + qx + r)k
kde k, l ∈ N, a negationslash= 0, p negationslash= 0, q2 − 4pr < 0, A negationslash= 0 a B2 + C2 negationslash= 0. Je zˇrejm´e, ˇze
pro integraci parci´aln´ıch zlomk˚u typu (I) m˚uˇzeme pouˇz´ıt substituci ax+b = t, kter´a
pˇrevede tento typ na tabulkov´y integr´al integraltext t−l dt.
Pˇr´ıklad 4.1.
integraldisplay 1
(3x + 4)5 dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
3x + 4 = t
3dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
1
3
integraldisplay 1
t5 dt = −
1
12(3x + 4)4 + c,
kde x ∈ (−∞,−4/3) nebo x ∈ (−4/3,∞).
Integrace parci´aln´ıch zlomk˚u tvaru (II) je jiˇz troˇsku n´aroˇcnˇejˇs´ı. Nejprve se budeme
zab´yvat pˇr´ıpadem, kdy k = 1. Pak je vhodn´e upravit integrand na tvar
Kf
prime(x)
f(x) + L
1
f(x),
kter´y jiˇz snadno integrujeme pomoc´ı prvn´ı substituˇcn´ı metody.
Pˇr´ıklad 4.2. Vypoˇctˇete integr´al
integraldisplay x
x2 + 3x + 3 dx.
ˇReˇsen´ı. Integrovan´a funkce je definovan´a pro vˇsechna x ∈ R a x2 + 3x + 3 > 0 na
R.
integraldisplay x
x2 + 3x + 3 dx =
integraldisplay 1
2(2x + 3)−
3
2
x2 + 3x + 3 dx
= 12
integraldisplay 2x + 3
x2 + 3x + 3 dx−
3
2
integraldisplay 1
x2 + 3x + 3 dx =
1
2I1 −
3
2I2
I1 =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
x2 + 3x + 3 = t, t > 0
(2x + 3)dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1
t dt = lnt = ln(x
2 + 3x + 3),
I2 =
integraldisplay 1
(x + 32)2 + 34 dx =
4
3
integraldisplay 1
(2x+3√3 )2 + 1 dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
2x+3√
3 = t
dx =
√3
2 dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
= 2
√3
3
integraldisplay dt
t2 + 1 =
2√3
3 arctgt =
2√3
3 arctgt =
2√3
3 arctg
2x + 3√
3 .
16
Celkem 1
2I1 −
3
2I2 =
1
2 ln(x
2 + 3x + 3)−√3arctg 2x + 3√
3 + c,
kde c ∈ R je libovoln´e.
Pˇr´ıklad 4.3.
integraldisplay 2x + 1
9x2 + 6x + 5 dx =
1
9
integraldisplay 18x + 6
9x2 + 6x + 5 dx +
1
3
integraldisplay 1
9x2 + 6x + 5 dx
= 19I1 + 13I2
I1 =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
9x2 + 6x + 5 = t
(18x + 6)dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1
t dt = ln|t|+ c1 = ln(9x
2 + 6x + 5) + c1
I2 =
integraldisplay 1
(3x + 1)2 + 4 dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
3x + 1 = t
3dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
1
3
integraldisplay 1
t2 + 4 dt
= 16 arctg 3x + 12 + c2.
Celkem dost´av´ame
integraldisplay 2x + 1
9x2 + 6x + 5 dx =
1
9 ln(9x
2 + 6x + 5) + 1
18 arctg
3x + 1
2 + c.
Zb´yv´a n´am integrace parci´aln´ıch zlomk˚u tvaru
Bx + C
(px2 + qx + r)k, k > 1, k ∈ N.
Nejdˇr´ıve uprav´ıme integrand na tvar
K f
prime(x)
(f(x))k + L
1
(f(x))k.
Prvn´ı sˇc´ıtanec integrujeme podle prvn´ı substituˇcn´ı metody, ve druh´em sˇc´ıtanci up-
rav´ıme v´yraz 1/(f(x))k na tvar 1/(t2 + a2)k a primitivn´ı funkci urˇc´ıme uˇzit´ım rekurentn´ıho
vztahu
integraldisplay 1
(t2 + a2)k+1 dt =
1
2ka2
parenleftBigg t
(t2 + a2)k + (2k −1)
integraldisplay 1
(t2 + a2)k dt
parenrightBigg
.
Nyn´ı si tento rekurentn´ı vztah odvod´ıme uˇzit´ım metody per partes. Oznaˇcme
Jk =
integraldisplay 1
(t2 + a2)k dt.
17
Jk =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
u(x) = 1(t2 + a2)k vprime (t) = 1
uprime (x) = − 2kt(t2 + a2)k+1 v(t) = t
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
= t(t2 + a2)k + 2k
integraldisplay t2
(t2 + a2)k+1 dt
= t(t2 + a2)k + 2k
integraldisplay (t2 + a2)−a2
(t2 + a2)k+1 dt
= t(t2 + a2)k + 2k
integraldisplay 1
(t2 + a2)k dt−2ka
2
integraldisplay 1
(t2 + a2)k+1 dt
a odtud dost´av´ame rovnici
Jk = t(t2 + a2)k + 2kJk −2ka2Jk+1.
Pˇr´ıklad 4.4. Vypoˇctˇete integr´al
integraldisplay 1
x3 −1 dx na (1,∞).
ˇReˇsen´ı.
integraldisplay 1
x3 −1 dx =
1
3
integraldisplay 1
x−1 dx−
1
3
integraldisplay x + 2
x2 + x + 1 dx =
1
3(I1 −I2)
I1 =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
x−1 = t, t > 0
dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1
t dt = lnt = ln(x−1),
I2 =
integraldisplay 1
2(2x + 1) +
3
2
x2 + x + 1 dx =
1
2
integraldisplay 2x + 1
x2 + x + 1 dx +
3
2
integraldisplay 1
x2 + x + 1 dx
= 12J1 + 32J2
J1 =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
x2 + x + 1 = t
(2x + 1)dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1
t dt = lnt = ln(x
2 + x + 1)
J2 =
integraldisplay 1
parenleftBig
x + 12
parenrightBig2
+ 34
dx = 43
integraldisplay 1
parenleftBig2x+1
√3
parenrightBig2
+ 1
dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
2x+1√
3 = t
dx =
√3
2 dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
= 2√3
integraldisplay 1
t2 + 1 dt =
2√
3 arctgt =
2√
3 arctg
2x + 1√
3 + c
Celkem tedy
integraldisplay 1
x3 −1 dx = ln
3√x−1−ln 6√x2 + x + 1− 1√
3 arctg
2x + 1√
3 + c
pro x ∈ (1,∞).
18
Pˇr´ıklad 4.5.
integraldisplay 2x + 3
(4x2 −4x + 3)2 dx =
1
4
integraldisplay 8x−4
(4x2 −4x + 3)2 dx + 4
integraldisplay 1
parenleftBig
(2x−1)2 + 2
parenrightBig2 dx
= 14I1 + 4I2,
I1 =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
4x2 −4x + 3 = t
(8x−4)dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1
t2 dt = −
1
t + c1 = −
1
4x2 −4x + 3 + c1
I2 =
integraldisplay 1
parenleftBig
(2x−1)2 + 2
parenrightBig2 dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
2x−1 = t
2dx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
1
2
integraldisplay 1
(t2 + 2)2 dt
= 12
parenleftBigg t
4(t2 + 2) +
1
4
integraldisplay 1
t2 + 2 dt
parenrightBigg
= 18
parenleftBigg t
t2 + 2 +
1√
2 arctg
t√
2
parenrightBigg
+ c2
= 18
parenleftBigg 2x−1
4x2 −4x + 3 +
1√
2 arctg
2x−1√
2
parenrightBigg
+ c2.
Z´avˇerem m´ame
integraldisplay 2x + 3
(4x2 −4x + 3)2 dx =
4x−3
4(4x2 −4x + 3) +
1
2√2 arctg
2x−1√
2 + c.
Cviˇcen´ı 4.1. Spoˇctˇete dan´e integr´aly na dan´ych oborech:
a)
integraldisplay 4x−1
x2 + 5x + 7 dx na R;
b)
integraldisplay 2x + 3
x2 + x−2 dx na (−2,1);
c)
integraldisplay x4 −6x2 + x−2
x4 −2x3 dx na (0,2);
d)
integraldisplay x2 −1
x3 + x2 + x dx na (0,∞);
e)
integraldisplay 3x2 −4x + 4
x3 −2x2 + 2x dx na (−∞,0);
f)
integraldisplay 5lnx + 3
x
parenleftBig
ln2 x−lnx + 1
parenrightBig dx na (0,∞).
19
5 Integrace goniometrick´ych funkc´ı.
Zavedeme nejprve pojem polynomu n promˇenn´ych:
P(u1,u2,...,un) =
m1summationdisplay
k1=0
m2summationdisplay
k2=0
···
mnsummationdisplay
kn=0
ak1k2...knuk11 uk22 ...uknn ,
kde n ∈ N, ak1k2...kn ∈ R, ki, mi jsou cel´a nez´aporn´a ˇc´ısla.
R(u1,u2,...,un) = P(u1,u2,...,un)Q(u
1,u2,...,un)
je racion´aln´ı funkce n promˇenn´ych.
Rozliˇs´ıme tˇri z´akladn´ı typy integr´al˚u.
5.1 Typ integraltext R(sin x,cos x) dx.
Necht’
R(u,v) = P(u,v)Q(u,v)
je racion´aln´ı funkce dvou promˇenn´ych u = sinx a v = cosx.
Integraci funkc´ı tohoto typu lze pˇrev´est na integraci funkc´ı racion´aln´ıch v promˇenn´e
t zaveden´ım n´asleduj´ıc´ıch substituc´ı:
1) Plat´ı-li R(−u,v) = −R(u,v), poloˇz´ıme cos x = t.
2) Plat´ı-li R(u,−v) = −R(u,v), poloˇz´ıme sin x = t.
3) Plat´ı-li R(−u,−v) = R(u,v), poloˇz´ıme tg x = t.
4) V ostatn´ıch pˇr´ıpadech poloˇz´ıme tg x2 = t.
Pˇri zaveden´ı substituce tg x = t vyuˇzijeme
tyto vztahy (pro lehk´e zapamatov´an´ı je
m˚uˇzeme z´ıskat z n´asleduj´ıc´ıho obr´azku:)
cosx = 1√1 + t2, sinx = t√1 + t2.
Pˇri substituci tg x2 = t dostaneme:
20
cosx = cos2 x2 −sin2 x2 = 1−t
2
1 + t2,
sinx = 2sin x2 cos x2 = 2t1 + t2.
Pˇr´ıklad 5.1. Vypoˇctˇete primitivn´ı funkci k funkci
cos3 x
1 + 4sin2 x
na intervalu R.
ˇReˇsen´ı.
R(u,−v) = (−v)
3
1 + 4u2 = −
v3
1 + 4u2 = −R(u,v).
Zvol´ıme substituci sin x = t.
integraldisplay cos3 x
1 + 4sin2 x dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
sin x = t
cos xdx = dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle =
integraldisplay 1−t2
1 + 4t2 dt
= 14
integraldisplay parenleftbigg
−1 + 51 + 4t2
parenrightbigg
dt = 14
parenleftbigg
−t + 52 arctg 2t
parenrightbigg
+ c
= 14
parenleftbigg
−sin x + 52 arctg(2sin x)
parenrightbigg
+ c.
Pˇr´ıklad 5.2. Vypoˇctˇete primitivn´ı funkci k funkci
1
4sin2 x−4sin xcos x + 7cos2 x
na intervalu (0,pi/2).
ˇReˇsen´ı.
R(−u,−v) = 14(−u)2 −4(−u)(−v) + 7(−v)2 = 14u2 −4uv + 7v2 = R(u,v).
Zvol´ıme substituci tg x = t.
integraldisplay 1
4sin2 x−4sin xcos x + 7cos2 x dx
=
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
tg x = t sin x = t√1+t2
x = arctg tdt cos x = 1√1+t2
dx = 11+t2dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
=
integraldisplay 1
parenleftBig
4 t21+t2 − 4t1+t2 + 7 11+t2
parenrightBig 11 + t2 dt =
integraldisplay 1
4t2 −4t + 7 dt =
integraldisplay 1
(2t−1)2 + 6 dt
= 12√6 arctg 2t−1√6 + c = 12√6 arctg 2tg x−1√6 + c
21
Pˇr´ıklad 5.3. Vypoˇctˇete primitivn´ı funkci k funkci
1
4−5sin x
na intervalu (0,pi/4).
ˇReˇsen´ı. Zvol´ıme substituci tg x2 = t.
integraldisplay 1
4−5sin x dx =
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
tg x2 = t sin x = 2t1+t2
x = 2arctg t cos x = 1−t21+t2
dx = 21+t2dt
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
=
integraldisplay 2
parenleftBig
4− 10t1+t2
parenrightBig
(1 + t2) dt =
integraldisplay 1
2t2 −5t + 2 dt
= 13
integraldisplay parenleftbigg 1
t−2 −
2
2t−1
parenrightbigg
dt + c = 13 (ln|t−2|−ln|2t−1|) + c
= 13 ln
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
tg x2 −2
2tg x2 −1
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle+ c.
5.2 Typ integraltext sinαxsinβx dx.
Necht’ α, β ∈ R. K v´ypoˇctu integr´al˚u
integraldisplay
sinαxsinβxdx,
integraldisplay
sinαxcosβxdx,
integraldisplay
cosαxcosβxdx.
pouˇzijeme vzorce
sinαxsinβx = 12 [cos(α−β)x−cos(α + β)x],
cosαxcosβx = 12 [cos(α−β)x + cos(α + β)x],
sinαxcosβx = 12 [sin(α + β)x + sin(α−β)x].
Pˇr´ıklad 5.4. Vypoˇctˇete primitivn´ı funkci k funkci sin3xcos2x na R.
ˇReˇsen´ı. Pouˇzijeme vzorec
sinαxcosβx = 12 [sin(α + β)x + sin(α−β)x].
integraldisplay
sin3xcos2xdx = 12
integraldisplay
(sin5x + sinx) dx = − 110 cos5x− 12 cosx + c.
22
5.3 Typ integraltext sinm xcosn x dx.
Necht’ m, n jsou cel´a nez´aporn´a a sud´a ˇc´ısla. Pro v´ypoˇcet integr´alu
integraldisplay
sinm xcosn xdx
pouˇzijeme vzorce
sin2 x = 12 (1−cos2x), cos2 x = 12 (1 + cos2x), sin2x = 2sinxcosx.
Pˇr´ıklad 5.5. Vypoˇctˇete primitivn´ı funkci k funkci sin4 xcos2 x na R.
ˇReˇsen´ı.
integraldisplay
sin4 xcos2 xdx = 14
integraldisplay
sin2 2xsin2 xdx = 116
integraldisplay
(1−cos4x)(1−cos2x) dx
= 116
integraldisplay
(1−cos2x−cos4x + cos4xcos2x) dx
= 116
parenleftbiggintegraldisplay
dx−
integraldisplay
cos4xdx−
integraldisplay
cos2xdx +
integraldisplay
cos4xcos2xdx
parenrightbigg
= 116
parenleftbigg
x− 14 sin4x− 12 sin2x + 14 sin2x + 112 sin6x
parenrightbigg
+ c
= 116
parenleftbigg
x− 14 sin4x− 14 sin2x + 112 sin6x
parenrightbigg
+ c.
Cviˇcen´ı 5.1. Spoˇctˇete dan´e integr´aly na dan´ych oborech:
a)
integraldisplay sinxcos2 x
cos2 x + 1 dx na R;
b)
integraldisplay sin5 xcosx
1 + 2cosx + cos2 x dx na (−pi,pi);
c)
integraldisplay sin3 x
2 + cosx dx na R;
d)
integraldisplay 1
1 + cos2 x dx na R;
e)
integraldisplay 1