Seminar 11
[Pages:4]Seminar 11 Primitive de forma R(sin x, cos x) dx
Func?tiile
R
sunt
func?tii
ra?tionale
de
forma
R(x)
=
P (x) Q(x)
,
unde
P, Q
sunt
polinoame
cu
coeficient, i reali.
Sa se calculeze integralele:
Problema 11.1.
a) (cos 5x - sin 2x) dx
c) sin 3x ? sin 2x dx
b) cos 2x ? sin 3x dx
d) cos x ? cos 3x dx
Problema 11.2. a) cos2 x dx,
c) sin4 x dx
b) sin2 3x dx
d) cos5 x dx
Problema 11.3.
sin3 x
a)
dx
2 + cos x
sin x b) 1 + cos x + cos2 x dx
Problema 11.4.
cos x
a)
dx
sin x
cos3 x b) sin4 x dx
Problema 11.5.
sin2 x
a)
dx
cos6 x
dx b)
(sin x + cos x)2
Problema 11.6. dx
a) 5 + 4 sin x
sin x dx c)
cos 2x dx
d) sin x + sin 2x
dx c)
cos x cos 2x sin 2x dx
d) (2 + sin x)2
dx c) cos4 x + sin4 x
dx d) sin2 x cos4 x
dx b)
5 - 3 cos x 1
2
SEMINAR 11. PRIMITIVE DE FORMA R(SIN X, COS X) DX
Indica?tii ?si raspunsuri
11.1. a) Se folosesc formulele
cos ax
sin ax dx = -
,
a
sin ax
cos ax dx =
, a = 0.
a
Ob?tinem
1
1
I = (cos 5x - sin 2x) dx = cos 5x dx - sin 2x dx = sin 5x + cos 2x + C.
5
2
b) Folosim formula
1 sin a ? cos b = [sin(a + b) + sin(a - b)]
2
pentru
a
=
3x
?si
b
=
2x.
Se
ob?tine
I
=
-
1 10
cos 5x -
1 2
cos x + C.
c) Se folose?ste formula
1 sin a ? sin b = [cos(a - b) - cos(a + b)]
2
?si
se
ob?tine
I
=
1 2
sin x
-
1 10
sin 5x
+
C.
d) Cu ajutorul formulei
1 cos a ? cos b = [cos(a - b) + cos(a + b)]
2
?si
se
ob?tine
I
=
1 4
sin 2x
+
1 8
sin 4x
+
C.
11.2. Cu a = b ^in formulele din exercit, iul anterior, rezulta noile formule
cos2
a
=
1
+
cos 2a ,
2
sin2
a
=
1
-
cos 2a .
2
a) Vom ob?tine
I=
cos2 x dx = 1
1 ? sin 2x ?
(1 + cos 2x) dx = x +
+ C.
2
2
2
b) Rezulta
I=
sin2 3x dx = 1
1 ? sin 6x ?
(1 - cos 6x) dx = x -
+ C.
2
2
6
c) Reducem gradul succesiv.
I=
sin4 x dx =
? 1 - cos 2x ?2 dx =
1 - 2 cos 2x + cos2 2x dx
2
4
1? = x - sin 2x +
cos2
? 2x dx
=
1
? x
-
sin 2x
+
1
? x
+
sin 4x??
+
C.
4
4
2
4
d) Pentru a integra func?tii trigonometrice la puteri impare procedam ^in felul urmator.
I = cos5 x dx = cos4 x ? cos x dx = (1 - sin2 x)2 ? cos x dx.
Cu schimbarea de variabila u = sin x, avem du = cos x dx ?si
I = (1 - u2)2 du = (1 - 2u2 + u4) du = u - 2u3 + u5 = sin x - 2 sin3 x + sin5 x + C.
35
3
5
3
11.3. Pentru integrale de forma R(sin x, cos x) dx, daca se verifica condi?tia
R(- sin x, cos x) = -R(sin x, cos x)
atunci se face schimbarea de variabila u = cos x. Toate integralele din acest exerci?tiu verifica aceasta condi?tie. a) Avem
sin3 x
(1 - u2)(- du)
u2 - 1
u2 - 4 + 3
I=
dx =
=
du =
du
2 + cos x
2+u
u+2
u+2
?
3?
u2
cos2 x
= u-2+
du = - 2u + 3 ln |u + 2| =
- 2 cos x + 3 ln(2 + cos x) + C.
u+2
2
2
b) Cu schimbarea de variabila u = cos x avem
sin x
du
I=
=-
=-
1 + cos x + cos2 x
u2 + u + 1
du
u
+
1 2
2
+
3 4
2 = -
arctg
u+
1 2
=
2 -
1 + 2 cos x arctg + C.
3
3
3
3
2
c) Aplic^and formula
dx x2-a2
=
1 2a
ln
x-a x+a
sin x
du
1
I=
=-
=-
cos 2x
2u2 - 1 2
du
2 2 cos x - 1
u2 -
1 2
=-
4
ln
2 cos x + 1
+ C.
d) Dupa schimbarea de variabila u = cos x rezulta
dx
dx
du
I=
=
=-
.
sin x + sin 2x
sin x(1 + 2 cos x)
(1 - u2)(1 + 2u)
Acum se descompune ^in frac?tii simple
-1
A
B
C
(1 - u2)(1
+
2u)
=
1-u
+
1+
u
+
1+
, 2u
cu A = -1/6, B = 1/2 ?si C = -4/3. Rezulta
1
1
2
I = ln(1 - cos x) + ln(1 + cos x) - ln(1 + 2 cos x) + C.
6
2
3
11.4. Ca^nd are loc condi?tia R(sin x, - cos x) = -R(sin x, cos x) atunci notam u = sin x. Toate integralele din acest exerci?tiu verifica aceasta condi?tie. a) Avem
cos x
du
I=
dx =
= ln |u| = ln | sin x| + C.
sin x
u
b)
cos3 x
1 - u2
I = sin4 x dx =
u4 du
=
u-4 du -
u-2 du
=
u-3 -3
-
u-1 -1
=
-1 3 sin3 x
+
1 sin x
+ C.
4
SEMINAR 11. PRIMITIVE DE FORMA R(SIN X, COS X) DX
c)
dx
du
I=
= cos x cos 2x
(1 - u2)(1 - 2u2) =
1 1 - sin x 2 2 sin x - 1
= ln
- ln
+ C.
2 1 + sin x 2
2 sin x + 1
du u2 - 1 -
2 du 2u2 - 1
d)
sin 2x dx
2u du
I = (2 + sin x)2 = (2 + u)2 = 2
4
= 2 ln(2 + sin x) +
+ C.
2 + sin x
u + 2 - 2 du (2 + u)2
11.5. Atunci ca^nd este ^indeplinita condi?tia R(- sin x, - cos x) = R(sin x, cos x) se face
substitu?tia
u=
tg x
?si
se
?tine
cont
ca
dx =
du 1+u2
,
sin x =
u 1+u2
?si
cos x =
1 1+u2
.
a)
I=
sin2 x cos6 x dx =
u2 1+u2
1 (1+u2)3
du 1 + u2
=
u2(1+u2) du =
(u2+u4) du
=
tg3
x tg5 +
x +C.
35
b)
dx
du
1
1
I=
(sin x + cos x)2 =
(u + 1)2
=
- u+1
=
- 1 + tg x
+ C.
c)
dx
(1 + u2) du
I = cos4 x + sin4 x =
= 1 + u4
1+ u2
1 u2
+
du
1 u2
.
Cu
schimbarea
de variabila
t
=
u
-
1 u
avem
dt =
1
+
1 u2
du
?si
t2
=
u2 +
1 u2
- 2.
A?sadar
I=
dt
= 1 arctg t
= 1
u- arctg
1 u
=
1
arctg tg x - ctg x + C.
t2 + 2
2
22
2
2
2
d)
dx
(1 + u2)2 du
1
u3
tg3 x
I = sin2 x cos4 x =
= - + 2u + = - ctg x + 2 tg x +
+ C.
u2
u
3
3
11.6. ^In cazul ^in care nici o condi?tie de la exerci?tiile anterioare nu este verificata se face
substitu?tia
u
=
tg
x 2
?si
se ?tine
cont
de
faptul
ca
dx =
2 du 1+u2
,
sin x =
2u 1+u2
?si
cos x =
. 1-u2
1+u2
a) Avem
dx
du
2
I=
=2
=
5 + 4 sin x
5 + 5u2 + 8u 5
u+
du
4 5
2+
9 25
=
2 arctg
3
4 + 5 tg 3
x 2
+ C.
b)
I=
dx =
5 - 3 cos x
2 du 5(1 + u2) - 3(1 - u2) =
du 1 + 4u2
=
arctg 2u 2
=
arctg
2 tg 2
x 2
+C.
................
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