3-misol.
2
8
xdx
x
+
∫
integralni hisoblang.
Bu kabi misollarni yechishda
o‘zgaruvchini almashtirish
qulay.
Agar
x
2
+8=
u
deyilsa,
du
= 2
xdx
,
1 4
2
xdx
=
du
bo‘ladi. U holda
2
2
1
1
1
ln
ln(
8)
.
8 2
2
2
xdx
du
u C
x
C
x
u
=
=
+ =
+ +
+
∫
∫
׀
u
׀ +
2
2
1
1
1
ln
ln(
8)
.
8 2
2
2
xdx
du
u C
x
C
x
u
=
=
+ =
+ +
+
∫
∫
Tekshirish:
Topilgan boshlang‘ich funksiyadan hosila olinsa, integral
ostidagi funksiya
2
8
x
x
+
hosil bo‘lishi kerak. Chindan ham,
2
2
2
2
2
2
1
1
1 1
1 2
ln(
8)
(ln(
8))
·
·( )
·
.
2
2
2
8
2
8
8
x
x
x
C
x
C
x
x
x
x
′
′
′
′
+ +
=
+
+
=
=
=
+
+
+
(
x
2
+8)ʹ
2
2
2
2
2
2
1
1
1 1
1 2
ln(
8)
(ln(
8))
·
·( )
·
.
2
2
2
8
2
8
8
x
x
x
C
x
C
x
x
x
x
′
′
′
′
+ +
=
+
+
=
=
=
+
+
+
Javob:
2
1 ln( 8) .
2
x
C
⋅
+ +
▲
4-misol.
sin
cos
x
e
xdx
∫
integralni hisoblang.
sin
x
=
t
almashtirish bajaramiz. U holda
dt
= cos
xdx
va berilgan
integ ral
t
e dt
∫
ko‘rinishni oladi. Integrallar jadvallarining 3-bandiga ko‘ra
t
t
e dt e C
= +
∫
bo‘ladi. Demak,
sin
sin
cos
.
x
e
xdx e
C
=
+
∫
sin
x
+C.
Tekshirish.
(
e
sin
x
+
C
)′ = (
e
sin
x
)′+
C
′=
e
sin
x
(sin
x
)'+0=
e
sin
x
cos
x
– berilgan
in teg ral ostidagi funksiyani hosil qildik.
Javob:
e
sin
x
+
C
.
▲
88
89
5-misol.
sin 5 cos3
x
xdx
⋅
∫
integralni hisoblang.
Bunda 2sin5
x
· cos3
x
= sin8
x
+sin2
x
ayniyat yordam beradi.
U holda
1
1
sin 5 cos3
sin8
sin 2
2
2
1
1
cos8
cos 2
( cos8 )
( cos 2 )
.
18
4
16
4
x
xdx
xdx
xdx
x
x
x
x C
C
=
+
=
=
−
+
−
± = −
−
+
∫
∫
∫
1
1
cos8
cos 2
( cos8 )
( cos 2 )
.
16
4
16
4
x
x
x
x C
C
=
−
+
−
+ = −
−
+
Javob:
cos8
cos 2
.
16
4
x
x C
−
−
+
▲
6*-misol.
sin 5 cos3
x
xdx
⋅
∫
cos
mx
cos
nxdx
integralni hisoblang.
cos
mx
cos
nx
=
1
2
(cos(
m
+
n
)
x
+cos(
m
–
n
)
x
) ayniyatga va integrallash
jad
valining 10-bandiga muvofiq:
1
1
cos
cos
cos(
)
cos(
)
2
2
1 sin(
)
1 sin(
)
.
2
2
mx
nxdx
m n xdxx
m n xdx
m n x
m n x C
m n
m n
=
+
+
−
=
+
−
= ⋅
+ ⋅
+
+
−
∫
∫
∫
1
1
cos
cos
cos(
)
cos(
)
2
2
1 sin(
)
1 sin(
)
.
2
2
mx
nxdx
m n xdxx
m n xdx
m n x
m n x C
m n
m n
=
+
+
−
=
+
−
= ⋅
+ ⋅
+
+
−
∫
∫
∫
1
1
cos
cos
cos(
)
cos(
)
2
2
1 sin(
)
1 sin(
)
.
2
2
mx
nxdx
m n xdxx
m n xdx
m n x
m n x C
m n
m n
=
+
+
−
=
+
−
= ⋅
+ ⋅
+
+
−
∫
∫
∫
Javob:
1
1
cos
cos
cos(
)
cos(
)
2
2
1 sin(
)
1 sin(
)
.
2
2
mx
nxdx
m n xdxx
m n xdx
m n x
m n x C
m n
m n
=
+
+
−
=
+
−
= ⋅
+ ⋅
+
+
−
∫
∫
∫
▲
7-misol.
2
5
6
dx
x
x
−
+
∫
integralni hisoblang.
Integral ostidagi funksiya uchun quyidagi tengliklar o‘rinlidir:
2
1
1
(
2) (
3)
1
1
5
6 (
2)(
3)
(
2)(
3)
3
2
x
x
x
x
x
x
x
x
x
x
− − −
=
=
=
−
−
+
−
−
−
−
−
−
.
Bundan
2
1
1
(
)
5
6
3
2
3
2
3
ln(
3) ln(
2)
ln
,
2
dx
dx
dx
dx
x
x
x
x
x
x
x
x
x
C
C
x
=
−
=
−
=
−
+
−
−
−
−
−
=
− −
− + =
+
−
∫
∫
∫
∫
= l n ׀
x
– 3 ׀ – l n ׀
x
– 2 ׀ +
C
=
2
1
1
(
)
5
6
3
2
3
2
3
ln(
3) ln(
2)
ln
,
2
dx
dx
dx
dx
x
x
x
x
x
x
x
x
x
C
C
x
=
−
=
−
=
−
+
−
−
−
−
−
=
− −
− + =
+
−
∫
∫
∫
∫
Javob:
ln│
2
1
1
(
)
5
6
3
2
3
2
3
ln(
3) ln(
2)
ln
,
2
dx
dx
dx
dx
x
x
x
x
x
x
x
x
x
C
C
x
=
−
=
−
=
−
+
−
−
−
−
−
=
− −
− + =
+
−
∫
∫
∫
∫
│+
C
.
▲
90
91
8-misol.
1 cos
dx
x
+
∫
integralni hisoblang.
Bu integralni hisoblash uchun 1+cos
x
=2cos
2
2
1 cos 2cos
2
x
+
=
va
2
cos
dx
tgx C
x
=
+
∫
tg
x
+
C
ekanidan foydalanamiz. U holda
2
1·2·tg
tg
.
1 cos
2
2 2
2
2cos
2
dx
dx
x C
x C
x
x
=
=
+
=
+
+
∫
∫
C
2
1·2·tg
tg
.
1 cos
2
2 2
2
2cos
2
dx
dx
x C
x C
x
x
=
=
+
=
+
+
∫
∫
Tekshirish
:
(tg
)
2
x C
′
+
=(tg
2
2
1
1
1
1
(
(
)
( ) 0
2
2
2
2
1 cos
cos
cos
2
2
x
x
x
tg
C
tg
C
x
x
x
′
′
′
′
+
=
+
=
⋅
+ = ⋅
=
+
integral ostidagi funksiya hosil bo‘ldi.
Javob:
tg
.
2
x
tg
C
+
▲
9-misol.
2
sin 2
xdx
∫
integralni hisoblang.
Integralni hisoblash uchun 2sin
2
2
x
=1– cos4
x
ayniyatdan foyda-
lanamiz.
2
1
sin 2
(1 cos4 )
2
xdx
x dx
=
−
∫
∫
2
1
1
1
1 1
1
sin 2
(1 cos )
cos4
sin 4
sin 4
.
2
2
2
2 2 4
2 8
x
x
x
sx dx
dx
xdx
x C
x C
=
−
=
−
= − ⋅
+ = −
+
∫
∫
∫
∫
Javob:
1 sin4
.
2 8
x
x C
−
+
▲
?
Savol va topshiriqlar
1. Integrallar jadvalidagi o‘zingiz xohlagan 4 ta misolni tanlang va uni
isbotlang.
2.
Integrallashning sodda qoidalarini bayon qiling. Misollarda
tushuntiring.
3. O‘zgaruvchi almashtirish usuli nima?
cos 2
sin 2
x
e
xdx
∫
integralni
hisoblashda shu usulni qo‘llang va misolni yechish jarayonini tushuntiring.
90
91
Mashqlar
Berilgan funksiyaning boshlang‘ich funksiyalaridan birini toping
(
16 – 18
):
16.
1)
5
3
3
4 ;
x
x
−
2)
7
4
8
5 ;
x
x
−
3)
2
4 4 ;
x x
−
4)
4
5
5
3 ;
x
x
+
5)
3
3
3 ;
x
x
+
6)
3
7
5 ;
x
x
−
7)
4
3
2
5
4
2 .
x
x
x
+
−
17.
1)
5cos
3sin ;
x
x
−
2)
7sin
4cos ;
x
x
+
3)
2cos
;
x
x a
−
4) 5
e
x
+2cos
x
+1; 5)
4+2·
e
–x
–7sin
x
;
6)
3
6
4
.
x
e
x
x
−
+ −
18.
1)
3
(
2) ;
x
−
2)
4
(
5) ;
x
+
3)
1
5
x
−
4)
3
6
7
x
+
;
5)
8
4cos(
5)
7
x
x
+ +
−
; 6)
4
2sin(
3)
2
x
x
− −
−
; 7)
4
5
1
(3
7)
.
x
x
+
+
Berilgan funksiyaning barcha boshlang‘ich funksiyalarini toping
(
19 – 20
):
19.
1)
cos(5
3);
x
+
2)
sin(7
6);
x
−
3)
2
cos(
1);
3
x
+
4)
5
sin(
2);
7
x
−
5)
e
2 3
4
x
e
+
;
6)
e
3–2
x
;
7)
2
4 ;
cos
x
8)
2
3
.
cos 4
x
;
9)
2
5
sin 5
x
.
20.
1)
3
5
4 (1 2 ) ;
x
x
− −
2)
4
6
1
(3
2)
;
x
x
+
−
3)
6
2
1;
cos
x
x
+
−
4)
2
3
2
6;
sin
x
x
−
+
5)
(1 3 )( 1);
x x
+
−
6)
3
2
1
2sin(3 1).
2
x
x
⋅
+
−
21.
Berilgan
f
(
x
) funksiya uchun grafigi
A
(
x
;
y
) nuqtadan o‘tadigan boshlan-
g‘ich funksiyani toping:
1)
( ) sin 4 ,
f x
x
=
( ; 7);
4
A
π
2)
f
(
x
)
=
cos5
x
,
( ; 4);
4
A
π
3)
2
2
( ) 3
,
2
f x
x
x
=
+
+
( 1;0);
A
−
4)
3
1
( ) 4
2
1
f x
x
x
=
−
−
(2;0)
A
;
|