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
;