Bài tập Tích phân đổi biến dành cho 12CB

BÀI TẬP TÍCH PHÂN CHỌN LỌC
DÀNH CHO LỚP 12 CB
DẠNG1: Sử dụng bảng nguyên hàm cơ bản
( )
( )
( )
2
2
0
0
2 2
2
2
0
0 0
2 2 2
2
2
2 2
0
0 0 0
2
4
2
0
1) sin osx =( 2 cos +sinx) = 3 2
2 2
1 1 sin 2
2) cos (1 cos 2 )
2 2 2 4
1 1 1 sin 4

3) cos sin sin 2 (1 cos 4 ) ( )
4 8 8 4 16
sin
4)
cos
x x
c dx
x
x dx x dx x
x
x x dx x dx x dx x
x
d
x
π
π
π π
π
π π π
π
π
π
π
 
+ − −
 ÷
 
 
= + = + =
 ÷

 
= = − = − =
∫
∫ ∫
∫ ∫ ∫
∫
( )
4
4
2
0
0
4 4 4
4
2 2 2 2
6
6 6 6
3 3
2 2
3
2
2 2
2 2
1 1
1
1 tan 1
cos 4
1 1 1 2
5) 4 2 (2 ) 2 cot(2 )
cos .sin sin 2 sin 2

3
1 1 5
6) ln 1 ln 2
1 1 2 2
1
7)
( 1) ( 1)
x dx x x
x
dx dx d x x
x x x x
x x x
dx x dx x
x x
dx x x
dx
x x x x
π
π
π π π
π
π
π π π
π
 
= − = − = −
 ÷
 
= = = − =
 

− +
 
= + = + − = +
 ÷
 ÷
− −
 
 
+ −
= =
+ +
∫
∫ ∫ ∫
∫ ∫
∫ ∫
2 2
2
1
1 1
2
2
2
0
0 0 0
2
2
1
x
1
x

0
0
1 1 2 1 4
ln ln ln ln
1 1 3 2 3
8) 1 sin cos cos cos sin sin 2
2 1
9) 2
ln 2 ln 2
x
dx dx
x x x
xdx x dx xdx xdx x x
dx
π
π
π π π
π
π
π
− = = − =
+ +
− = = − = − =
= =
∫ ∫
∫ ∫ ∫ ∫
∫
( )
2 2
2

2
2
4
4 4
2
2 2 2
6 6 6
2
6
8 8
2
8
3
3
1
3 2
1 1
0
1
10) cot 1 cot 1
sin 4
1 sin 2 cos 2 2sin cos 2cos
11) 2 cos
sin cos sin cos
2sin 1
12) 3 3
1
13) sin cos3 sin 4 s
2
xdx dx x x

x
x x x x x
dx dx xdx
x x x x
x
dx
x dx x
x
x xdx x
π π
π
π
π π
π π π
π π π
π
π
π
π
−
 
= − = − + = −
 ÷
 
+ + +
= =
+ +
= =
= = =
= −

∫ ∫
∫ ∫ ∫
∫ ∫
∫
( )
0
0
2 2
3 3 3
2 2
2 2 2
3 3 3 3 3
3 3
2 2
2 2
2 2 2 2 2
2 3
3 2 2
in 2
1 cos 4 cos 2 1
2 4 2 4
1 1 1
14)
( 1) ( 1) 1
1 1
2 2 2
( 1) ( 1) 1 1
1 1 2
( 2 ln 1 2ln ) 2ln
1 3 1

x
x x
dx
dx dx
x x x x x x
dx dx dx dx dx
x x x x x x x x
x
x x
x x x
π
π
−
 
= + =
 ÷
 
 
 
= = −
 
 
− − −
 
 
= − + = − − − +
− − − −
 
= + + − − + = +
 ÷

− −
 
∫
∫ ∫ ∫
∫ ∫ ∫ ∫ ∫
( )
3
2
0 0 0 0
0
1 1 1
3 2 3 2 3 4
0 0 0
2 3
2ln
3 4
1 1 1
15) sin 2 cos sin 2 (1 cos 2 ) sin 2 sin 4
2 2 4
1 1
cos 2 cos 4 0
4 16
1
16) (1 ) (1 3 3 ) 3 3
20
x xdx x x dx xdx xdx
x x
x x dx x x x x dx x x x x dx
π π π π
π

= +
= + = +
 
= − − =
 ÷
 
− = − + − = − + − =
∫ ∫ ∫ ∫
∫ ∫ ∫
Đặt x = asint hoặc x = a tant để tính tích phân
4
2
0
1) 16 x dx
−
∫
Đặt x = 4sint , t
;
2 2
t
π π
−
 
∈
 
 

2 2
0 0
2

0 0
4
2
2
2
0
0
2
2
2
0
2
4 4
2
0 0
4 ostdt 4 ost 4 ostdt=16 os
sin 2
8 (1 cos 2 ) 8( ) 4
2
2) . 2 tan , ; 2(1 tan )
4 2 2
0 0; 2
4
2(1 tan ) 1
4(1 tan ) 2
x t
x t
dx c I c c c tdt
t
t dt t

dx
x t t dx t dt
x
x t x t
t dt
I dt
t
π π
π
π
π
π π
π
π π
π
= ⇒ =
= ⇒ =
= ⇒ =
= + = + =
 
= ∈ − ⇒ = +
 ÷
+
 
= ⇒ = = ⇒ =
+
= = =
+
∫ ∫
∫

∫
∫
4
0
3
2
3
3 3 3
3
2
3
3 3 3
1
2 8
3) . 2sin , ; 2 ostdt
2 2
4
3 ; 3
3 3
2 cos 2cos 2
2 ost 3
4(1 sin )
t
dx
I x t t dx c
x
x t x t
tdt tdt
I dt t
c

t
π
π π π
π
π
π π π
π
π π
π π
π
−
−
− − −
=
 
= = ∈ − ⇒ =
 
 
−
= − ⇒ = − = ⇒ =
= = = = =
−
∫
∫
∫ ∫ ∫
3
2 2
0
3
2

2
1
2
1
2
0
1
2
4
0
1
2
2
2
2
2
2
0
81
4) 9 3sin KQ:
16
2 3
5) sin KQ: ln
3
1
6) sin KQ:
4
1
7) tan :
1 8

1
8) sin KQ:1
4
2
9)
2
x x dx x t
dx
x t
x x
dx
x t
x x
xdx
x t KQ
x
x
dx x t
x
x
d
x
π
π
π
π
− =
+
=
−

=
+ −
=
+
−
= −
+
−
∫
∫
∫
∫
∫
∫
3
2 2
1
4 2 2
2cos KQ:
4
10) tan
1
x x t
dx
x t
x x
π
+ −
=
=

+
∫
Dạng2:Đặt t = u(x) với các hàm số
( ) ( ). '( )f x g u u x
=
3
4
2
2
7
2
2
0
3
3
0
1
1
3
4
8
0
1
4 2
0
1 7
1) . 9 : ln
6 4
9
17

2) s inxcosx(1+cosx) osx KQ:
12
1 106
3) 1 :
15
1
1 ln 14
4) 1 ln :
3
5) :
1 16
6)
1
e
dx
t x KQ
x x
dx t c
x
dx t x KQ
x
x
dx t x KQ
x
x dx
t x KQ
x
xdx
x x
π

π
= +
+
= →
+
= + →
+
+
= + →
= →
+
+ +
∫
∫
∫
∫
∫
∫
2
1 1 1
2 2
2 2 2
0 0 0
1 1
2 2
2
0 0
:
6 3
1 3 1 3

7) =
3 3 3 3 3
1 1 1 1 3
: ln
3 3 3 3 6 4
x x
x x x
x
x
t x KQ
dx dx e e dx
e e e
e dx e
dx KQ
e
π
= →
+ −
=
+ + +
+
= − → −
+
∫ ∫ ∫
∫ ∫