T×m hä nguyªn hµm cña gx.
Trang 11 2
1
x x 0
(2x 1)e− − dx
2 Với x 0;
4
π
∈ xác định a,b sao cho 1 a cos x bcos x
cos x 1 sin x 1 sin x= +
3 Tính
/ 4
3 0
cos x cos x
π
4
/ 2
0
sin x cos x 1
dx sin x 2cos x 3
5
1
3 0
(3x 1)dx
(x 3)
+
+
6
1
3
0
xdx
(x 1)+
7
1 2
4
0
dx
−
+
8 2x 2
0
e sin xdx
π
9
/ 2
0
cos xdx
2 cos 2x
π
+
10
1
2
1
dx
x 2x cos 1 ,(0< < )
−
α π
11
2a
2 2
a
x −a dx ,(a>0)
12
/ 2 3
0
4sin xdx
1 cos x
π
+
13
a
2 2
0
x +a dx
14
2
0
1 sin xdx
π
+
15
3 / 8
/ 8
dx sin x cos x
π
Trang 216
2
1
dx
x 1+ + x 1−
17 Gpt
x
2 0
(u x )du sin x− =
18
b
2
1
x ln xdx
19
/ 2
2 0
x cos xdx
π
20
2
2
2 / 3
dx
x x −1
21
0
cos x sin xdx
π
22.Cho hµm sè: f(x) sin x.sin 2x.cos5x=
a T×m hä nguyªn hµm cña g(x)
b. TÝnh tÝch ph©n: 2 x
2
f(x)
π
−π
=
+
23
ln 2 2x
x
0
e
dx
e +1
24
1 2
0
dx
x 1
−
+
25
/ 4
0
cos x 2sin x
dx 4cos x 3sin x
+
26
1
3
0
3dx
1 x+
27
1
0
dx
x +4x +3
28
/ 3
/ 6
tg x cot g x 2dx
π
π
Trang 329
/ 3
/ 6
dx sin x sin(x / 6)
π
30
/ 4
x / 4
sin x cos x
dx
π
−π
+ +
31
2
2
1
ln(x 1)
dx x
+
32
/ 2
3
sin xdx sin x cos x
π
+
0
x 1 x dx+
34
1/ 9
3x
0
4x 1 sin (2x 1)
− +
35
7 / 3
3
0
x 1
dx 3x 1
+
+
∫
x
2
4
2
(10 sin x)dx
−
36
x
5 4x
−
−
37
/ 2
2 / 2
x cos x
dx
4 sin x
π
−π
+
−
38
/ 2
3 0
5cosx 4sin x
dx (cosx sin x)
+
39
/ 2 4
0
cos x
dx cos x sin x
π
+
40
/ 3 2
6
/ 4
sin x
dx cos x
π
41
2 2
2 2
dx
−
−
+ +
42
2 0
sin x cos x
dx
1 cos x
π
+
Trang 41 4
x 1
x
dx
1 2
0
x sin x cos xdx
π
45
/ 2
0
I=π∫ cos x cos 2xdx
/ 2
0
46
/ 3
2 0
x sin x
dx cos x
∫
2 0
x
dx
x+ x +1
2
2
sin 4x
1 cos x
+
+
47
2
0
1 sin xdx
π
+
48
dx
x 3 (x 3x 2)
+
49
/ 4
3 0
cos2x
dx sin x cosx 2
π
50
1 3 2
2
0
dx
1 2
2
0
x 3x 10
dx
x 2x 9
∫
51
/ 4
0
sin 4x
dx sin x cos x
π
+
52
2 2
−
53
1
5 3 6
0
x (1 x ) dx−
Trang 5/ 4
0
dx
1 5
0
x J=
π
=
+
55
1
0
x 1 xdx−
3 2
1 e
1 x J=
−
=
+ +
57
ln 2
x
0
dx
e +5
58
4
2
1
dx
x (1 x)+
59
/ 2
3 0
4sin x
dx (sin x cos x)
π
+
60 11
0
sin xdx
π
61
/ 4
0
sin x cos xdx
π
62
e
2 1/ 2
ln x
dx (1 x)+
63
/ 4
2
0
cos x cos 4xdx
π
64
7 / 3
3
0
x 1
dx 3x 1
+
+
65
1
2 2 0
(1 x x ) dx− −
66
/ 2
x 2
0
e cos xdx
π
67
0
1 cos 2xdx
π
+
68
x x 1 J= x(x 1)
+
=
Trang 669 / 4 ( )
0
ln 1 tgx dx
π
+
70
/ 2
0
3sin x 4cos x
dx 3sin x 4cos x
+
3
0
x −2x +xdx
∫
71
/ 4
0
sin x.cosx
dx sin 2x cos2x
π
+
72
/ 2
0
sin x cos x
dx a,b 0
a cos x b sin x
;
π
≠ +
73
2 / 2 2
2 0
x
dx
1 x−
74
/ 4
2 0
x(2cos x 1)dx
π
−
75
/ 3
2 / 4
1 4
0
x
π
π
/ 2
4 3
sin x 7cos x 6
dx x cos x sin xdx 4sin x 3cos x 5
76
1
4 2
0
x
dx
x +x +1
77
/ 2
2
0
(x 1)sin xdx
π
+
78
/ 2 3
0
4sin x
dx
1 cos x
π
+
79
2
2x x
dx
1 x dx
−
−
+
80
4 / 3 dx
x sin
2
π
π
81
4
2
x
x 7x 12
π
Trang 73
2
2
x −1dx
83
1
2
0
x +1dx
84
/ 4
2 0
dx
2 cos x
π
−
85
1
2 3 0
(1 x ) dx−
86
2 / 2
x / 2
x sin x
1 2
π
−π
π
=
+
87
/ 2
x 0
1 sin x
e dx
1 cos x
+
88
10
2
1
x lg xdx
89
x
2 x
dx
x.e dx
e 1
−
+
90
3
dx
−
91
1/ 2
0
dx
1 cos x+
92
/ 2
2 / 2
cos x ln(x 1 x )dx
π
−π
dx
x 1 sin x cos x
π π
+
+
93
1
2
0
xtg xdx
94
1
2 0
xdx
(x 1)+
95
/ 4 3
4 0
4sin x
dx
1 cos x
π
+
Trang 896 / 2 3 3
/ 3
sin x sin x
cot gxdx sin x
π
π
−
97
1
2 1
dx
98
/ 2
0
cos x ln(1 cos x)dx
π
+
1/ 3
0
dx (2x +1) x +1
∫
99
2 b
2 2 0
a x
dx
a x
−
+
∫ (a, b lµ sè thùc d¬ng cho tríc) (HV KTQS_01A)
100
a
2 2 2 0
x x +a dx ,a 0>
0
x sin xdx
2 cos x
π
+
102
4
dx (cos x sin x)dx
cos x
π
+
1
0
0
π
103
4
2 7
dx
x x +9
3sin xdx x x 1dx
π
+
105
2
2 1
(x ln x) dx
106
3
1
ln 2 ln x
dx x
+
107
/ 4
2 0
1 sin 2x
dx cos x
Trang 9108
1
3
0
3dx
1 x+
109
1
2 2x 0
(1 x) e dx+
110
2 x
(2x 1)cos xdx
1 sin 2x
π π −
+ +
111
dx
e 3 x
+ +
112
2x
1 sin 2x cos 2x (1 e )
sin x cos x 1 e
π
π
113
2x
cos xdx
e sin 3xdx
1 cos x
+
114
1
19 0
x(1 x) dx−
115
2 3
dx
xtg xdx x(x 1)
π
+
116
6
/ 2
4
/ 4
cos x
dx sin x
π
117
2
1
ln(1 x)dx +
118
1 4
2
1
x sin x
dx
−
+
+
119
/ 2
0
dx
2 sin x cos x
π
120
1
2
0
x sin xdx
121
a
2 2 2
0
x a −x dx (a 0)>
122
1
0
x 1 x dx−
Trang 102
2
1
xdx
0
x sin xdx
π
/ 2
0
dx sin x cos x
π
+
∫
125
1
0
dx
1+ x
126
2
1 cos x
π
+ + +
4 x− 4 x−
127
e 1 cos x
π
+ +
128. Tính
sin x 3 cosx sin x 3 cosx
Từ đó suy ra:
5 / 3
3 / 2
cos2x
dx cosx 3 sin x
π
129
x
2cos xdx 5e sin 2xdx
3 2sin x
+
130 Cho f(x) liên tục trên R : f (x) f ( x)+ − = 2 2cos2x− ∀ ∈x R Tính
3 / 2
3 / 2
f (x)dx
π
131
/ 2
0
(sin x sin x cos x sin x)dx
π
dx
x 1 −
+
+
133
(sin x 2cos x)
3sin x cos x
π +
−
+
Trang 11134 2 2
0
sin x cos xdx
π
135
1 cos x x(1 x )
1 x
1 2
−
−
+
1
1
(e sin x e x )dx
−
+
138
2
0
t
dt
t + +2t 1
139
2 4
1 x
1 x
+
140
1 2
2
0
(x x)dx
+
+
141
/ 4 3
dx sin x cos3xdx
1 tgx
+
142
2
2
1
ln x
dx
x
143
/ 2 6
0
sin x
dx sin x cos x
π
+
144
2
7
dx
2 x 1+ +
145
/ 2
2
dx
1 sin x
1 cos x
+ +
146
4
dx
x ln xdx cos x
π
147
dx sin x cos x 1 2cos x
π
−
148
1 2
2 0
x x arctgx
dx
1 x
+ +
+
Trang 122 10 3
x 1
dx (1 3x)(1 2x 3x ) dx 3x 2
+
150
2
π
+
151
2 3
x 1 + +
+
sin x cos x sin x cos x sin x cos x
2
3 x 2
x(ln x 1)+ +
152
sin 2x(1 sin x) dx sin x cos x(1 cos x) dx
2
2
x 1
x 2x (x 1)sin xdx dx
π
+
+ +
153
2 2
sin xdx
cos x 3
+
4
3
xdx cos 2xdx
(2x 1)
π
+
154
1 2
x sin x
9 4cos x
π
− +
155
2
x
-sin xdx
1 sin xdx
1 3
− +
x ln xdx x 1 x dx−
x sin xdx
arctg(cos x)dx
1 cos x
+
4 2
dx cos xdx sin x cos x x 5x 6
157
1 x
3 x
e
dx x sin xdx x sin xdx
1 e
−
−
+
Trang 131/ 2 4 / 2
dx
x 1 1 sin x
π
4 4
2x 1
1 cos x
+ +
sin x cos xdx e sin ( x)dx
π
π
158
2
1
1 x
x
x 1 dx− e dx
159
x
(1 e )
e +
2x x
x e 3e 2
160
3
1
2
0
x
dx
x +1
161
x
ln 2
3 x
0
e
dx
e +1
2x 3
1
−
163
/ 2
0
1 cos x.sin x.cos xdx
π
−
164
2 3
2 5
dx
x x +4
165
/ 4
0
xdx
1 cos2x
π
+
166
1
0
x 1 x dx−
167
2 / 4
0
1 2sin x
dx
1 sin 2x
+
Trang 142x
ln5
x
ln 2
e
dx
e −1
169. Cho hµm sè: f(x) a 3 bxex
(x 1)
+ , t×m a, b biÕt r»ng:
f '(0)= −22 vµ
1 0
f(x)dx 5=
170
2
2
0
x −x dx
1
3 x
0
x e dx
172
2
e
1
ln xdx x
+
173
2
1
x
dx
1+ x 1−
174
e
1
1 3ln x.ln x
dx x
+
175 3 ( )
2
2
ln x −x dx
176
/ 2
0
sin 2x sin x
dx
1 3cosx
+
177
/ 2
0
sin 2x.cosx
dx
1 cosx
π
+
sin x
0
e cosx cosxdx
π
+
179
7
3
0
x 2
dx
x 1
+
+
180
/ 2
2
0
sin xtgxdx
π
181
/ 2
cosx
0
e sin 2xdx
π
182
2
2
0
dx
+
Trang 15183 / 4( )
sin x 0
tgx e cosx dx
π
+
184
e
2
1
x ln xdx
185
/ 2
0
sin 2x
dx cos x 4sin x
π
+
186
6
2
dx 2x 1+ + 4x 1+
187 1( ) 2x
0
x 2 e dx−
188
/ 2
0
(x 1)sin 2xdx
π
+
189 2( )
1
x 2 ln xdx−
190
ln5
ln3
dx
dx
e +2e− −3
191
10
5
dx
x 2 x 1− −
192
e
1
3 2 ln x
dx
x 1 2 ln x
−
+
193
3
2
0
x 2x
dx
+
+
3
−
+ − −
195
4
2
5
0
x
dx
x +1
196
3
3 1
dx
x x+
197
ln8
ln3
e +1.e dx
198
2
0
x.sin xdx
π
Trang 161
0
x 1 xdx−
200
3
1
ln x
dx
x ln x 1+
201
/ 2
2 0
(2x 1)cos xdx
π
−