333 BÀI TOÁN TÍCH PHÂN LUYỆN THI ĐẠI HỌC

1/ Cho hàm số : f(x)= x.sinx+x
2
. Tìm nguyên hàm của hàm số g(x)= x.cosx
biết rằng nguyên hàm này triệt tiêu khi x=k
π
2/Định m để hàm số: F(x) = mx
3
+(3m+2)x
2
-4x+3 là một nguyên hàm của hàm số:
f(x) = 3x
2
+10x-4.
3/Tìm họ nguyên hàm của hàm số: f(x)= cos
3
x.sin8x.
TÍNH :
4/I =
3
2
4
3tg x dx
π
π
∫

5/I =
4
2

3
0
π
(2cos
2
x-3sin
2
x)dx
9 / I =
2
2
sin( x)
4
dx
sin( x)
4
π
−π
π
−
π
+
∫

10 / I =
∫
−
3
6
π

−
∫

14/I =
2
4
0
sin x dx
π
∫

15/I =
∫
3
4
22
2
cos
2
sin
1
π
π
xx
dx
16/I =
∫
4
6
π

dx
x 4 x−
∫
19/ I =
∫
2
4
4
sin
1
π
π
x
dx
20/ I =
∫
4
0
6
cos
1
π
x
dx
21/I =
dxxxnsix )cos(2cos
44
2
0
+

5 2
0
x 1 x dx+
∫
26/I =
1
0
x
dx
2x 1+
∫
27/I =
1
x
0
1
dx
e 4+
∫
28/I =
2
x
1
1
dx
1 e
−
−
∫
29/I =

3
3
0
x 1
dx
3x 1
+
+
∫
33/I =
2
3
2
0
(x 3) x 6x 8dx− − +
∫
.
35/I =
4
2
2
1
dx
x 16 x−
∫
36*/I =
6
2
2 3
1

2
2
x 1
dx
x x 1
−
−
+
+
∫
41/I =
ln 2
x
0
e 1dx−
∫
42/I =
1
0
1
dx
3 2x−
∫
43/I =
2
5
0
sin xdx
π
∫

2
6
1
dx
sin x cot gx
π
π
∫
48/I =
3
2
e
1
ln x 2 ln x
dx
x
+
∫
.
49/I =
e
1
sin(ln x)
dx
x
∫
50/I =
1
3 4 5
0

(1 x ) dx−
∫
55*/I =
1
2x
0
1
dx
e 3+
∫
56/I =
x
ln3
x 3
0
e
dx
(e 1)+
∫
57/I =
0
2x
3
1
x(e x 1)dx
−
+ +
∫
58/I =
2

dx
e 1−
∫
62/I =
2
e
1
x 1
.ln xdx
x
+
∫
63/I =
2
1
0
x
dx
(x 1) x 1+ +
∫
64/I =
2
0
sin x.sin 2x.sin3xdx
π
∫
65/I =
2
4 4
0

∫
69/I =
9
3
1
x. 1 xdx−
∫
70/I =
2
3
0
x 1
dx
3x 2
+
+
∫
71*/I =
6
0
x
sin dx
2
π
∫
72*/I =
2
0
x
dx

cos(ln x)dx
π
∫
77*/I =
2
2
0
4 x dx+
∫
78/I =
2
1
x
dx
1 x 1+ −
∫
.
79/I =
e
1
1 3ln x ln x
dx
x
+
∫
80/I =
3
2
2
ln(x x)dx−

3
2
3
1
dx
x 3+
∫
86/I =
1
2
0
1
dx
4 x−
∫
87/I =
2
4
0
sin xdx
π
∫
88/I =
3
2
6
ln(sin x)
dx
cos x
π

93/I =
3
3
2
1
x
dx
x 16−
∫
.
94/I =
6
2
0
cos x
dx
6 5sin x sin x
π
− +
∫
95*/I =
2
e
2
e
1 1
( )dx
ln x
ln x
−

∫
100/I =
2
0
1 sin xdx
π
+
∫
101/I =
3
4
4
sin 2x dx
π
π
∫
102/I =
0
1 sin xdx
π
−
∫
103/I =
1
3
2
1
ln(x x 1) dx
−
 

1 2
−
+
∫
107/I =
2
4
0
xsin xdx
π
∫
108/I =
2
4
0
x cos xdx
π
∫
109/I =
6
2
0
x.sin xcos xdx
π
∫
110*/I =
2 x
1
2
0

2
0
1 x
x.ln dx
1 x
+
−
∫
115/I =
2
t
1
ln x
dx I 2
x
 
⇒ <
 ÷
 
∫
116/I =
3
0
sin x.ln(cosx)dx
π
∫
117/I =
2
e
2

2
sin x 3
0
e .sin x cos xdx
π
∫
122/I =
2
4
0
sin 2x
dx
1 cos x
π
+
∫
123/I =
1
2
0
3
dx
x 4x 5− −
∫
124/I =
2
2
1
5
dx

0
2
2
sin 2x
dx
(2 sin x)
−π
+
∫
129/I =
1
2
0
x 3
dx
(x 1)(x 3x 2)
−
+ + +
∫
130/I =
1
3
0
4x
dx
(x 1)+
∫
131/I =
1
4 2

6
1
dx
cos x.sin x
π
π
∫
135/I =
3
0
sin x.tgxdx
π
∫
136/I =
3
4
1
dx
sin 2x
π
π
∫
.
137/I =
3
4
2 2 5
0
sin x
dx

0
1 sin x
dx
1 3cos x
π
+
+
∫
141/I =
2
0
cos x
dx
sin x cos x 1
π
+ +
∫
142/I =
4
2
1
1
dx
x (x 1)+
∫
143/I =
1
3
3
1

2
1
1
dx
x 2x 9
−
+ +
∫
148/I =
3
2
1
1
dx
4x x−
∫
149/I =
2
2
1
4x x 5 dx
−
− +
∫
150/I =
2
2
2
2x 5
dx

dx
x 9 x+
∫

154/I =
2
x 2
0
e sin xdx
π
∫
155/I =
4
2
4 4
0
cos x
dx
cos x sin x
π
+
∫
156/I =
1
0
3
dx
x 9 x+ −
∫
157/I =

0
x cos x dx
π
∫
163/I =
2
0
x cos xsin xdx
π
∫
164/I =
6
2
0
x cos xsin xdx
π
∫
165/I =
4
x
1
e dx
∫
166/I =
4
3x
0
e sin 4x dx
π
∫

1
ln xdx
∫
172/I =
e
1
x(2 ln x)dx−
∫
173/I =
2
e
2
e
1 1
( )dx
ln x
ln x
−
∫
174/I =
2
2
1
(x x)ln x dx+
∫
175/I =
2
2
1
1

∫
179/I =
2
3
cos x.ln(1 cos x)dx
π
π
−
∫
180/
2
2
sin x 3
0
e sin x cos x dx
π
∫
181/I=
2
4
0
sin 2x
dx
1 sin x
π
+
∫
.
182/I =
2

1
dx
x (x 1)+
∫
186/I =
1
2
0
ln(1 x)
dx
x 1
+
+
∫
187/I
4
1
6
0
1 x
dx
1 x
+
+
∫
188/I =
1
15 8
0
x 1 x dx+

dx
1 cos x
π
+
∫
193/I =
2
0
sin 2x sin x
dx
1 3cos x
π
+
+
∫
194/I =
2
4
0
1 2sin x
dx
1 sin 2x
π
−
+
∫
195/I =
5 3
3
2

4
2
0
x.tg x dx
π
∫
199/I =
5
3
( x 2 x 2 )dx
−
+ − −
∫
200/I =
4
1
2
dx
x 5 4
−
+ +
∫
201/I =
2
1
x
dx
x 2 2 x+ + −
∫
202/I =

0
sin x.ln(1 cos x)dx
π
+
∫
206/I =
2
3
2
1
x 1
dx
x
+
∫
207/I =
3
4
2
0
sin x
dx
cos x
π
∫
208/I =
2
2
0
cos x.cos4x dx

0
x
dx
4 x−
∫
213/I =
1
2
0
x
dx
4 x−
∫
214/I =
1
4
2
2
0
x
dx
x 1−
∫
215/I =
2
0
sin3x
dx
cos x 1
π

dx
1 x+
∫
219/I =
x
ln 2
x
0
1 e
dx
1 e
−
+
∫
220/I =
1
0
x 1 x dx−
∫
221/I =
1
2
0
x 1dx+
∫
222/I =
2
3 3
0
(cos x sin x)dx

7
3
3
0
x 1
dx
3x 1
+
+
∫
.
227/I =
2
6
1 sin 2x cos2x
dx
cos x sin x
π
π
+ +
+
∫
228/I =
x 2
1
2x
0
(1 e )
dx
1 e

∫
232*/I =
2
0
xsin x.cos xdx
π
∫
233/I =
2
0
cos x
dx
cos2x 7
π
+
∫
234/I =
4
2
1
1
dx
x (x 1)+
∫
235/I =
2
2 3
0
sin 2x(1 sin x) dx
π

cos x cos x cos xdx
π
π
−
−
∫
240*/I =
1
2
1
ln( x a x)dx
−
+ +
∫
241/I =
2
x
0
1 sin x
dx
(1 cos x)e
π
−
+
∫
242/I =
2
0
sin 2x sin x
dx

0
x
dx
1 x−
∫
246/I =
2
1
2
2
2
1 x
dx
x
−
∫
247/I =
2
1
2
0
x
dx
4 x−
∫
248/I =
2
2
2
3

2
1
1
dx
(1 x)x+
∫
253/I =
2
3
0
x 1
dx
3x 2
+
+
∫
254*/I =
3
4
cos x sin x
dx
3 sin 2x
π
π
+
+
∫
.
255/I =
2

0
(1 x ) dx−
∫
259/I =
4
2
0
x.tg xdx
π
∫
260/I=
2
2 2
0
1
dx
(4 x )+
∫
261/I =
2
1
3
0
3x
dx
x 2+
∫
262*/I =
5
2

6
0
sin x sin x
dx
cos2x
π
+
∫
265/I =
2
3
1
dx
sin x 1 cosx
π
π
+
∫
266/I =
3
6 2
1
1
dx
x (1 x )+
∫
.
267/I =
2
2

π
−
+ +
∫
271/I =
4 4
4
0
sin x cos x
dx
sin x cos x 1
π
−
+ +
∫
272/I =
2
0
sin xcosx cosx
dx
sin x 2
π
+
+
∫
273/I =
1
1
x
3

dx
x 1+
∫
277*/I =
4
1
6
0
x 1
dx
x 1
+
+
∫
278/I =
1
3
0
x
dx
(2x 1)+
∫
279/I =
7
2
1
dx
2 x 1+ +
∫
280/I =

0
x ln(x 1)dx+
∫
284/I =
3
2
2
1
3x
dx
x 2x 1+ +
∫
285/I =
1
3 2
0
4x 1
dx
x 2x x 2
−
+ + +
∫
286/I =
1
2
2
1
2
1
dx

∫
290/I =
2
3 3
0
(cos x sin x)dx
π
+
∫
291/I =
2
5 4
0
cos x sin xdx
π
∫
292/I =
2
4 4
0
cos2x(sin x cos x)dx
π
+
∫
293/I =
2
0
1
dx
2 sin x

∫
297*/I =
2
3
1
1
dx
x 1 x+
∫
298/I =
3
1
2
0
x
dx
x 1 x+ +
∫
299/I =
1
2
1
1
dx
1 x 1 x
−
+ + +
∫
300/I =
3

dx
sin x 2
π
+
∫
304/I =
3
2
0
cos x
dx
cos x 1
π
+
∫
305/I =
2
0
1
dx
2cos x sin x 3
π
+ +
∫
306/I =
2
2
3
cos x
dx

+
∫
310*/I =
2
0
sin x
dx
cos x sin x
π
+
∫
311/I =
4
2
4 4
0
sin x
dx
cos x sin x
π
+
∫
312*/I =
2
2
0
tgx
dx
1 ln (cosx)
π

1
2
0
x
dx
x 4+
∫
317*/I =
3
2
4 2
0
cos x
dx
cos 3cos x 3
π
− +
∫
318*/Tìm x> 0 sao cho
2 t
x
2
0
t e
dt 1
(t 2)
=
+
∫
319*/I =

∫
323/I =
3
4
4
tg x dx
π
π
∫
324*/I =
4
0
1
dx
2 tgx
π
+
∫
325/I =
5
2
0
sin x
dx
cos x 1
π
+
∫
326/I =
3

3
3
2
4
1
x x
dx
x
−
∫
330/I =
x
ln3
x x
0
e
dx
(e 1) e 1+ −
∫
331/I =
1
4
e
2
1
e
1
dx
x cos (ln x 1)
π