Bài 3: Tính các tích phân sau :
1 I =
1
0
(3 + cos 2 )
2 I =
1
0
( + )
∫ x x e dxx
3 I = 2
0
(1 sin )cos
+
π
4 I = 2
1
0
( + sin )
∫ x ex x dx
0
t anx cos
= ∫
x
π
0
(1 2sin ) cos
π
+
7 I=
3
2
0 + 1
∫ x xdx
8 J= 2
2
2
0( + 2)
∫ x xdx
3 0
sin cos sin
∏
0
cos sin
π
11 J = 2
0
( + 1)sin
π
12
1 3
2 0
x
1 +
x
13 6( )
0
1 − sin 3
π
14
1
5 0
(1 )
= ∫ −
0
sin 6 sin 2 − 6
π
16 I =
2
5 1
(1 − )
17 J = 2
0
(2 − 1).cos
π
0
cos sin
π
19
1 0
(2 1)
= ∫ + x
20
2 2 1
2 1
=
+
J
21
1
2 3 4 1
(1 )
−
= ∫ −
22
1 0
(4 1)
= ∫ + x
I x e dx
23 6( )
0
1 − sin 3
π
24
1
5 0
(1 )
= ∫ −