Tính các tích phân sau:
4 0
sin 2x
1 cos x
π
=
+
2 2 2
3
1
=
−
2 2 1
ln(x 1)
x
+
2 0
sin x
1 cos x
π
=
+
2
2 2
x cosx
4 sin x
π
π
+
=
−
0
x sin x
cos x
π
= ∫
2
3
0
I x.cos x.sinx.dx
π
1
1
=
+
0
4 x
4 x
−
+
∫ 2
2 1
1
x 1 ln x
=
−
0
e
1 e
=
+
1
1
x(x 1)
=
+
∫ 0
x
1 sin x
π
=
+
1
1
−
=
0
sin 2x
(2 sin x)
π
=
+
∫ 4
0
x
=
0
4sin x
1 cosx
π
= +
∫
1
1 2
1 x dx I
1 x x
−
−
=
+
∫ 8
0
2sin 2x 3sin x
6cos x 2
π
+
=
−
0
sin x cos x
sin x 2cos x
π
−
=
+
1
3 2ln x
x 1 2ln x
−
=
+
∫ 4
0
x
1 cos 2x
π
=
+
∫
7 3 0
x 2
x 1
+
=
+
0
4 x
+
=
−
∫ 2
2 0
2sin x.cos x
13 5cos x
π
=
−
0
cos x.sin x
1 sin x
π
=
+
4
tan x
cosx 1 cos x
π π
=
+
∫ 1
2 1
3
2
=
−
0
I ( sin x cos x )dx
π
6
sin x cos x
2009 1
π
−π
+
=
+
∫
6 3
0
tan x
cos2x
π
= ∫