大学定积分的计算题，求高人指点，最好写出详细的计算过程

Question

如题所述

Answer 1

先求出不定积分，需用万能换元法
令z
=
tan(x/2)，dx
=
2dz/(1
+
z²)，sinx
=
2z/(1
+
z²)
∫
1/(2
+
sinx)
dx
=
∫
[2/(1
+
z²)]/[2
+
(2z)/(1
+
z²)]
dz
=
∫
1/[(1
+
z²)
+
z]
dz
=
∫
1/[(z
+
1/2)²
+
3/4]
dz
=
(2/√3)arctan[(z
+
1/2)
·
2/√3]
+
C
=
(2/√3)arctan[(2tan(x/2)
+
1)/√3]
+
C
分区间，注意若ƒ(x)
=
1/(2
+
sinx)，ƒ(0)
=
ƒ(π)
=
ƒ(2π)
F(x)
=
(2/√3)arctan[(2tan(x/2)
+
1)/√3]
∫(0→2π)
1/(2
+
sinx)
dx
=
∫(0→π)
1/(2
+
sinx)
dx
+
∫(π→2π)
1/(2
+
sinx)
dx
=
[F(π)
-
F(0)]
-
[F(2π)
-
F(π)]，x
=
π是间断点，分左右极限做
=
[lim(x→0)
F(x)
-
lim(x→π⁺)
F(x)]
-
[lim(x→2π)
F(x)
-
lim(x→π⁻)
F(x)]
=
[π/(3√3)
-
(-
π/√3)]
-
[π/(3√3)
-
π/√3]
=
2π/√3