关于高等数学带绝对值定积分的题目

这绝对值是全看sinx的图像而取决的。
y
=
(x
+
|sinx|)/(1
+
cosx)、-
π/2
≤
x
≤
π/2
当-
π/2
≤
x
≤
0、|sinx|
=
-
sinx、因为在[-
π/2，0]的sinx
≤
0
y
=
(x
-
sinx)/(1
+
cosx)
当0
≤
x
≤
π/2、|sinx|
=
sinx、因为在[0，π/2]的sinx
≥
0
y
=
(x
+
sinx)/(1
+
cosx)
∫[-
π/2→π/2]
(x
+
|sinx|)/(1
+
cosx)
dx
=
∫[-
π/2→0]
(x
-
sinx)/(1
+
cosx)
dx
+
∫[0→π/2]
(x
+
sinx)/(1
+
cosx)
dx
=
A
+
B
A
=
∫[-
π/2→0]
(x
-
sinx)/(1
+
cosx)
dx
=
∫[-
π/2→0]
x/(1
+
cosx)
dx
-
∫[-
π/2→0]
sinx/(1
+
cosx)
dx
=
∫[-
π/2→0]
x/(1
+
cosx)
dx
-
[xsinx/(1
+
cosx)]
|[-
π/2→0]
+
∫[-
π/2→0]
x/(1
+
cosx)
dx
=
-
[xtan(x/2)]
+
2∫[-
π/2→0]
x/[2cos²(x/2)]
dx
=
(-
π/2)tan(-
π/4)
+
2∫[-
π/2→0]
x
d[tan(x/2)]
=
π/2
+
2[xtan(x/2)]
|[-
π/2→0]
-
4∫[-
π/2→0]
tan(x/2)
d(x/2)
=
π/2
-
2(-
π/2)tan(-
π/4)
+
4ln[cos(x/2)]
|[-
π/2→0]
=
π/2
-
π
+
2ln[1]
-
4ln[1/√2]
=
2ln[2]
-
π/2
B
=
∫[0→π/2]
(x
+
sinx)/(1
+
cosx)
dx
=
∫[0→π/2]
x/(1
+
cosx)
dx
+
∫[0→π/2]
sinx/(1
+
cosx)
dx
=
∫[0→π/2]
x/(1
+
cosx)
dx
+
[xsinx/(1
+
cosx)]
|[0→π/2]
-
∫[0→π/2]
x/(1
+
cosx)
dx
=
[xtan(x/2)]
|[0→π/2]
=
(π/2)tan(π/4)
=
π/2
原式
=
2ln[2]
-
π/2
+
π/2
=
2ln[2]

绝对值|sinx|在区间[-π/2,π/2]符号变化，分成两段求积分就好了

结果是2ln2。
解答见图，点击放大：