1/sinx的不定积分如何求？

∫
1/sinx
dx
=
∫
cscx
dx
=
∫
cscx
*
(cscx
-
cotx)/(cscx
-
cotx)
dx
=
∫
(-
cscxcotx
+
csc²x)/(cscx
-
cotx)
dx
=
∫
d(cscx
-
cotx)/(cscx
-
cotx)
=
ln|cscx
-
cotx|
+
C
∫
1/sinx
dx
=
∫
1/[2sin(x/2)cos(x/2)]
dx
=
∫
1/[cos²(x/2)tan(x/2)]
d(x/2)
=
∫
1/[tan(x/2)]
d[tan(x/2)]
=
ln|tan(x/2)|
+
C
∫
1/sinx
dx
=
∫
sinx/sin²x
dx
=
∫
1/(cos²x
-
1)
d(cosx)
=
(1/2)∫
[(cosx
+
1)
-
(cosx
-
1)]/[(cosx
+
1)(cosx
-
1)]
d(cosx)
=
(1/2)∫
[1/(cosx
-
1)
-
1/(cosx
+
1)]
d(cosx)
=
(1/2)ln|(cosx
-
1)/(cosx
+
1)|
+
C
=
(1/2)ln|[2sin²(x/2)]/[2cos²(x/2)]|
+
C
=
(1/2)
*
2ln|tan(x/2)|
+
C
=
ln|tan(x/2)|
+
C
万能代换：令y
=
tan(x/2)、dx
=
2dy/(1
+
y²)、sinx
=
2y/(1
+
y²)
∫
1/sinx
dx
=
∫
1/[2y/(1
+
y²)]
*
2dy/(1
+
y²)
=
∫
(1
+
y²)/(2y)
*
2dy/(1
+
y²)
=
∫
1/y
dy
=
ln|y|
+
C
=
ln|tan(x/2)|
+
C
这几个答案都可以互相转换的。
其中ln|tan(x/2)|
=
ln|sin(x/2)/cos(x/2)|
=
ln|[2sin(x/2)cos(x/2)]/[2cos²(x/2)]|
=
ln|sinx/(1
+
cosx)|
=
ln|[sinx(1
-
cosx)]/[(1
+
cosx)(1
-
cosx)]|
=
ln|(sinx
-
sinxcosx)/sin²x|
=
ln|1/sinx
-
cosx/sinx|
=
ln|cscx
-
cotx|