1/x√(1-x^2) 的不定积分

Answer 1

（1）解：令x=sint,dx=costdt
∫dx/[x√(1-x^2)]=∫costdt/[sintcost]=∫1/sint*dt=∫sint/(sint)^2*dt=∫sint/[1-(cost)^2]*dt
=-∫d(cost)/[1-(cost)^2]=-1/2*∫[1/(cost+1)-1/(cost-1)]d(cost)
=-1/2*(ln|cost+1|-ln|cost-1)+C
=-1/2*ln|(cost+1)/(cost-1)|+C
=-ln|(sint/(cost-1)|+C
=-ln|x/[√(1-x^2)-1]|+C
=ln|[√(1-x^2)-1]/x|+C
（2）解法1：
解：原式=
∫{
[(sin
x)^2
+(cos
x)^2
]
/[(sin
x)^2
(cos
x)^2
]
}dx
=
∫[
(sec)^2
]dx
+∫[
(csc)^2
]dx
=
tan
x
-cot
x
+C
=
sin
x
/cos
x
-cos
x
/sin
x
+C
=
[
(sin
x)^2
-(cos
x)^2
]
/
(cos
x
sin
x)
+C
=
-cos
2x
/
[
(1/2)sin
2x
]
+C
=
-2
cot
2x
+C,
(C为任意常数).
解法2：
原式=
∫dx
/
[(1/4)
(sin
2x)^2]
=
4
∫[
(csc
2x)^2
]
dx
=
2
∫[
(csc
2x)^2
]
d(2x)
=
-2
cot
2x
+C,
(C为任意常数).
满意请采纳，谢谢~