e*******n
发帖数: 174 | 1 No Chinese input here. Sorry about it.
I've got an integral that I was able to integrate with Mathematica. It does
not appear to be difficult. So I tried to solve it manually. I am not able
to figure it out. For the sake of my curiosity, I wonder if anybody could
help me out. Any hints would be greatly appreciated.
The integral has the following form:
Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, 0, 2*Pi}]
where "a" is a real positive number less than 1.
Again, thanks for any hint | s******1
发帖数: 969 | 2 积分分成[0,Pi]，[Pi,2Pi],
Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
= Integrate[Cos[x]^2*Log[a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x,0, Pi}]
然后：
Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
＝Integrate[Cos[x]^2*Log[1-a^2], {x,0, Pi}] | e*******n
发帖数: 174 | 3
~~~~~~~~~~~~~~~~~~~~~~~~~~?
How did you do this? I am lost. :)
【在 s******1 的大作中提到】
: 积分分成[0,Pi]，[Pi,2Pi],
: Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
: = Integrate[Cos[x]^2*Log[a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x,0, Pi}]
: 然后：
: Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
: ＝Integrate[Cos[x]^2*Log[1-a^2], {x,0, Pi}]
| s******1
发帖数: 969 | 4 A==Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]:
Introduce y=x-Pi so the range is [0,Pi].
Cos[x]=Cos[y+Pi]=-Cos[y]. Sin[x]=Sin[x+Pi]=-Sin[x].
A=Integrate[Cos[y]^2*Log[a*Cos[y] +Sqrt[1-a^2*Sin[y]^2]], {y,0, Pi}]
=Integrate[Cos[x]^2*Log[a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x,0, Pi}];
B==Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, 0, Pi}];
Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
= A+B
=Integrate[Cos[x]^2*{Log[a*Cos[x] +Sqrt[1- | n*n
发帖数: 202 | 5 晕了....
[1
【在 s******1 的大作中提到】
: A==Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]:
: Introduce y=x-Pi so the range is [0,Pi].
: Cos[x]=Cos[y+Pi]=-Cos[y]. Sin[x]=Sin[x+Pi]=-Sin[x].
: A=Integrate[Cos[y]^2*Log[a*Cos[y] +Sqrt[1-a^2*Sin[y]^2]], {y,0, Pi}]
: =Integrate[Cos[x]^2*Log[a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x,0, Pi}];
: B==Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, 0, Pi}];
: Integrate[Cos[x]^2*Log[-a*Cos[x] +Sqrt[1-a^2*Sin[x]^2]], {x, Pi, 2*Pi}]
: = A+B
: =Integrate[Cos[x]^2*{Log[a*Cos[x] +Sqrt[1-
|
|
