avatar
e*n
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
avatar
s*1
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}]
avatar
e*n
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}]

avatar
s*1
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-
avatar
n*n
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-

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