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Help me with this logarithmic integral

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


[itex] \displaystyle \int_0^{2 \pi} x \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]


Homework Equations



The Attempt at a Solution



Using property of definite integral
2I = [itex]\displaystyle \int_0^{2 \pi} 2 \pi \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


[itex] \displaystyle \int_0^{2 \pi} x \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]


Homework Equations



The Attempt at a Solution



Using property of definite integral
2I = [itex]\displaystyle \int_0^{2 \pi} 2 \pi \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]
I'm not sure what "property of definite integral" would give you that. You'll have to spell it out. Here's a hint. Change the variable to u=2pi-x and see what happens.
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
You can also rewrite the log of a quotient into something simpler.
 
  • #4
utkarshakash
Gold Member
855
13
I'm not sure what "property of definite integral" would give you that. You'll have to spell it out. Here's a hint. Change the variable to u=2pi-x and see what happens.
[itex]\int_0^a f(x) = \int_0^a f(a-x)[/itex]. This is what I've used.

Then I added both integrals to get rid of x outside log.
 
  • #5
utkarshakash
Gold Member
855
13
You can also rewrite the log of a quotient into something simpler.
Can you please elaborate? I didn't get you.
 
  • #6
180
8
[itex]\int_0^a f(x) = \int_0^a f(a-x)[/itex]. This is what I've used.

Then I added both integrals to get rid of x outside log.
Integrating from one point to another finds the area under the curve between those two points. Think about the graphs of f(x) and f(a-x). f(a-x) is going to be flipped over the x-axis and shifted to the left by "a" units.. Most of the time, the value of the integral wouldn't be the same for both functions. Sorry to tell ya, but I think the property of definite integral may only apply to periodic functions. I don't think this function is periodic

Can you please elaborate? I didn't get you.
[itex]log(\frac{a}{b})=log(a)-log(b)[/itex]
[itex]log(a*b)=log(a)+log(b)[/itex]
 
  • #7
3,812
92
Using property of definite integral
2I = [itex]\displaystyle \int_0^{2 \pi} 2 \pi \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]
Okay. Now observe that
$$\int_0^{2 \pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx=2\int_0^{\pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx$$

Use the same property you used before.
 
  • #8
utkarshakash
Gold Member
855
13
Okay. Now observe that
$$\int_0^{2 \pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx=2\int_0^{\pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx$$

Use the same property you used before.
Well done.
 

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