How Do You Solve This Logarithmic Integral Involving Cosine?

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[Dick]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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.

 

[SteamKing]

You can also rewrite the log of a quotient into something simpler.

 

[utkarshakash]

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.

\int_0^a f(x) = \int_0^a f(a-x). This is what I've used.

Then I added both integrals to get rid of x outside log.

 

[utkarshakash]

You can also rewrite the log of a quotient into something simpler.

Can you please elaborate? I didn't get you.

 

[Hertz]

\int_0^a f(x) = \int_0^a f(a-x). 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.

log(\frac{a}{b})=log(a)-log(b)
log(a*b)=log(a)+log(b)

 

[Saitama]

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

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.

 

[utkarshakash]

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.