Question

Here's a fun problem: Integrate
\[\int_{-\pi}^\pi \ln(2+\cos x)\,dx\]

Answer 1

2 \(\int_0^\pi ..\)

Answer 2

a fun problem

Answer 3

ans is 0

Answer 4

I think

Answer 5

\(= 2 \int_0^\pi \ln (2-\cos x) dx\)

Answer 6

Not zero, this here is an even function.

Answer 7

\[\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{2a}f(2a-x)dx\]

Answer 8

did I mess up??

Answer 9

im thinking of parameter \[I(a) = \int_{-\pi}^\pi \ln(2+\cos(a x))\,dx\]

Answer 10

hey @ganeshie8

Answer 11

SITH WHATS THE ANSWR???

Answer 12

http://www.wolframalpha.com/input/?i=integral++log%28+2+%2B+cos%28x%29%29

Answer 13

@ganeshie8 I've been making slow progress with a slightly different parameter,
\[I(b)=\int_{-\pi}^\pi \ln(2+b\cos x)\,dx\]
There's definite an answer though, I'm curious to see its closed form.

Answer 14

*if it has one

Answer 15

DUDE I got it

Answer 16

Here what Mathematica gave
as an answer
\[
-2 \pi \ln \left(4-2 \sqrt{3}\right)
\]

Answer 17

@eliassaab Mma won't always be there to help ;)

Answer 18

FORGET MY PREVIOUS COMMENT

Answer 19

I know, but just to check your answer if you get one

Answer 20

but I am comin close

Answer 21

For those interested, my parameterization gives me the following integral along the way:
\[\int\frac{1}{b}\sqrt{\frac{2+b}{2-b}}(2+b)\,db\]
which I'm told (thanks to WA) that there is an elementary antiderivative.

Answer 22

conjugate

Answer 23

\[\int \frac{\sqrt{\frac{b+2}{2-b}} (b+2)}{b} \,
db=\\
\frac{\sqrt{\frac{b+2}{2-b}}
\left(\sqrt{b+2} (b-2)+8 \sqrt{2-b} \sin
^{-1}\left(\frac{\sqrt{b+2}}{2}\right)-4
\sqrt{2-b} \tanh
^{-1}\left(\frac{\sqrt{b+2}}{\sqrt{2-b}}\right)\right)}{\sqrt{b+2}}
\]

Answer 24

whoa

Answer 25

hey ARE U SURE ITS NOT LOG(1-cosx)???

Answer 26

or log(1+cosx)??

Answer 27

cmon man pls

Answer 28

I solved this one

Answer 29

lmao

Answer 30

Yes, I came up with the problem. \(\ln(1\pm\cos x)\) is a bit strange, WA says both integrals converge despite the singularities:

\(\ln(1+\cos x)\):
http://www.wolframalpha.com/input/?i=Integrate%5BLog%5B1%2BCos%5Bx%5D%5D%2C%7Bx%2C-pi%2Cpi%7D%5D

\(\ln(1-\cos x)\):
http://www.wolframalpha.com/input/?i=Integrate%5BLog%5B1-Cos%5Bx%5D%5D%2C%7Bx%2C-pi%2Cpi%7D%5D

Answer 31

I SOLVED IT

Answer 32

U GET \[2\pi{\log1/2}\]

Answer 33

I dont use wolfram lmao

Answer 34

YA BCOZ when u use 2 .,.............u get stuck at 4-\[\cos^2x\]

Answer 35

SITH is ur problem solvable???...what does wolfram say??

Answer 36

Your answer is incorrect, the area of the region described by the integral lies above the x-axis, it can't possibly be negative \(\bigg(\)or zero, depending on if you meant \(2\pi\log\dfrac{1}{2}\) or \(\left.2\pi\dfrac{\log1}{2}\right)\).

Answer 37

NOPE ITS NOT................IF U USE 1 instead of 2

Answer 38

wait lemme check for errors

Answer 39

The problem isn't integrating \(\ln(1+\cos x)\)...

Answer 40

SITH..........I QUIT FROM THIS FUN PROBLEM.........

Answer 41

Interestingly,\[\int_{-\pi}^\pi\ln(1+\cos x)\,dx=\int_{-\pi}^\pi\ln(1-\cos x)\,dx\]
but that's irrelevant, seeing as there's a \(\bf2\) in the original problem.

Answer 42

@ganeshie8 upon differentiating your parameterization I think we'll have to implement some complex contour, I remember seeing something of this form before:
\[I'(a)=\int_{-\pi}^\pi\frac{-x\sin ax}{2+\cos ax}\,dx\]

Answer 43

yeah i got scared of that and trying if power series works smooth

let me pause on power series a bit, this does look interesteing :
\[\int_{-\pi}^\pi\ln(2+\cos x)\,dx=\int_{-\pi}^\pi\ln(2-\cos x)\,dx\]
this is tryue because of that f(a-x) thingy..
messing wid this a bit we get
\[I = \int_0^{\pi} \ln(4-\cos^2x) dx\]

then its the dead end i guess

Answer 44

Kudos to @eliassaab for the elegant solution, as seen here: http://openstudy.com/users/eliassaab#/updates/54a358eae4b0b8a54fb07d0b#/updates/54a358eae4b0b8a54fb07d0b