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Integral of cosine function 
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#1
Jan1411, 02:10 PM

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Hi. I have been experimenting a little to come up with the following "conjecture"
[tex] \int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi=0 [/tex] where a and b are arbitrary constants and f(x) is any function. Is this true? I guess it can be shown by expanding f in a power series of cosines? 


#2
Jan1411, 03:23 PM

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hi daudaudaudau!
doesn't work for f = √, a = b 


#3
Jan1411, 03:35 PM

PF Gold
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Yes, I'm getting that that this is true. It can be proven using the substitution
[tex]x=\cos{\phi}[/tex] [tex]dx=\sin{\phi}d\phi[/tex] But this substitution is not 1to1: Each x value corresponds to 2 phi values on [0,2pi]. So you need to break the region of integration into [0,pi] and [pi,2pi]. If you look at the graph of the cos function, you will see it is 1to1 on these two intervals and goes from 1 to 1 on [0,pi] and from 1 to 1 on [pi,2pi]. So the integral becomes: [tex] \int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi= \int_1^{1}f(a+bx)dx + \int_{1}^{1}f(a+bx)dx = \int_{1}^{1}f(a+bx)dx + \int_{1}^{1}f(a+bx)dx = 0 [/tex] Tinytim: I get this even in the example you gave. 


#4
Jan1411, 03:55 PM

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Integral of cosine function
hmm … i think i've been misled by the ambiguity of the √ function
yes, we can get it directly from the original integral … if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]_{φ=0}^{2π} 


#6
Jan1411, 04:08 PM

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#8
Jan1411, 04:50 PM

P: 296

clever :)



#9
Jan1411, 09:31 PM

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The integral from pi to pi is zero, because you are integrating an odd function.
And the integral from pi to pi equals the integral from 0 to 2pi. QED. 


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