- #1

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[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?

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- Thread starter daudaudaudau
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- #1

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[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

tiny-tim

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hi daudaudaudau!

doesn't work for f = √, a = b

doesn't work for f = √, a = b

- #3

LeonhardEuler

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[tex]x=\cos{\phi}[/tex]

[tex]dx=-\sin{\phi}d\phi[/tex]

But this substitution is not 1-to-1: 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 1-to-1 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]

Tiny-tim: I get this even in the example you gave.

- #4

tiny-tim

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yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]

- #5

LeonhardEuler

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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}

Happens to everyone

That is a better way of proving it.

- #6

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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}

What happened to the sine function?

- #7

tiny-tim

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chain rule

- #8

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clever :-)

- #9

AlephZero

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And the integral from -pi to pi equals the integral from 0 to 2pi.

QED.

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